Question

2 cos theta; 3 sin theta - 4 sin cube theta = 1; prove

Answer 1

Answer:

4 = Cos^{6}\theta - 4Cos^{4}\theta + 8Cos^{2}\theta

Step-by-step explanation:

In the question  

We have  

Sin\theta + Sin^{2}\theta+ Sin^{3}\theta = 1

Sin\theta + Sin^{3}\theta = 1 - Sin^{2}\theta

Sin\theta (1 +Sin^{2}\theta) = Cos^{2}\theta

Sin\theta(1 + 1 - Cos^{2}\theta) = Cos^{2}\theta

Sin\theta (2 - Cos^{2}\theta) = Cos^{2}\theta

Squaring on both sides

[Sin\theta(2 - Cos^{2}\theta)]^2 =[Cos^2}\theta]^2

Sin^{2}\theta (2 - Cos^{2}\theta)^2 = Cos^{4}\theta

(1 - Cos^{2}\theta) (4 + Cos^{4}\theta - 4Cos^{2}\theta) = Cos^{4}\theta

4 + Cos^{4}\theta  - 4Cos^{2}\theta  - 4Cos^{2}\theta  - Cos^{6}\theta  + 4Cos^{4}\theta = Cos^{4}\theta

4 = Cos^{6}\theta + 8Cos^{2}\theta - 4Cos^{4}\theta

4 = Cos^{6}\theta - 4Cos^{4}\theta + 8Cos^{2}\theta

 Hence the result.