Question

if cos theta + cos square theta + cos cube theta equal to 1 then sin power 6 theta - 4 sin power 4 theta + 8 sin square theta + 3 =?​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The value is 4

Step-by-step explanation:

$\theta=\frac{\pi}{3} \text { or } 60^{\circ}$

Explanation:

Okay. We've got:

$\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4$

Let's ignore the RHS for now.

$\begin{aligned}&\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta} \\&\frac{\cos \theta(1+\sin \theta)+\cos \theta(1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)} \\&\frac{\cos \theta((1-\sin \theta)+(1+\sin \theta))}{1-\sin ^{2} \theta} \\&\frac{\cos \theta(1-\sin \theta+1+\sin \theta)}{1-\sin ^{2} \theta}\end{aligned}$

$\frac{2 \cos \theta}{1-\sin ^{2} \theta}$

According to the Pythagorean Identity,

$\begin{aligned}&\sin ^{2} \theta+\cos ^{2} \theta=1 \text {. So: } \\&\cos ^{2} \theta=1-\sin ^{2} \theta\end{aligned}$

Now that we know that, we can write:

$\frac{2 \cos \theta}{\cos ^{2} \theta}$

$\frac{2}{\cos \theta}=4 \frac{\cos \theta}{2}=\frac{1}{4} \cos \theta=\frac{1}{2} \theta=\cos ^{-1}\left(\frac{1}{2}\right)$