How would I solve this. I'm kinda stuck... $$\cos(y) + \cos(z) = -1$$ $$\sin(y) + \sin(z) = 0$$
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Hint: make a substitution of $y = u + v$ and $z = u - v$. Then try dividing the equations (taking care not to divide by $0$). – Theo Bendit Mar 02 '19 at 03:21
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By the sum to product identities, $$\cos(y)+\cos(z) = 2\cos(\frac{y+z}2)\cos(\frac{y-z}2)=-1$$$$\sin(y)+\sin(z) = 2\sin(\frac{y+z}2)\cos(\frac{y-z}2)=0$$
Since the first product isn't zero, neither of the multiplicands is zero. Hence, $\cos(\frac{y-z}2)\neq0\to\sin(\frac{y+z}2)=0\to y+z=2\pi k$ for some $k$. WLOG assume $y+z=2\pi$. So, $2\cos(\frac{y+z}2)=-2$, so $\cos(\frac{y-z}2)=\frac12$. Hence, $\frac{y-z}2=\pm\frac\pi3\to y-z=\pm\frac{2\pi}3$. You now have two equations, two variables. Can you solve from here?
Rushabh Mehta
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I would square the equations and add them and subtract them. You get some expressions for $\cos(y\pm z)$. Just remember to check the result with the original equations, since squaring them will transform $-1$ into $+1$
Andrei
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