Subjects trigonometry

Exact Trig Values 762052

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1. **Problem statement:** Find the exact values of the following trigonometric ratios: (i) $\sin\left(\frac{5\pi}{6}\right)$ (ii) $\cos\left(\frac{4\pi}{3}\right)$ (iii) $\tan\left(\frac{3\pi}{4}\right)$ (iv) $\sin\left(\frac{11\pi}{6}\right)$ (v) $\tan\left(-\frac{\pi}{4}\right)$ (vi) $\cos\left(-\frac{2\pi}{3}\right)$ 2. **Formula and rules:** - Use unit circle values for sine, cosine, and tangent. - Sine is positive in Q1 and Q2, cosine positive in Q1 and Q4, tangent positive in Q1 and Q3. - Negative angles rotate clockwise, so use symmetry properties: $\sin(-\theta) = -\sin(\theta)$, $\cos(-\theta) = \cos(\theta)$, $\tan(-\theta) = -\tan(\theta)$. 3. **Calculations:** (i) $\sin\left(\frac{5\pi}{6}\right)$ is in Q2 where sine is positive. Reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$. $$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$$ (ii) $\cos\left(\frac{4\pi}{3}\right)$ is in Q3 where cosine is negative. Reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$. $$\cos\left(\frac{4\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$$ (iii) $\tan\left(\frac{3\pi}{4}\right)$ is in Q2 where tangent is negative. Reference angle is $\pi - \frac{3\pi}{4} = \frac{\pi}{4}$. $$\tan\left(\frac{3\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1$$ (iv) $\sin\left(\frac{11\pi}{6}\right)$ is in Q4 where sine is negative. Reference angle is $2\pi - \frac{11\pi}{6} = \frac{\pi}{6}$. $$\sin\left(\frac{11\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$$ (v) $\tan\left(-\frac{\pi}{4}\right)$ uses the negative angle identity: $$\tan\left(-\frac{\pi}{4}\right) = -\tan\left(\frac{\pi}{4}\right) = -1$$ (vi) $\cos\left(-\frac{2\pi}{3}\right)$ uses the even property of cosine: $$\cos\left(-\frac{2\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right)$$ $\frac{2\pi}{3}$ is in Q2 where cosine is negative, reference angle $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$. $$\cos\left(\frac{2\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$$ **Final answers:** (i) $\frac{1}{2}$ (ii) $-\frac{1}{2}$ (iii) $-1$ (iv) $-\frac{1}{2}$ (v) $-1$ (vi) $-\frac{1}{2}$