Question

1+cos^2(2theta) = 2(cos^4(theta)+sin^4(theta))

Answer 1

$$\begin{gathered} 1+{\cos }^2\left(2\theta \right)=2({\cos }^4\theta +{\sin }^4\theta ) \\ LHS\hspace{0.17em}=1+{\cos }^2\left(2\theta \right) \\ =1+(\cos 2\theta {)}^2 \\ =1+({\cos }^2\theta -{\sin }^2\theta {)}^2 \\ =1+({\cos }^4\theta +{\sin }^4\theta -2{\sin }^2\theta {\cos }^2\theta ) \\ AndRHS\hspace{0.17em}=2({\cos }^4\theta +{\sin }^4\theta )=2{\cos }^4\theta +2{\sin }^4\theta \\ So1+({\cos }^4\theta +{\sin }^4\theta -2{\sin }^2\theta {\cos }^2\theta )=2{\cos }^4\theta +2{\sin }^4\theta \\ Hence1-2{\sin }^2\theta {\cos }^2\theta ={\cos }^4\theta +{\sin }^4\theta \\ Or1={\cos }^4\theta +{\sin }^4\theta +2{\sin }^2\theta {\cos }^2\theta \\ Or({\cos }^2\theta +{\sin }^2\theta {)}^2=1 \\ Henceproved \end{gathered}$$

If you want the value of theta then at every theta the expression is satisfied.