Answer to Question #311451 in Trigonometry for bookaddict

Question

Prove:

a. (sin𝑥+cos𝑥)^2 ≡1+2sin𝑥cos𝑥

b. (1+sin𝑥+ cos𝑥)^2 ≡2 (1+sin𝑥) (1+cos𝑥)

c. (6−cos^2𝜃)/(sin^2𝜃+5) ≡1

Answer 1

a. (sinx+cosx)² = (sinx+cosx)(sinx+cosx) = sinx(sinx+cosx) + cosx(sinx+cosx) = sin²x + sinxcosx + cosxsinx + cos²x = sin²x + 2sinxcosx + cos²x = (sin²x + cos²x) + 2sinxcosx = 1 + 2sinxcosx

b. (1+ sinx + cosx)² = (1 + sinx + cosx)(1 + sinx + cosx) = (1 + sinx + cosx) + sinx(1 + sinx + cosx) + cosx(1+ sinx + cosx) = 1 + sinx + cosx + sinx + sin²x +sinxcosx + cosx +cosxsinx + cos²x = 1 + (sin²x + cos²x) + 2sinx + 2cosx + 2sinxcosx = 2 + 2sinx(1+cosx) + 2cosx = 2(1 + cosx) + 2sinx(1 + cosx) = 2(1 + cosx)(1 + sinx) = 2(1 + sinx)(1 + cosx)

c. $\dfrac{6-cos^2\theta}{sin^2\theta+5} = \dfrac{6sin^2\theta+6cos^2\theta-cos^2\theta}{sin^2\theta+5sin^2\theta +5cos^2\theta} = \dfrac{6sin^2\theta+5cos^2\theta}{6sin^2\theta +5cos^2\theta} =1$