## Trigonometry (11th Edition) Clone

$$(1+\sin x+\cos x)^2=2(1+\sin x)(1+\cos x)$$ The expression is an identity.
$$(1+\sin x+\cos x)^2=2(1+\sin x)(1+\cos x)$$ We would examine the left side first. $$A=(1+\sin x+\cos x)^2$$ $$A=[1+(\sin x+\cos x)]^2$$ $$A=1+2(\sin x+\cos x)+(\sin x+\cos x)^2$$ $$A=1+2(\sin x+\cos x)+(\sin^2 x+2\sin x\cos x+\cos^2 x)$$ $$A=1+2\sin x+2\cos x+\sin^2 x+2\sin x\cos x+\cos^2x$$ $$A=1+2\sin x+2\cos x+2\sin x\cos x+(\sin^2x+\cos^2x)$$ $$A=1+2\sin x+2\cos x+2\sin x\cos x+1$$ (for $\sin^2 x+\cos^2x=1$) $$A=2\sin x+2\cos x+2\sin x\cos x+2$$ Now from the right side, $$B=2(1+\sin x)(1+\cos x)$$ $$B=2(1+\cos x+\sin x+\sin x\cos x)$$ $$B=2+2\cos x+2\sin x+2\sin x\cos x$$ Therefore, $A=B$. The expression is an identity..