Question

Explaination fast.

Wrong answer will be reported

Answer 1

$\displaystyle\longrightarrow\sf{\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^2(2\alpha)=\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}(2\sin\alpha\cos\alpha)^2}$

$\displaystyle\longrightarrow\sf{\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^2(2\alpha)=\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\cdot4\sin^2\alpha\cos^2\alpha}$

$\displaystyle\longrightarrow\sf{\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^2(2\alpha)=\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha}$

$\displaystyle\longrightarrow\sf{\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^2(2\alpha)=(\sin^2\alpha)^2+2\sin^2\alpha\cos^2\alpha+(\cos^2\alpha)^2}$

$\displaystyle\longrightarrow\sf{\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^2(2\alpha)=(\sin^2\alpha+\cos^2\alpha)^2}$

$\displaystyle\longrightarrow\sf{\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^2(2\alpha)=1^2}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\sin^4\alpha+\cos^4\alpha+\dfrac{1}{2}\sin^2(2\alpha)=1}}}$