using trigonometric identities

Question

For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$

I tried to use that:

\begin{align} \sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\ &=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\ &=1^2-\frac{1}{2}(2\sin x\cos x)^2\\ &=1-\frac{1}{2}\sin^2 (2x)\\ &=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\ &=\frac{3}{4}+\frac{1}{4}\cos 4x \end{align}

but i can't try more

Answer 1

$\displaystyle \sin^4x+\cos^2x=\frac{(1-\cos2x)^2}{4}+\frac{1+\cos 2x}{2}=\frac{3+\cos^22x}{4}=\frac{3+\frac{1+\cos 4x}{2}}{4}=\frac{7+\cos4x}{8}$

$\displaystyle \sin^2x+\cos^4x=\sin^4\left(\frac \pi2-x\right)+\cos^2\left(\frac \pi2-x\right)=\frac{7+\cos4\left(\frac \pi2-x\right)}{8}=\frac{7+\cos4x}{8}$

Answer 2

Note that $$\begin{align}\sin^4x +\cos^2x &= \sin^2x(1-\cos^2x) +\cos^2x\\ &=\sin^2x+\cos^2x(1-\sin^2x)=\sin^2x +\cos^4x. \end{align}$$ Hence, according to your work, $$\begin{align} 2(\sin^4x +\cos^2x)&=(\sin^4x +\cos^2x) +(\sin^2x +\cos^4x)\\ &=\sin^4x +\cos^4x+1=\frac{7+\cos 4x}{4}.\end{align}$$ Can you take it from here?

Answer 3

$c^4-s^4=(c^2+s^2)(c^2-s^2)=c^2-s^2$

$\iff c^4+s^2=c^2+s^4=P$(say) where $c=\cos x,s=\sin x$

$P+P=1+c^4+s^4=1+(c^2+s^2)^2-2c^2s^2=2-\dfrac{\sin^22x}2=2-\dfrac{1-\cos4x}4=?$

using $\sin2x=2\sin x\cos x,\cos2y=2-2\sin^2y$

We need $$\dfrac1P+\dfrac2P=?$$