Answer:
See below for proof.
Step-by-step explanation:
Given trigonometric identity:
[tex]\cos2x + 2 \sin^2x - \sin2x = (\cos x - \sin x)^2[/tex]
To prove the identity, we can use the following Double Angle Formulas:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Double Angle Formulas}}\\\\\sin (2 \theta)= 2 \sin \theta \cos \theta\\\\\cos (2 \theta)=\cos^2 \theta - \sin^2 \theta\end{array}}[/tex]
Rewrite the left side using these:
[tex]\cos^2x-\sin^2x+2\sin^2x-2\sin x \cos x[/tex]
Simplify:
[tex]\cos^2x+\sin^2x-2\sin x \cos x \\\\\\\cos^2x-2\sin x \cos x+\sin^2 x[/tex]
Now, factor by applying the perfect square formula, (a - b)² = a² - 2ab + b². In this case a = cos(x) and b = sin(x), so:
[tex](\cos x - \sin x)^2[/tex]
Hence, we have proved that:
[tex]\cos2x + 2 \sin^2x - \sin2x = (\cos x - \sin x)^2[/tex]
[tex]\dotfill[/tex]
As one calculation:
[tex]\;\;\;\:\cos2x + 2 \sin^2x - \sin2x\\\\=\cos^2x-\sin^2x+2\sin^2x-2\sin x \cos x\\\\=\cos^2x+\sin^2x-2\sin x \cos x \\\\=\cos^2x-2\sin x \cos x+\sin^2 x\\\\= (\cos x - \sin x)^2[/tex]