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Could anyone help me understand half angle formula and double angle formula?
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I'm having trouble fully understanding these two problems. I know I have to use the half angle formula for one, and the double for another. The first question goes as this
Given that $$cos(x) =frac-1517$$ and $x$ in quadrant II find the exact values of:
$$sin left(frac x2right)$$ and $$cos left(frac x2right)$$
For this problem the biggest question I do have is what should I input for $x$ exactly? It states $cos(x)$ is a fraction, so should I simply add that as $x$ in $$frac x2 $$? Or if I need to find what $$sinleft(frac x2right)$$ is how could I find that?
Secondly, what would the best course of action be to see if it would be a negative or positive since in QII it could be either or.
Second is Given $$sin(x) = frac-725$$ and $x$ in quadrant III, find the value of $sin(2x)$,$cos(2x)$,$tan(2x)$.
This one I am kinda lost, would I need to find the value of what sin is and just plug it into what x is (For each corresponding formula.)? Or tackle it differently?
This is somewhat a loaded question, and I am sorry for that. But I do appreciate any help anyone could give.
trigonometry
edited 8 mins ago
Y.Lin
606
asked 2 hours ago
Carl S
111
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up vote
2
down vote
favorite
I'm having trouble fully understanding these two problems. I know I have to use the half angle formula for one, and the double for another. The first question goes as this
Given that $$cos(x) =frac-1517$$ and $x$ in quadrant II find the exact values of:
$$sin left(frac x2right)$$ and $$cos left(frac x2right)$$
For this problem the biggest question I do have is what should I input for $x$ exactly? It states $cos(x)$ is a fraction, so should I simply add that as $x$ in $$frac x2 $$? Or if I need to find what $$sinleft(frac x2right)$$ is how could I find that?
Secondly, what would the best course of action be to see if it would be a negative or positive since in QII it could be either or.
Second is Given $$sin(x) = frac-725$$ and $x$ in quadrant III, find the value of $sin(2x)$,$cos(2x)$,$tan(2x)$.
This one I am kinda lost, would I need to find the value of what sin is and just plug it into what x is (For each corresponding formula.)? Or tackle it differently?
This is somewhat a loaded question, and I am sorry for that. But I do appreciate any help anyone could give.
trigonometry
edited 8 mins ago
Y.Lin
606
asked 2 hours ago
Carl S
111
New contributor
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Do you know the half angle formulas for sine and cosine? Write them out at the start of the question so we can see what you know. Perhaps the answer will be obvious if you actually write them.
– David K
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm having trouble fully understanding these two problems. I know I have to use the half angle formula for one, and the double for another. The first question goes as this
Given that $$cos(x) =frac-1517$$ and $x$ in quadrant II find the exact values of:
$$sin left(frac x2right)$$ and $$cos left(frac x2right)$$
For this problem the biggest question I do have is what should I input for $x$ exactly? It states $cos(x)$ is a fraction, so should I simply add that as $x$ in $$frac x2 $$? Or if I need to find what $$sinleft(frac x2right)$$ is how could I find that?
Secondly, what would the best course of action be to see if it would be a negative or positive since in QII it could be either or.
Second is Given $$sin(x) = frac-725$$ and $x$ in quadrant III, find the value of $sin(2x)$,$cos(2x)$,$tan(2x)$.
This one I am kinda lost, would I need to find the value of what sin is and just plug it into what x is (For each corresponding formula.)? Or tackle it differently?
This is somewhat a loaded question, and I am sorry for that. But I do appreciate any help anyone could give.
trigonometry
edited 8 mins ago
Y.Lin
606
asked 2 hours ago
Carl S
111
New contributor
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm having trouble fully understanding these two problems. I know I have to use the half angle formula for one, and the double for another. The first question goes as this
Given that $$cos(x) =frac-1517$$ and $x$ in quadrant II find the exact values of:
$$sin left(frac x2right)$$ and $$cos left(frac x2right)$$
For this problem the biggest question I do have is what should I input for $x$ exactly? It states $cos(x)$ is a fraction, so should I simply add that as $x$ in $$frac x2 $$? Or if I need to find what $$sinleft(frac x2right)$$ is how could I find that?
Secondly, what would the best course of action be to see if it would be a negative or positive since in QII it could be either or.
Second is Given $$sin(x) = frac-725$$ and $x$ in quadrant III, find the value of $sin(2x)$,$cos(2x)$,$tan(2x)$.
This one I am kinda lost, would I need to find the value of what sin is and just plug it into what x is (For each corresponding formula.)? Or tackle it differently?
This is somewhat a loaded question, and I am sorry for that. But I do appreciate any help anyone could give.
trigonometry
trigonometry
edited 8 mins ago
Y.Lin
606
asked 2 hours ago
Carl S
111
New contributor
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 mins ago
Y.Lin
606
asked 2 hours ago
Carl S
111
New contributor
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 mins ago
Y.Lin
606
edited 8 mins ago
Y.Lin
606
edited 8 mins ago
Y.Lin
606
606
asked 2 hours ago
Carl S
111
New contributor
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Carl S
111
asked 2 hours ago
Carl S
111
111
New contributor
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Carl S is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Do you know the half angle formulas for sine and cosine? Write them out at the start of the question so we can see what you know. Perhaps the answer will be obvious if you actually write them.
– David K
2 hours ago
add a comment |Â
1
Do you know the half angle formulas for sine and cosine? Write them out at the start of the question so we can see what you know. Perhaps the answer will be obvious if you actually write them.
– David K
2 hours ago
1
1
Do you know the half angle formulas for sine and cosine? Write them out at the start of the question so we can see what you know. Perhaps the answer will be obvious if you actually write them.
– David K
2 hours ago
Do you know the half angle formulas for sine and cosine? Write them out at the start of the question so we can see what you know. Perhaps the answer will be obvious if you actually write them.
– David K
2 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
Guide:
We do not plug in a value of $x$, we solve for the trigonometry value without having the need to find its value explicitly.
If $x$ is in the second quadrant, then $fracx2$ is in the first quadrant.
$$cos(x) = -frac1517$$
is $$2cos^2left( fracx2right)-1=-frac1517$$
Solve for $cosleft( fracx2right)$.
For the second question, given $sin(x)$ and knowing which quadrant it is in, you should be able to find $cos(x)$ using Pythagoras Theorem, and use double angle formula to recover $sin(2x), cos(2x)$. Then use the definition of tangent.
answered 2 hours ago
Siong Thye Goh
90.1k1460112
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up vote
3
down vote
It is very helpful to know the following identities.
1) $sin ^2 x + cos ^2 x =1 $
2) $sin ^2 x = (1/2)(1-cos 2x)$
3) $cos ^2 x = (1/2 )(1+ cos 2x)$
4) $sin 2x = 2sin x cos x$
5) $cos 2x = cos ^2 x - sin ^2x $
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
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up vote
0
down vote
- You have formulas, namely:
$sin(2x) = 2*cos(x)*sin(x)$
$cos(2x) = cos^2(x) - sin^2(x)$
$sin^2(x) + cos^2(x) = 1$
There are plenty more, but generally speaking everything can be derived from each other.
- You have the unit circle, which helps you determine the sign of the function. You need it, because when you try to derive a, for example, sin from the trig identity, you end up with expression: $ |sin(x)| = sqrt1 - cos^2(x)$ and at this point the quadrant information helps you recover the sign of sin(x)
For the first question you are given that $cos(x) = -15/17$ and $x$ is in the 2nd quadrant. This tells you, that $x in [pi/2;pi]$. Respectively, $x/2 in [pi/4;pi/2] = $ first quadrant, which means, that cos(x/2) and sin(x/2) are positive.
Now you just use the formula, for example, the second one I listed, to find the cos(x/2)
$cos(x) = cos^2(x/2) - sin^2(x/2) = 2cos^2(x/2) - 1 => |cos(x/2)| = cos(x/2)$(verified sign already to be positive)$ = sqrtfrac1 + cos(x)2$
I guess you can do the math.
answered 1 hour ago
Makina
3039
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up vote
0
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I always forget the formulas for stuff like $sinleft(fracx2right)$. But you don't need to remember them. You only need to know two important formulas.
beginalign
1 = cos^2(x) + sin^2(x) tag1 \ \
cos(2x) = cos(x + x) = cos(x)cos(x) - sin(x)sin(x) = &cos^2(x) - sin^2(x) tag2
endalign
so we end up with
beginalign
1 &= cos^2(x) + sin^2(x) tag1 \ \
cos(2x) & = cos^2(x) - sin^2(x) tag2
endalign
But in these equations we didn't have to use $x$, we could have used $y$ or $a$ or $5$. They are all true no matter what real value we stick in there. So watch what happens if instead of putting $x$ in there we put $x/2$.
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(not2fracxnot2right) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
So here we are
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(xright) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
To get the half angle formulas all we have to do is add or substract these two equations. For instance, if we subtract $(2)$ from $(1)$ we get
beginalign
1 - cos(x) = 2sin^2left(fracx2right) tag3
endalign
Now, I'm going to plug in what you said that $cos(x) = frac-1517$ into $(3)$ just above:
beginalign
1 - frac-1517 &= 2sin^2left(fracx2right) \ \
1 + frac1517 & = 2sin^2left(fracx2right) \ \
frac3217 &= 2sin^2left(fracx2right) \ \
frac1617 &= sin^2left(fracx2right) \ \
pmfrac4sqrt17 &= sinleft(fracx2right) \ \
pmfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Now, we have to ask ourselves if the answer should be positive or negative, and why? If $x$ is in quadrant two, that means if we cut its angle in half, it will move to $x/2$ in quadrant 1. Try it yourself on a piece of paper. Or you can think about how the biggest angle in quadrant 2 is 180 degrees, so half of that is 90, so anything smaller than 180 must be cut in half to smaller than 90.
In quadrant 1, sin values are positive so we know the answer has to be
beginalignfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Edit: don't forget the plus or minus when taking the square root.
answered 1 hour ago
Zduff
1,452819
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Guide:
We do not plug in a value of $x$, we solve for the trigonometry value without having the need to find its value explicitly.
If $x$ is in the second quadrant, then $fracx2$ is in the first quadrant.
$$cos(x) = -frac1517$$
is $$2cos^2left( fracx2right)-1=-frac1517$$
Solve for $cosleft( fracx2right)$.
For the second question, given $sin(x)$ and knowing which quadrant it is in, you should be able to find $cos(x)$ using Pythagoras Theorem, and use double angle formula to recover $sin(2x), cos(2x)$. Then use the definition of tangent.
answered 2 hours ago
Siong Thye Goh
90.1k1460112
add a comment |Â
up vote
3
down vote
Guide:
We do not plug in a value of $x$, we solve for the trigonometry value without having the need to find its value explicitly.
If $x$ is in the second quadrant, then $fracx2$ is in the first quadrant.
$$cos(x) = -frac1517$$
is $$2cos^2left( fracx2right)-1=-frac1517$$
Solve for $cosleft( fracx2right)$.
For the second question, given $sin(x)$ and knowing which quadrant it is in, you should be able to find $cos(x)$ using Pythagoras Theorem, and use double angle formula to recover $sin(2x), cos(2x)$. Then use the definition of tangent.
answered 2 hours ago
Siong Thye Goh
90.1k1460112
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Guide:
We do not plug in a value of $x$, we solve for the trigonometry value without having the need to find its value explicitly.
If $x$ is in the second quadrant, then $fracx2$ is in the first quadrant.
$$cos(x) = -frac1517$$
is $$2cos^2left( fracx2right)-1=-frac1517$$
Solve for $cosleft( fracx2right)$.
For the second question, given $sin(x)$ and knowing which quadrant it is in, you should be able to find $cos(x)$ using Pythagoras Theorem, and use double angle formula to recover $sin(2x), cos(2x)$. Then use the definition of tangent.
answered 2 hours ago
Siong Thye Goh
90.1k1460112
Guide:
We do not plug in a value of $x$, we solve for the trigonometry value without having the need to find its value explicitly.
If $x$ is in the second quadrant, then $fracx2$ is in the first quadrant.
$$cos(x) = -frac1517$$
is $$2cos^2left( fracx2right)-1=-frac1517$$
Solve for $cosleft( fracx2right)$.
For the second question, given $sin(x)$ and knowing which quadrant it is in, you should be able to find $cos(x)$ using Pythagoras Theorem, and use double angle formula to recover $sin(2x), cos(2x)$. Then use the definition of tangent.
answered 2 hours ago
Siong Thye Goh
90.1k1460112
answered 2 hours ago
Siong Thye Goh
90.1k1460112
answered 2 hours ago
Siong Thye Goh
90.1k1460112
answered 2 hours ago
Siong Thye Goh
90.1k1460112
90.1k1460112
add a comment |Â
add a comment |Â
up vote
3
down vote
It is very helpful to know the following identities.
1) $sin ^2 x + cos ^2 x =1 $
2) $sin ^2 x = (1/2)(1-cos 2x)$
3) $cos ^2 x = (1/2 )(1+ cos 2x)$
4) $sin 2x = 2sin x cos x$
5) $cos 2x = cos ^2 x - sin ^2x $
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
add a comment |Â
up vote
3
down vote
It is very helpful to know the following identities.
1) $sin ^2 x + cos ^2 x =1 $
2) $sin ^2 x = (1/2)(1-cos 2x)$
3) $cos ^2 x = (1/2 )(1+ cos 2x)$
4) $sin 2x = 2sin x cos x$
5) $cos 2x = cos ^2 x - sin ^2x $
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
add a comment |Â
up vote
3
down vote
up vote
3
down vote
It is very helpful to know the following identities.
1) $sin ^2 x + cos ^2 x =1 $
2) $sin ^2 x = (1/2)(1-cos 2x)$
3) $cos ^2 x = (1/2 )(1+ cos 2x)$
4) $sin 2x = 2sin x cos x$
5) $cos 2x = cos ^2 x - sin ^2x $
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
It is very helpful to know the following identities.
1) $sin ^2 x + cos ^2 x =1 $
2) $sin ^2 x = (1/2)(1-cos 2x)$
3) $cos ^2 x = (1/2 )(1+ cos 2x)$
4) $sin 2x = 2sin x cos x$
5) $cos 2x = cos ^2 x - sin ^2x $
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
answered 2 hours ago
Mohammad Riazi-Kermani
37.3k41957
37.3k41957
add a comment |Â
add a comment |Â
up vote
0
down vote
- You have formulas, namely:
$sin(2x) = 2*cos(x)*sin(x)$
$cos(2x) = cos^2(x) - sin^2(x)$
$sin^2(x) + cos^2(x) = 1$
There are plenty more, but generally speaking everything can be derived from each other.
- You have the unit circle, which helps you determine the sign of the function. You need it, because when you try to derive a, for example, sin from the trig identity, you end up with expression: $ |sin(x)| = sqrt1 - cos^2(x)$ and at this point the quadrant information helps you recover the sign of sin(x)
For the first question you are given that $cos(x) = -15/17$ and $x$ is in the 2nd quadrant. This tells you, that $x in [pi/2;pi]$. Respectively, $x/2 in [pi/4;pi/2] = $ first quadrant, which means, that cos(x/2) and sin(x/2) are positive.
Now you just use the formula, for example, the second one I listed, to find the cos(x/2)
$cos(x) = cos^2(x/2) - sin^2(x/2) = 2cos^2(x/2) - 1 => |cos(x/2)| = cos(x/2)$(verified sign already to be positive)$ = sqrtfrac1 + cos(x)2$
I guess you can do the math.
answered 1 hour ago
Makina
3039
add a comment |Â
up vote
0
down vote
- You have formulas, namely:
$sin(2x) = 2*cos(x)*sin(x)$
$cos(2x) = cos^2(x) - sin^2(x)$
$sin^2(x) + cos^2(x) = 1$
There are plenty more, but generally speaking everything can be derived from each other.
- You have the unit circle, which helps you determine the sign of the function. You need it, because when you try to derive a, for example, sin from the trig identity, you end up with expression: $ |sin(x)| = sqrt1 - cos^2(x)$ and at this point the quadrant information helps you recover the sign of sin(x)
For the first question you are given that $cos(x) = -15/17$ and $x$ is in the 2nd quadrant. This tells you, that $x in [pi/2;pi]$. Respectively, $x/2 in [pi/4;pi/2] = $ first quadrant, which means, that cos(x/2) and sin(x/2) are positive.
Now you just use the formula, for example, the second one I listed, to find the cos(x/2)
$cos(x) = cos^2(x/2) - sin^2(x/2) = 2cos^2(x/2) - 1 => |cos(x/2)| = cos(x/2)$(verified sign already to be positive)$ = sqrtfrac1 + cos(x)2$
I guess you can do the math.
answered 1 hour ago
Makina
3039
add a comment |Â
up vote
0
down vote
up vote
0
down vote
- You have formulas, namely:
$sin(2x) = 2*cos(x)*sin(x)$
$cos(2x) = cos^2(x) - sin^2(x)$
$sin^2(x) + cos^2(x) = 1$
There are plenty more, but generally speaking everything can be derived from each other.
- You have the unit circle, which helps you determine the sign of the function. You need it, because when you try to derive a, for example, sin from the trig identity, you end up with expression: $ |sin(x)| = sqrt1 - cos^2(x)$ and at this point the quadrant information helps you recover the sign of sin(x)
For the first question you are given that $cos(x) = -15/17$ and $x$ is in the 2nd quadrant. This tells you, that $x in [pi/2;pi]$. Respectively, $x/2 in [pi/4;pi/2] = $ first quadrant, which means, that cos(x/2) and sin(x/2) are positive.
Now you just use the formula, for example, the second one I listed, to find the cos(x/2)
$cos(x) = cos^2(x/2) - sin^2(x/2) = 2cos^2(x/2) - 1 => |cos(x/2)| = cos(x/2)$(verified sign already to be positive)$ = sqrtfrac1 + cos(x)2$
I guess you can do the math.
answered 1 hour ago
Makina
3039
- You have formulas, namely:
$sin(2x) = 2*cos(x)*sin(x)$
$cos(2x) = cos^2(x) - sin^2(x)$
$sin^2(x) + cos^2(x) = 1$
There are plenty more, but generally speaking everything can be derived from each other.
- You have the unit circle, which helps you determine the sign of the function. You need it, because when you try to derive a, for example, sin from the trig identity, you end up with expression: $ |sin(x)| = sqrt1 - cos^2(x)$ and at this point the quadrant information helps you recover the sign of sin(x)
For the first question you are given that $cos(x) = -15/17$ and $x$ is in the 2nd quadrant. This tells you, that $x in [pi/2;pi]$. Respectively, $x/2 in [pi/4;pi/2] = $ first quadrant, which means, that cos(x/2) and sin(x/2) are positive.
Now you just use the formula, for example, the second one I listed, to find the cos(x/2)
$cos(x) = cos^2(x/2) - sin^2(x/2) = 2cos^2(x/2) - 1 => |cos(x/2)| = cos(x/2)$(verified sign already to be positive)$ = sqrtfrac1 + cos(x)2$
I guess you can do the math.
answered 1 hour ago
Makina
3039
answered 1 hour ago
Makina
3039
answered 1 hour ago
Makina
3039
answered 1 hour ago
Makina
3039
3039
add a comment |Â
add a comment |Â
up vote
0
down vote
I always forget the formulas for stuff like $sinleft(fracx2right)$. But you don't need to remember them. You only need to know two important formulas.
beginalign
1 = cos^2(x) + sin^2(x) tag1 \ \
cos(2x) = cos(x + x) = cos(x)cos(x) - sin(x)sin(x) = &cos^2(x) - sin^2(x) tag2
endalign
so we end up with
beginalign
1 &= cos^2(x) + sin^2(x) tag1 \ \
cos(2x) & = cos^2(x) - sin^2(x) tag2
endalign
But in these equations we didn't have to use $x$, we could have used $y$ or $a$ or $5$. They are all true no matter what real value we stick in there. So watch what happens if instead of putting $x$ in there we put $x/2$.
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(not2fracxnot2right) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
So here we are
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(xright) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
To get the half angle formulas all we have to do is add or substract these two equations. For instance, if we subtract $(2)$ from $(1)$ we get
beginalign
1 - cos(x) = 2sin^2left(fracx2right) tag3
endalign
Now, I'm going to plug in what you said that $cos(x) = frac-1517$ into $(3)$ just above:
beginalign
1 - frac-1517 &= 2sin^2left(fracx2right) \ \
1 + frac1517 & = 2sin^2left(fracx2right) \ \
frac3217 &= 2sin^2left(fracx2right) \ \
frac1617 &= sin^2left(fracx2right) \ \
pmfrac4sqrt17 &= sinleft(fracx2right) \ \
pmfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Now, we have to ask ourselves if the answer should be positive or negative, and why? If $x$ is in quadrant two, that means if we cut its angle in half, it will move to $x/2$ in quadrant 1. Try it yourself on a piece of paper. Or you can think about how the biggest angle in quadrant 2 is 180 degrees, so half of that is 90, so anything smaller than 180 must be cut in half to smaller than 90.
In quadrant 1, sin values are positive so we know the answer has to be
beginalignfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Edit: don't forget the plus or minus when taking the square root.
answered 1 hour ago
Zduff
1,452819
add a comment |Â
up vote
0
down vote
I always forget the formulas for stuff like $sinleft(fracx2right)$. But you don't need to remember them. You only need to know two important formulas.
beginalign
1 = cos^2(x) + sin^2(x) tag1 \ \
cos(2x) = cos(x + x) = cos(x)cos(x) - sin(x)sin(x) = &cos^2(x) - sin^2(x) tag2
endalign
so we end up with
beginalign
1 &= cos^2(x) + sin^2(x) tag1 \ \
cos(2x) & = cos^2(x) - sin^2(x) tag2
endalign
But in these equations we didn't have to use $x$, we could have used $y$ or $a$ or $5$. They are all true no matter what real value we stick in there. So watch what happens if instead of putting $x$ in there we put $x/2$.
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(not2fracxnot2right) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
So here we are
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(xright) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
To get the half angle formulas all we have to do is add or substract these two equations. For instance, if we subtract $(2)$ from $(1)$ we get
beginalign
1 - cos(x) = 2sin^2left(fracx2right) tag3
endalign
Now, I'm going to plug in what you said that $cos(x) = frac-1517$ into $(3)$ just above:
beginalign
1 - frac-1517 &= 2sin^2left(fracx2right) \ \
1 + frac1517 & = 2sin^2left(fracx2right) \ \
frac3217 &= 2sin^2left(fracx2right) \ \
frac1617 &= sin^2left(fracx2right) \ \
pmfrac4sqrt17 &= sinleft(fracx2right) \ \
pmfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Now, we have to ask ourselves if the answer should be positive or negative, and why? If $x$ is in quadrant two, that means if we cut its angle in half, it will move to $x/2$ in quadrant 1. Try it yourself on a piece of paper. Or you can think about how the biggest angle in quadrant 2 is 180 degrees, so half of that is 90, so anything smaller than 180 must be cut in half to smaller than 90.
In quadrant 1, sin values are positive so we know the answer has to be
beginalignfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Edit: don't forget the plus or minus when taking the square root.
answered 1 hour ago
Zduff
1,452819
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I always forget the formulas for stuff like $sinleft(fracx2right)$. But you don't need to remember them. You only need to know two important formulas.
beginalign
1 = cos^2(x) + sin^2(x) tag1 \ \
cos(2x) = cos(x + x) = cos(x)cos(x) - sin(x)sin(x) = &cos^2(x) - sin^2(x) tag2
endalign
so we end up with
beginalign
1 &= cos^2(x) + sin^2(x) tag1 \ \
cos(2x) & = cos^2(x) - sin^2(x) tag2
endalign
But in these equations we didn't have to use $x$, we could have used $y$ or $a$ or $5$. They are all true no matter what real value we stick in there. So watch what happens if instead of putting $x$ in there we put $x/2$.
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(not2fracxnot2right) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
So here we are
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(xright) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
To get the half angle formulas all we have to do is add or substract these two equations. For instance, if we subtract $(2)$ from $(1)$ we get
beginalign
1 - cos(x) = 2sin^2left(fracx2right) tag3
endalign
Now, I'm going to plug in what you said that $cos(x) = frac-1517$ into $(3)$ just above:
beginalign
1 - frac-1517 &= 2sin^2left(fracx2right) \ \
1 + frac1517 & = 2sin^2left(fracx2right) \ \
frac3217 &= 2sin^2left(fracx2right) \ \
frac1617 &= sin^2left(fracx2right) \ \
pmfrac4sqrt17 &= sinleft(fracx2right) \ \
pmfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Now, we have to ask ourselves if the answer should be positive or negative, and why? If $x$ is in quadrant two, that means if we cut its angle in half, it will move to $x/2$ in quadrant 1. Try it yourself on a piece of paper. Or you can think about how the biggest angle in quadrant 2 is 180 degrees, so half of that is 90, so anything smaller than 180 must be cut in half to smaller than 90.
In quadrant 1, sin values are positive so we know the answer has to be
beginalignfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Edit: don't forget the plus or minus when taking the square root.
answered 1 hour ago
Zduff
1,452819
I always forget the formulas for stuff like $sinleft(fracx2right)$. But you don't need to remember them. You only need to know two important formulas.
beginalign
1 = cos^2(x) + sin^2(x) tag1 \ \
cos(2x) = cos(x + x) = cos(x)cos(x) - sin(x)sin(x) = &cos^2(x) - sin^2(x) tag2
endalign
so we end up with
beginalign
1 &= cos^2(x) + sin^2(x) tag1 \ \
cos(2x) & = cos^2(x) - sin^2(x) tag2
endalign
But in these equations we didn't have to use $x$, we could have used $y$ or $a$ or $5$. They are all true no matter what real value we stick in there. So watch what happens if instead of putting $x$ in there we put $x/2$.
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(not2fracxnot2right) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
So here we are
beginalign
1 &= cos^2left(fracx2right) + sin^2left(fracx2right) tag1 \ \
cosleft(xright) & = cos^2left(fracx2right) - sin^2left(fracx2right) tag2
endalign
To get the half angle formulas all we have to do is add or substract these two equations. For instance, if we subtract $(2)$ from $(1)$ we get
beginalign
1 - cos(x) = 2sin^2left(fracx2right) tag3
endalign
Now, I'm going to plug in what you said that $cos(x) = frac-1517$ into $(3)$ just above:
beginalign
1 - frac-1517 &= 2sin^2left(fracx2right) \ \
1 + frac1517 & = 2sin^2left(fracx2right) \ \
frac3217 &= 2sin^2left(fracx2right) \ \
frac1617 &= sin^2left(fracx2right) \ \
pmfrac4sqrt17 &= sinleft(fracx2right) \ \
pmfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Now, we have to ask ourselves if the answer should be positive or negative, and why? If $x$ is in quadrant two, that means if we cut its angle in half, it will move to $x/2$ in quadrant 1. Try it yourself on a piece of paper. Or you can think about how the biggest angle in quadrant 2 is 180 degrees, so half of that is 90, so anything smaller than 180 must be cut in half to smaller than 90.
In quadrant 1, sin values are positive so we know the answer has to be
beginalignfrac4sqrt1717 &= sinleft(fracx2right)
endalign
Edit: don't forget the plus or minus when taking the square root.
answered 1 hour ago
Zduff
1,452819
edited 1 hour ago
answered 1 hour ago
Zduff
1,452819
answered 1 hour ago
Zduff
1,452819
answered 1 hour ago
Zduff
1,452819
1,452819
add a comment |Â
add a comment |Â
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1
Do you know the half angle formulas for sine and cosine? Write them out at the start of the question so we can see what you know. Perhaps the answer will be obvious if you actually write them.
– David K
2 hours ago