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!

  • $\begingroup$ I would write $\sin$ and $\cos$ with the complex exponential and pray for some simplifications, but maybe there's a more clever way to follow $\endgroup$
    – tommy1996q
    Oct 14, 2018 at 8:53

1 Answer 1


$$\frac{\cos4a-1}{\sin{a}-\sin{3a}}=\frac{-2\sin^22a}{-2\sin{a}\cos2a}=\frac{\sin2a}{\sin{a}}\cdot\frac{\sin2a}{\cos2a}$$ $$=\frac{2\sin{a}\cos{a}}{\sin{a}}\cdot\tan2a=2\cos{a}\tan2a.$$ 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 $$\frac{\sin\alpha}{\cos\alpha}=\tan\alpha.$$

  • $\begingroup$ Ohhh right, Werner and prosthaphaeresis, I always forget them $\endgroup$
    – tommy1996q
    Oct 14, 2018 at 8:59
  • $\begingroup$ Thanks! Could you maybe provide a bit more steps so I see which formulas you've used? Still can't wrap my head around it... $\endgroup$
    – binoculars
    Oct 14, 2018 at 9:06
  • $\begingroup$ @binoculars I added something. See now. $\endgroup$ Oct 14, 2018 at 9:13
  • $\begingroup$ Got it! I didn't catch cos(4a)-1 could be seen as -(1-cos(4a)) at first. Thanks! $\endgroup$
    – binoculars
    Oct 14, 2018 at 9:22
  • $\begingroup$ You are welcome! $\endgroup$ Oct 14, 2018 at 9:29

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