Sin Cos Identity 2A595B

1. **State the problem:** We need to prove that $$(\sin x + \cos x)^2 + (\sin x - \cos x)^2 = 2$$ for any angle $x$. 2. **Recall the formula:** The square of a sum and difference are given by: $$ (a+b)^2 = a^2 + 2ab + b^2 $$ $$ (a-b)^2 = a^2 - 2ab + b^2 $$ 3. **Apply the formula:** Let $a = \sin x$ and $b = \cos x$. $$ (\sin x + \cos x)^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x $$ $$ (\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x $$ 4. **Add the two expressions:** $$ (\sin x + \cos x)^2 + (\sin x - \cos x)^2 = (\sin^2 x + 2\sin x \cos x + \cos^2 x) + (\sin^2 x - 2\sin x \cos x + \cos^2 x) $$ 5. **Simplify by combining like terms:** $$ = \sin^2 x + 2\sin x \cos x + \cos^2 x + \sin^2 x - 2\sin x \cos x + \cos^2 x $$ $$ = (\sin^2 x + \sin^2 x) + (2\sin x \cos x - 2\sin x \cos x) + (\cos^2 x + \cos^2 x) $$ $$ = 2\sin^2 x + 0 + 2\cos^2 x $$ 6. **Use the Pythagorean identity:** $$ \sin^2 x + \cos^2 x = 1 $$ So, $$ 2\sin^2 x + 2\cos^2 x = 2(\sin^2 x + \cos^2 x) = 2 \times 1 = 2 $$ 7. **Conclusion:** We have shown that $$ (\sin x + \cos x)^2 + (\sin x - \cos x)^2 = 2 $$ which completes the proof.