To find sin(45-x) +sin(45+x).

Solution:

No they do not sum up to zero.

We know sin (a+b) = sinacosb+cosasinb ...(1)

sin (a-b) = sinacosb - cosa sinb... (2)

Therefore sin(a+b) +sin(a-b) = 2sinacosb , the other terms cancel.

Therefore sin (45-x) = sin45cosx -cos45sinx...(3)

sin(45+x) = sin45cosx +cos45sinx...(4).

Addin eq(3) and (4) we get: sin(45-x)+sin(45+x) = 2sin45cosx = 2(1/sqrt2) cosx = 2(sqrt2) cosx/(sqrt2*sqrt2) = (sqrt2)cosx.

So 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 sum to be calculated:

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

We'll combine and eliminate like terms:

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

We'll simplify and we'll get:

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

**We can notice that the result of the sum is not zero, but sqrt2*cosx.**