Sin 5a cos 2a - sin 6a cos a upon sin 2a sin a minus sin 2a cos 3a
Answers
Step-by-step explanation:
We are asked to show that (sin3a+sina+sin5a)/(cos3a+cosa+cos5a)=tan3a.
Note that sin3a=sina(-1+4cos^2a) and sin5a=sina(1-12cos^2a+16cos^4a)
Similarly cos3a=cosa(1-4sin^2a) and cos5a=cosa(1-12sin^2a+16sin^4a)
Consider the numerator. Substituting and factoring out the common sina yields:
sina((-1+4cos^2a)+1+(1-12cos^2a+16cos^4a)) or
sina(1-8cos^2a+16cos^4a)=sina(4cos^2a-1)^2
The denominator, after substituting and factoring out a common cosa, yields:
cosa((1-4sin^2a)+1+(1-12sin^2a+16sin^4a)) or
cosa(3-16sin^2a+16sin^4a)=cosa(-4sin^2a+3)(-4sin^2a+1)
So the rational expression can be expressed as the product of three rational expressions:
sina/cosa * (4cos^2a-1)/(-4sin^2a+3) * (4cos^2a-1)/(-4sin^2a+1)
4cos^2a-1 = -4sin^2+3, so the middle factor is 1. The first factor is tana.
The last factor:
(4cos^2a-1)/(-4sin^2+1)=(2cos^2a+cos2a)/(cos2a-2sin^2a)
=(2cos^2a+cos^2a-sin^2a)/(cos^2a-sin^2a-2sin^2a)
=(3cos^2a-sin^2a)/(cos^2a-3sin^2a)
=(3-sin^2a/cos^2a)/(1-3sin^2a/cos^2a)
=(3-tan^2a)/(1-3tan^2a)
Thus we have tana * 1 * (3-tan^2a)/(1-3tan^2a)
=(3tana-tan^3a)/(1-3tan^2a)
=tan(3a) as required.