When solving for $sin$ or $cos(2t)$ given $sin$ or $cos(t)$ is the quadrant relevant?
$begingroup$
I notice some of the homework problems in my book ask me to find the sine or cosine of twice an angle given the sine or cosine of the angle. They also mention $P(t)$ is in some given quadrant.
My question is, wouldn't the answer be the same regardless of which quadrant $P(t)$ is in? I suspect this is just the book's way of trying to throw me off.
trigonometry
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
$endgroup$
add a comment |
$begingroup$
I notice some of the homework problems in my book ask me to find the sine or cosine of twice an angle given the sine or cosine of the angle. They also mention $P(t)$ is in some given quadrant.
My question is, wouldn't the answer be the same regardless of which quadrant $P(t)$ is in? I suspect this is just the book's way of trying to throw me off.
trigonometry
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
$endgroup$
add a comment |
$begingroup$
I notice some of the homework problems in my book ask me to find the sine or cosine of twice an angle given the sine or cosine of the angle. They also mention $P(t)$ is in some given quadrant.
My question is, wouldn't the answer be the same regardless of which quadrant $P(t)$ is in? I suspect this is just the book's way of trying to throw me off.
trigonometry
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
$endgroup$
I notice some of the homework problems in my book ask me to find the sine or cosine of twice an angle given the sine or cosine of the angle. They also mention $P(t)$ is in some given quadrant.
My question is, wouldn't the answer be the same regardless of which quadrant $P(t)$ is in? I suspect this is just the book's way of trying to throw me off.
trigonometry
trigonometry
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
edited Apr 16 '16 at 21:48
Jennifer
8,41721737
8,41721737
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
asked Apr 16 '16 at 21:45
moonman239moonman239
338114
338114
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
The quadrant is relevant. Consider the following examples.
Example 1. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a first quadrant angle.
Solution. Since $theta$ is a first quadrant angle, $costheta > 0$. Thus,
$$costheta = sqrt{1 - sin^2theta} = frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(frac{4}{5}right) = frac{24}{25}$$
Example 2. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a second quadrant angle.
Solution. Since $theta$ is a second quadrant angle, $costheta < 0$. Thus,
$$costheta = -sqrt{1 - sin^2theta} = -frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(-frac{4}{5}right) = -frac{24}{25}$$
Notice that the only difference in the statement of examples 1 and 2 is the quadrant of the angle $theta$. Hence, the reason we obtained values with different signs was due solely to the quadrant of the angle.
On the other hand, the double angle formulas for cosine
begin{align*}
cos(2theta) & = cos^2theta - sin^2theta\
& = 2cos^2theta - 1\
& = 1 - 2sin^2theta
end{align*}
depend on the squares of sine and/or cosine, so the value of $cos(2theta)$ does not depend on the quadrant of the angle.
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
$endgroup$
add a comment |
$begingroup$
It does depend on the quadrant. A simplified way to think about it is that $cos(x)$ has the sign of the $x$-coordinate and $sin(x)$ has the sign of the $y$-coordinate. For example, the sign of cosine in quadrant one is positive but the sign of cosine in the third quadrant is negative due to the sign of the $x$-axis in each quadrant.
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
oldest
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votes
$begingroup$
The quadrant is relevant. Consider the following examples.
Example 1. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a first quadrant angle.
Solution. Since $theta$ is a first quadrant angle, $costheta > 0$. Thus,
$$costheta = sqrt{1 - sin^2theta} = frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(frac{4}{5}right) = frac{24}{25}$$
Example 2. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a second quadrant angle.
Solution. Since $theta$ is a second quadrant angle, $costheta < 0$. Thus,
$$costheta = -sqrt{1 - sin^2theta} = -frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(-frac{4}{5}right) = -frac{24}{25}$$
Notice that the only difference in the statement of examples 1 and 2 is the quadrant of the angle $theta$. Hence, the reason we obtained values with different signs was due solely to the quadrant of the angle.
On the other hand, the double angle formulas for cosine
begin{align*}
cos(2theta) & = cos^2theta - sin^2theta\
& = 2cos^2theta - 1\
& = 1 - 2sin^2theta
end{align*}
depend on the squares of sine and/or cosine, so the value of $cos(2theta)$ does not depend on the quadrant of the angle.
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
$endgroup$
add a comment |
$begingroup$
The quadrant is relevant. Consider the following examples.
Example 1. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a first quadrant angle.
Solution. Since $theta$ is a first quadrant angle, $costheta > 0$. Thus,
$$costheta = sqrt{1 - sin^2theta} = frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(frac{4}{5}right) = frac{24}{25}$$
Example 2. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a second quadrant angle.
Solution. Since $theta$ is a second quadrant angle, $costheta < 0$. Thus,
$$costheta = -sqrt{1 - sin^2theta} = -frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(-frac{4}{5}right) = -frac{24}{25}$$
Notice that the only difference in the statement of examples 1 and 2 is the quadrant of the angle $theta$. Hence, the reason we obtained values with different signs was due solely to the quadrant of the angle.
On the other hand, the double angle formulas for cosine
begin{align*}
cos(2theta) & = cos^2theta - sin^2theta\
& = 2cos^2theta - 1\
& = 1 - 2sin^2theta
end{align*}
depend on the squares of sine and/or cosine, so the value of $cos(2theta)$ does not depend on the quadrant of the angle.
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
$endgroup$
add a comment |
$begingroup$
The quadrant is relevant. Consider the following examples.
Example 1. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a first quadrant angle.
Solution. Since $theta$ is a first quadrant angle, $costheta > 0$. Thus,
$$costheta = sqrt{1 - sin^2theta} = frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(frac{4}{5}right) = frac{24}{25}$$
Example 2. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a second quadrant angle.
Solution. Since $theta$ is a second quadrant angle, $costheta < 0$. Thus,
$$costheta = -sqrt{1 - sin^2theta} = -frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(-frac{4}{5}right) = -frac{24}{25}$$
Notice that the only difference in the statement of examples 1 and 2 is the quadrant of the angle $theta$. Hence, the reason we obtained values with different signs was due solely to the quadrant of the angle.
On the other hand, the double angle formulas for cosine
begin{align*}
cos(2theta) & = cos^2theta - sin^2theta\
& = 2cos^2theta - 1\
& = 1 - 2sin^2theta
end{align*}
depend on the squares of sine and/or cosine, so the value of $cos(2theta)$ does not depend on the quadrant of the angle.
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
$endgroup$
The quadrant is relevant. Consider the following examples.
Example 1. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a first quadrant angle.
Solution. Since $theta$ is a first quadrant angle, $costheta > 0$. Thus,
$$costheta = sqrt{1 - sin^2theta} = frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(frac{4}{5}right) = frac{24}{25}$$
Example 2. Find $sin(2theta)$ if $sintheta = dfrac{3}{5}$ and $theta$ is a second quadrant angle.
Solution. Since $theta$ is a second quadrant angle, $costheta < 0$. Thus,
$$costheta = -sqrt{1 - sin^2theta} = -frac{4}{5}$$
Hence,
$$sin(2theta) = 2sinthetacostheta = 2left(frac{3}{5}right)left(-frac{4}{5}right) = -frac{24}{25}$$
Notice that the only difference in the statement of examples 1 and 2 is the quadrant of the angle $theta$. Hence, the reason we obtained values with different signs was due solely to the quadrant of the angle.
On the other hand, the double angle formulas for cosine
begin{align*}
cos(2theta) & = cos^2theta - sin^2theta\
& = 2cos^2theta - 1\
& = 1 - 2sin^2theta
end{align*}
depend on the squares of sine and/or cosine, so the value of $cos(2theta)$ does not depend on the quadrant of the angle.
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
edited Nov 24 '18 at 10:47
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
answered Apr 16 '16 at 22:19
N. F. TaussigN. F. Taussig
43.7k93355
43.7k93355
add a comment |
add a comment |
$begingroup$
It does depend on the quadrant. A simplified way to think about it is that $cos(x)$ has the sign of the $x$-coordinate and $sin(x)$ has the sign of the $y$-coordinate. For example, the sign of cosine in quadrant one is positive but the sign of cosine in the third quadrant is negative due to the sign of the $x$-axis in each quadrant.
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
$endgroup$
add a comment |
$begingroup$
It does depend on the quadrant. A simplified way to think about it is that $cos(x)$ has the sign of the $x$-coordinate and $sin(x)$ has the sign of the $y$-coordinate. For example, the sign of cosine in quadrant one is positive but the sign of cosine in the third quadrant is negative due to the sign of the $x$-axis in each quadrant.
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
$endgroup$
add a comment |
$begingroup$
It does depend on the quadrant. A simplified way to think about it is that $cos(x)$ has the sign of the $x$-coordinate and $sin(x)$ has the sign of the $y$-coordinate. For example, the sign of cosine in quadrant one is positive but the sign of cosine in the third quadrant is negative due to the sign of the $x$-axis in each quadrant.
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
$endgroup$
It does depend on the quadrant. A simplified way to think about it is that $cos(x)$ has the sign of the $x$-coordinate and $sin(x)$ has the sign of the $y$-coordinate. For example, the sign of cosine in quadrant one is positive but the sign of cosine in the third quadrant is negative due to the sign of the $x$-axis in each quadrant.
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
edited Nov 24 '18 at 3:20
Rócherz
2,7762721
2,7762721
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
answered Apr 16 '16 at 21:56
A.RiesenA.Riesen
18819
18819
add a comment |
add a comment |
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