The question asks to prove that - \(\displaystyle{\frac{{{\sin{{x}}}}}{{{\cos{{3}}}{x}}}}+{\frac{{{\sin{{3}}}{x}}}{{{\cos{{9}}}{x}}}}+{\frac{{{\sin{{9}}}{x}}}{{{\cos{{27}}}{x}}}}={\frac{{{1}}}{{{2}}}}{\left({\tan{{27}}}{x}-{\tan{{x}}}\right)}\)

Trent Fuller

Trent Fuller

Answered question

2022-04-10

The question asks to prove that -
sinxcos3x+sin3xcos9x+sin9xcos27x=12(tan27xtanx)

Answer & Explanation

cutimnm135imsa

cutimnm135imsa

Beginner2022-04-11Added 21 answers

In general, we have
k=1nsin(3k1x)cos(3kx)=tan(3nx)tan(x)2
We can prove this using induction. n=1 is trivial. All we need to make use of is that
tan(3t)=3tan(t)tan3(t)13tan2(t)
For the inductive step, we have
k=1nsin(3k1x)cos(3kx)=tan(3nx)tan(x)2
Adding the last term, we obtain
k=1n=1sin(3k1x)cos(3kx)=tan(3nx)tan(x)2+sin(3nx)cos(3n+1x)
=tan(3n+1x)tan(x)2

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?