# Proove sin(45+x) + sin(45-x)=2^1/2*cosx.

*print*Print*list*Cite

### 2 Answers

We know sin (A+B) = sinAcosB + cosAsinB

sin (A-B) = SinAcosB - cosAsinB.

Adding Sin(A+B) +sin (A-b) = 2sinAcosB

Therefore sin90 = 1.

Also sin45 = sin(90-45) = cos45.

sin90 = sin(45+45) = sin45*cos45+cos45sin45= 2sin^2(45)

sin90 = 1 = 2(sin45)^2.

So sin45 = 1/sqrt2 = (sqrt2)/2.

Therfore sin(45+x)+sin(45-x) = 2sin45 cosx

sin(45+x) +sin(45-x) = 2 (1/sqrt2)*cosx , as sin45 = (sqrt2)/2

sin(45+x) +sin (45-x) = (sqrt2)cosx

We'll apply the sine function to the sum and the difference of angles 45 and x:

sin (a-b) = sin a*cos b - sin b*cos a

sin (a+b) = sin a*cos b + sin b*cos a

We'll put a = 45 and b = x.

sin (45+x) = sin 45*cos x + sin x*cos 45

sin (45+x) = sqrt2*cos x/2 + sqrt2*sin x/2 (1)

sin (45-x) = sin 45*cos x - sin x*cos 45

sin (45-x) = sqrt2*cos x/2 - sqrt2*sin x/2 (2)

Now, we'll substitute (1) and (2) in the given identity:

sin(45+x) + sin(45-x) = sqrt2*cosx

sqrt2*cos x/2 + sqrt2*sin x/2 + sqrt2*cos x/2 - sqrt2*sin x/2 = sqrt2*cosx

We'll combine and eliminate like terms:

2*sqrt2*cos x/2 = sqrt2*cosx

We'll simplify and we'll get:

sqrt2*cosx = sqrt2*cosx

**The identity sin(45+x) + sin(45-x) = sqrt2*cosx is true.**