addition formula for trigonometry problem

[hiroya]

1st post

sinx + sin3x + sin5x + sin7x = tan4x
cosx + cos3x + cos5x + cos7x

i can't seem to start this. first i tried the left side by using the general addition formula turning sin(3x) to sin(x + 2x) and sin5x to sin(3x+2) but i can't finish it cause its very long. can anyone help me with this?

 

[soroban]

Hello, hiroya!

Welcome aboard!

We need two lesser-known "addition" formulas:

. . \(\displaystyle \L\sin A\,+\,\sin B\:=\:2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\cos\left(\frac{A-B}{2}\right)\)

. . \(\displaystyle \L\cos A\,+\,\cos B\:=\:2\cdot\cos\left(\frac{A+B}{2}\right)\cdot\cos\left(\frac{A-B}{2}\right)\)

\(\displaystyle \L\frac{\sin x\,+\,\sin3x\,+\,\sin5x\,+\,\sin7x}{\cos x\,+\,\cos3x\,+\,\cos5x\,+\,\cos7x}\:=\:\tan4x\)

The numerator is:
. . . .\(\displaystyle \\sin x\,+\,\sin7x)\,+\,(\sin3x\,+\,5x)\)
. . \(\displaystyle =\:2\cdot\sin4x\cdot\cos3x\,+\,2\cdot\sin4x\cdot\cos x\)
. . \(\displaystyle =\:2\cdot\sin4x[\cos3x\,+\,\cos x]\)

The denominator is:
. . . . \(\displaystyle (\cos x\,+\,\cos7x)\,+\,(\cos3x\,+\,\cos5x)\)
. . \(\displaystyle = \:2\cdot\cos4x\cdot\cos3x\,+\,2\cdot\cos4x\cdot\cos x\)
. . \(\displaystyle = \:2\cdot\cos4x[\cos3x\,+\,\cos x]\)


The fraction becomes: \(\displaystyle \L\:\frac{2\cdot\sin4x[\cos3x\,+\,\cos x]}{2\cdot\cos4x[\cos3x\,+\,\cos x]} \:=\:\frac{\sin4x}{\cos4x}\:=\:\tan4x\)

 

[hiroya]

thanks, it really helped me a lot.

btw can i ask another question?

 

[jwpaine]

Don't ask to ask. Just ask. If your question does not have to do with this current threads topic, go ahead and create another thread for your question.

Either way, just ask and if it's not approperate to the thread you started, a moderator will split it.

Cheers,
John.