Question

$\cos \theta = -\frac{3}{5}$, $\theta$ in Quadrant III $\sin \theta = \square$

          $\cos \theta = -\frac{3}{5}$, $\theta$ in Quadrant III
$\sin \theta = \square$
        
cosθ = -(3)/(5), θ in Quadrant III
sinθ = □

Added by Pedro B.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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cos heta =-(3)/(5), heta in Quadrant III sin heta = 1 cos0= 315 O in Quadrant III sin0=
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Transcript

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00:01 For this problem, we want to find the value of cosine of theta minus v.
00:04 Given the value of cosine of theta, which is equal to 3 over 5, with theta in the fourth quadrant, and tangent fee equal to negative square to 15, where fee is in quadrant 2.
00:16 Now we recall cosine of theta minus fee is equal to cosine of theta times cosine of fee plus sine of theta times sine of f.
00:27 In this identity, we only have to, have a value for cosine theta.
00:32 So we still have to find cosine fee, sine theta, and a sign of fee.
00:38 Now for sign of theta, we will use the information of cosine theta, which is equal to 3 over 5, where theta is in quadrant 4.
00:51 If we are to draw this one, it should look like this.
00:55 This being the terminal side, and then this will be our reference angle theta.
01:00 And since cosine theta is x over r, the radius will be 5 and the x value will be positive 3.
01:08 So we still have to find y, and to do that, we will use pythagorean theorem.
01:15 So in pythagorean theorem, that's 3 squared plus y squared equal to 5 squared, which means y squared is equal to 16...
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