# Thread: Trigonometric Part of HSC 2016 Q 16 a) i)

1. ## Trigonometric Part of HSC 2016 Q 16 a) i)

Hi,

I was curious for Q 16 a) i), why does sin(theta)=sin(-alpha) mean that theta = - alpha? Why don't you have to consider it in terms of general solutions, with theta = 2*pi*k - (-1)^k alpha, and then some how work it out from there?

If -pi < x <= pi, -pi < y < pi, and sin(x) = 1/2 and sin(y) = -1/2, then couldn't x = pi/6, 5pi/6 and y = -pi/6, -5pi/6 . Then, couldn't x = pi/6 and y = -5pi/6, then x does not equal -y?

Thanks

2. ## Re: Trigonometric Part of HSC 2016 Q 16 a) i)

Substitute z and w into the equation given and take real and complex parts

(1) $\cos{\alpha}+\cos{\theta}+1=0$
(2) $\sin{\alpha}+\sin{\theta}=0$

Solutions to (2) are $\theta = -\alpha$ and $(-\pi+\alpha)$ The latter does not solve (1), as upon substitution gives 1=0, as $\cos{\alpha}=-\cos{(-\pi+\alpha)}$ leaving $\theta = -\alpha$ as the only valid solution

3. ## Re: Trigonometric Part of HSC 2016 Q 16 a) i)

Awesome! Thank you so much dan964 . I thought I was going crazy wondering why when only considering (2) theta could only equal - alpha.

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