Question

0-theta ...if sin0 +sin^2 0 = 1 ...then proove cos^2 0+cos^4 0 =1

Answer 1

So you mean if
\[\sin\theta+\sin^2\theta=1\]
show that
\[\cos^2\theta+\cos^4\theta=1~~?\]
The first equation can be solved for \(\theta\):
\[\begin{align*}\sin\theta+\sin^2\theta&=1\\
\sin^2\theta+\sin\theta-1&=0\\
\sin^2\theta+\sin\theta+\frac{1}{4}-\frac{5}{4}&=0\\
\left(\sin\theta+\frac{1}{2}\right)^2&=\frac{5}{4}\\
\sin\theta+\frac{1}{2}&=\pm\frac{\sqrt5}{2}\\
\sin\theta&=\frac{-1\pm\sqrt5}{2}
\end{align*}\]
In fact, you don't have to solve directly for \(\theta\) to continue. \(\sin\theta\) will do fine.

Recall the identity: \(\sin^2\theta+\cos^2\theta=1\). The second equation is then
\[\begin{align*}\cos^2\theta+\cos^4\theta&=\left(1-\sin^2\theta\right)+\left(1-\sin^2\theta\right)^2
\end{align*}\]
You have two values for \(\sin\theta\) that give
\[\sin^2\theta=\frac{3-\sqrt5}{2},~\frac{3+\sqrt5}{2}\]
Plugging into the equation above, you have
\[\begin{align*}\cos^2\theta+\cos^4\theta&=\left(1-\frac{3-\sqrt5}{2}\right)+\left(1-\frac{3-\sqrt5}{2}\right)^2\\
&\text{and}\\
\cos^2\theta+\cos^4\theta&=\left(1-\frac{3+\sqrt5}{2}\right)+\left(1-\frac{3+\sqrt5}{2}\right)^2
\end{align*}\]
Verify that both right hand sides reduce to 1.