Got some questions from my niece who is studying for her exams. This one, I couldn't figure out:
Simplify the following: $$\frac{\cos(4a)-1}{\sin(a)-\sin(3a)}$$
I'm stuck at the $4a$ and $3a$... should I split $4a$ in $2a+2a$? Then what about the $3a$?
Everything that can help us get it right is much appreciated!
$$\frac{\cos4a-1}{\sin{a}-\sin{3a}}=\frac{-2\sin^22a}{-2\sin{a}\cos2a}=\frac{\sin2a}{\sin{a}}\cdot\frac{\sin2a}{\cos2a}$$ $$=\frac{2\sin{a}\cos{a}}{\sin{a}}\cdot\tan2a=2\cos{a}\tan2a.$$ I used the following. $$1-\cos\alpha=2\sin^2\frac{\alpha}{2};$$ $$\sin\alpha-\sin\beta=2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2};$$ $$\sin2\alpha=2\sin\alpha\cos\alpha$$ and $$\frac{\sin\alpha}{\cos\alpha}=\tan\alpha.$$