Question

Prove that,
(1+sin theta + cos theta) ^2 = 2(1+sin theta) (1+cos theta) ​

Answer 1

Answer:

Proof below

Step-by-step explanation:

Trigonometric Identities

Before we prove the given identity, we need to recall:

$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc$

We have to prove that:

$(1+\sin\theta+\cos\theta)^2=2(1+\sin\theta)(1+\cos\theta)$

Let's expand the square on the left side as indicated in the formula above:

$(1+\sin\theta+\cos\theta)^2=1^2+\sin^2\theta+\cos^2\theta+2\sin\theta+2\cos\theta+2\sin\theta\cos\theta$

Recall the fundamental trigonometric identity:

$\sin^2\theta+\cos^2\theta=1$

Substituting and operating:

$(1+\sin\theta+\cos\theta)^2=2+2\sin\theta+2\cos\theta+2\sin\theta\cos\theta$

Factoring by 2:

$(1+\sin\theta+\cos\theta)^2=2(1+\sin\theta+\cos\theta+\sin\theta\cos\theta)$

Factoring cos θ in the last two terms:

$(1+\sin\theta+\cos\theta)^2=2[1+\sin\theta+\cos\theta(1+\sin\theta)]$

Factoring $1+\sin\theta$

$(1+\sin\theta+\cos\theta)^2=2(1+\sin\theta)(1+\cos\theta)$

Proven