I was working through a trigonometry problem, and was having some difficulty so I decided to look at the solution. Here are the steps:

$$\frac{\sin(2x+50^\circ)+\sin(150^\circ)}{\sin(2x+50^\circ)-\sin(150^\circ)}=\frac{\cos(50^\circ)-\cos(2x+50^\circ)}{\cos(50^\circ)+\cos(2x+50^\circ)}$$ $$\frac{\sin(2x+50^\circ)}{\sin150^\circ}=\frac{-\cos50^\circ}{\cos(2x+50^\circ)}$$

(Image that replaced text.)

I am not exactly how the solution got from the first step to the second one. I would just like some clarification on the intermediate step.







This is similar to Siong's answer, but written differently using ratios: \begin{align} \frac{a+b}{a-b} - 1 &= \frac{c-d}{c+d} -1 \\[2ex] \require{cancel} \frac{2b}{a-b} &= -\frac{2d}{c+d} \\[2ex] \frac{a-b}{b} &= -\frac{c+d}{d} \\[2ex] \frac{a}{b} - 1 &= -\frac{c}{d} -1 \\ \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.