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Prove the identity.\newline(sinx+cosx)2=1+sin2x(\sin x+\cos x)^{2}=1+\sin 2x

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Q. Prove the identity.\newline(sinx+cosx)2=1+sin2x(\sin x+\cos x)^{2}=1+\sin 2x
  1. Expand using binomial formula: We start by expanding the left side of the equation using the binomial formula (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2.\newline(\sin x + \cos x)^\(2 = (\sin x)^22 + 22(\sin x)(\cos x) + (\cos x)^22
  2. Apply Pythagorean identity: Next, we recognize that (sinx)2(\sin x)^2 and (cosx)2(\cos x)^2 are part of the Pythagorean identity, which states that (sinx)2+(cosx)2=1(\sin x)^2 + (\cos x)^2 = 1.\newline(sinx)2+(cosx)2=1(\sin x)^2 + (\cos x)^2 = 1
  3. Use double angle formula: We also know that 2(sinx)(cosx)2(\sin x)(\cos x) is the double angle formula for sine, which states that sin2x=2(sinx)(cosx)\sin 2x = 2(\sin x)(\cos x).\newline2(sinx)(cosx)=sin2x2(\sin x)(\cos x) = \sin 2x
  4. Substitute into expanded equation: Now, we substitute the Pythagorean identity and the double angle formula into our expanded equation. \newline(sinx+cosx)2=1+sin2x(\sin x + \cos x)^2 = 1 + \sin 2x
  5. Prove the identity: We have shown that the left side of the equation, when expanded and simplified using known trigonometric identities, equals the right side of the equation.\newlineTherefore, the identity (sinx+cosx)2=1+sin2x(\sin x + \cos x)^2 = 1 + \sin 2x is proven.