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# ​Trig identities again. I don’t get this?? The one with 2 stars is apparently harder but I don’t even get the easy one.

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1 Trig identities again. I don’t get this?? The one with 2 stars is apparently harder but I don’t even get the easy one.

Oct 1, 2018

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This doesn't involve proving trig identities, but using them to find the value(s) of $$\theta$$  and x that make the equations true.

By definition: $$\sec\theta=\frac{1}{\cos\theta}$$

Using this in a and rearranging:   $$\frac{1}{\cos^2\theta}=2\tan\theta$$

Multiply both sides by $$\cos^2\theta$$ and using the fact that $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ we get, on simplifying: $$1=2\sin\theta\cos\theta$$

The RHS is just $$\sin{2\theta}$$  so we have $$1=\sin{2\theta}$$

This results in $$2\theta=\frac{\pi}{2}\text{ or }\theta=\frac{\pi}{4}$$

For b rewrite x as $$2\theta$$ and use $$\cos{2\theta}=\cos^2\theta-\sin^2\theta\text{ and }\sin{}2\theta=2\sin\theta\cos\theta\text{ together with }\cos^2\theta+\sin^2\theta=1$$

You should eventually find there are two values of x between 0 and 2pi that satisfy the equation.

Oct 2, 2018
edited by Alan  Oct 2, 2018