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How should one go about solving this problem?

Use identities to find the exact values at $\alpha$ for the remaining five trigonometric functions.

$\cos\alpha = -\sqrt{2}/4$ and $\alpha$ is in quadrant III.

Thanks

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closed as off-topic by Clarinetist, Em., colormegone, Leucippus, choco_addicted May 17 '16 at 2:59

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For $\sin\alpha$, use the Trigonometry Fundamental Theorem, $$\sin^2\alpha+\cos^2\alpha=1,$$ to obtain $\sin\alpha=\pm\sqrt{14}/4.$ Because $\alpha$ is in QIII, its sine must be negative.

For the other four, use $\tan\alpha=\frac{\sin\alpha}{\cos\alpha},$ $\sec\alpha=\frac{1}{\cos\alpha},$ $\csc\alpha=\frac{1}{\sin\alpha},$ $\cot\alpha=\frac{\cos\alpha}{\sin\alpha}.$

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Outline:

In QIII, both sine and cosine are negative.

Use the Pythagorean identity: $\sin^2 \alpha+\cos^2 \alpha = 1$ to find the value of $\sin\alpha$.

Then use quotient and/or reciprocal identities to find the other trig values of $\alpha$.

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