Show that \sin(2x)+\sin(4x)+...+\sin(2nx)=\frac{\sin(nx)\sin((n+1)x}{\sin(x)}

etudiante9c2

etudiante9c2

Answered question

2022-05-01

Show that sin(2x)+sin(4x)+...+sin(2nx)=sin(nx)sin((n+1)xsin(x)

Answer & Explanation

Frederick Greer

Frederick Greer

Beginner2022-05-02Added 17 answers

Let
1rnsin2rx=sinnxsin(n+1)xsinx
1rn+1sin2rx=sinnxsin(x+1)xsinx+sin2(n+1)x
=2sinnxsin(n+1)x+2sin2(n+1)xsinx2sinx
Applying 2sinAsinB=cos(AB)cos(A+B)
2sinnxsin(n+1)x+2sin2n+1)xsinx
=cosxcos(2n+1)x+cos(2n+1)xcos(2n+3)x
=cosxcos(2n+3)x
Now apply cosCcosD=2sinC+D2sinDC2 to get
cosxcos(2n+3)x=2sin(n+2)xsin(n+1)x

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?