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Suppose cos(4)=-0.3. Use the trig identity sin(4) + cos² (4) = 1 to find sin(4) in quadrant

II. Round to ten-thousandth.
A. 0.9539
B. 0.9154
C.-0.9276
D.-0.9348

1 Answer

3 votes

The answer is C. -0.9276.

We know that $\sin^2(\theta) + \cos^2(\theta) = 1$ for all angles $\theta$. Substituting $\theta = 4$, we get

$$\sin^2(4) + \cos^2(4) = 1.$$

Since $\cos(4) = -0.3$, we have

$$\sin^2(4) + (-0.3)^2 = 1.$$

Simplifying the right side, we get $\sin^2(4) = 0.91$. Taking the square root of both sides, we get

$$\sin(4) = \pm \sqrt{0.91}.$$

Since angle 4 is in quadrant II, the sine of the angle is negative, so $\sin(4) = -\sqrt{0.91}$. Rounding to ten-thousandths, we get $\sin(4) = \boxed{-0.9276}$.

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