(Sin + Cos)² + (Sin - Cos)² = 2?

Question

Using Trigonometric Identities.

Find the answer 2.

Answer 1

Q = (sin² + cos² + 2sin.cos) + (sin² + cos² - 2sin.cos).
= 2(sin² + cos² )
= 2 {you would have noticed that [sin² + cos² = 1] is a Trigonometric Identity.

Answer 2

(sin x + cos x)² + (sin x - cos x)² = 2
(sin x)^2 + sin x cos x + (cos x)^2 + (sin x)^2 - sin x cos x + (cos x)^2 = 2

2 (sin x)^2 + 2 (cos x)^2 = 2

2[ (sin x)^2 + (cos x)^2 ] = 2

2 [ 1 ] = 2

2 = 2

Also, you may note that some here are using both left and right hand sides of the given identity in an attempt to prove. You must use only one side and convert to the other. I have taken the left side and manipulated and simplified until I arrived at 2 (which is the right hand side).

Answer 3

(sin +cos)^2 + )sin - cos)^2 = sin^2 + 2sin cos + cos^2 + sin^2 - 2sin cos + cos^2

= 2 sin^2 + 2 cos^2 = 2(sin^2 + cos^2 = 2 since sin^2 + cos^2 = 1

Answer 4

(sin x + cos x)² = sin² x + cos² x + 2 sin x cos x
(sin x - cos x)² = sin² x + cos² x - 2 sin x cos x
Sum = 2 (sin ² x + cos ² x)
Sum = 2

Answer 5

the answer is the pythagorean identity
sin^2x+cos^2x=1
so 1=1 and 2=2

Answer 6

(sin x + cos x)^2 + (sin x - cos x)^2 = 2

Expand the squares.

(sin x)^2 + (2)(sin x)(cos x) + cos x^2 + (sin x)^2 - (2)(sin x)(cos x) + (cos x)^2 = 2

(2)(sin x)(cos x) and (-2)(cos x)(sin x) cancel each other.

You are left with

2(sin^2 x + cos^2 x) = 2

sin^2 x + cos^2 x = 1

Thus this is proved.

adj^2 + opp^2 = hyp^2

(adj/hyp)^2 + (opp/hyp)^2 = (hyp/hyp)^2

adj/hyp = cos x

opp/hyp = sin x

hyp/hyp = 1

(cos x)^2 + (sin x)^2 = 1

A common identity seen.

Answer 7

sin2 + cos2 + 2sincos + sin2+ cos2-2sincos
sin2+cos2=1 and 2sincos cancel each other
so ans is 2

Answer 8

And the answer is

Do your own homework.