This relates to the shift method in trigonometry and I was trying do the same question with all the different variations


I tried solving it using $\sin(\theta-\alpha)$ and it worked. And so then I tried using $\sin(\theta+\alpha)$ by changing the equation into $\sin\theta\cos\alpha-(-\sin\alpha\cos\theta)$ and this also worked. But when I tried solving it using $\cos(\theta+\alpha)$ I get an answer saying that $\theta=0$ and i'm fairly sure that shouldn't happen.

$\rightarrow \cos(\theta+\alpha)=\cos\alpha\cos\theta-\sin\alpha\sin\theta$

$=-\sin\alpha\sin\theta-(-\cos\alpha\cos\theta)$ (am I allowed to do that??)

$\rightarrow r\sin\alpha=-3$ and $r\cos\alpha=-2$

$\rightarrow \tan\alpha=\frac{3}{2}$ , $a=56^\circ19'$

$\rightarrow \sqrt{13}\cos(\theta+56^\circ19')=2$

$\rightarrow \cos(\theta+56^\circ19')=\frac{2}{\sqrt{13}}$

$\rightarrow \theta+56^\circ19'=56^\circ19'$


Since $r$ should be positive, $\sin\alpha$ and $\cos\alpha$ are both negative, so you are in the third quadrant, so the angle is not $56^\circ19'$ but rather $56^\circ19'+180^\circ$. $\theta$ turns out to be $\pi+2n\pi$ (or $-\pi+2n\pi$)

  • $\begingroup$ Oh ok i see. thank you $\endgroup$ – kjhg Jul 29 '16 at 12:36

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