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High School
Geometry
Trigonometry
Trigonometric equations and identities

Gregory Jones

Answered question

2022-01-03

How can I prove that:

{(1 + \sin x + \cos x)}^{2} = 2 (1 + \cos x) (1 + \sin x)

Ive

Answer & Explanation

stomachdm

Beginner2022-01-04Added 33 answers

To achieve your goal, you want to factor out

(1 + \sin x)

{(1 + \sin x)}^{2} + \cos^{2} x + 2 (1 + \sin x) \cos x =

{(1 + \sin x)}^{2} + (1 - \sin^{2} x) + 2 (1 + \sin x) \cos x =

[(1 + \sin x) + (1 - \sin x) + 2 \cos x] (1 + \sin x) =

[2 + 2 \cos x] (1 + \sin x) = 2 (1 + \cos x) (1 + \sin x)

macalpinee3

Beginner2022-01-05Added 29 answers

Just continue expanding like you have, to obtain

1 + \sin^{2} (x) + 2 \sin (x) + \cos^{2} (x) + 2 \cos (x) + 2 \sin (x) \cos (x)

which then beomes

2 + 2 \sin (x) + 2 \cos (x) + 2 \sin (x) \cos (x)

From there, factor out the two and notice that 1+a+b+ab=(1+a)(1+b).

Vasquez

Expert2022-01-08Added 669 answers

$(1 + \sin x + \cos x)^{2} = 1 + \sin^{2} x + \cos^{2} x + 2 \sin x + 2 \cos x + 2 \sin x \cos x$
$= 2 + 2 \sin x + 2 \cos x + 2 \sin x \cos x$
$= 2 (1 + \sin x) (1 + \cos x)$

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