Math CentralQuandaries & Queries


Question from matt, a student:

how does sin2x break down (not with identities) and how would sin3x be created. My
prob. is sin 2x/ sin 3x and I want to know how the double(or triple angle) would break
down. I want to be able to cancel out sins. Thanks!

Hi Matt,

The only way I know to break down this expression is to use the multiple angle expressions for sine and cosine.

sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
cos(A + B) = cos(A) cos(B) - sin(A) sin(B)

Simplify the numerator with the sine expression by taking A = B = x and simplify the denominator using A = x and B = 2x. The resulting denominator will have sin(2x) and cos(2x) which you can write in terms of sin(x) and cos(x) using the expressions above. Finally if you use the fact that sin2(x) + cos2(x) = 1 you can write sin(2x)/sin(3x) as a function of cos(x) alone.



I was hoping to understand how sin2x was formed. ie sinx * sinx = sin2 x. I didn't find that out.


The similar expression for sin(2x) is sin(2x) = 2 sin(x) * cos(x). For sin(3x) it's
sin(3x) = 3 cos2(x) * sin(x) - sin3(x)

That's about the best you can do.



Can you break 3 cos2(x) * sin(x) = sin(3x) into the steps it would take to bring it back to sin(3x)? Thanks


I am going to start with sin(3x) but you certainly can follow the steps in reverse order if you wish. I will use the expressions for sin(A + B) and cos(A + B) above.

= sin(2x + x)
= sin(2x) cos(x) + cos(2x) sin(x)
= 2 sin(x) cos2(x) + [cos2(x) - sin2(x)]sin(x)
= 2 sin(x) cos2(x) + cos2(x) sin(x) - sin3(x)
= 3 sin(x) cos2(x) - sin3(x)


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