Ytukyg

Helping my son with his trigonometry review. We know that
$$sin(a+b) = sin a cos b + cos a sin b$$
We also know that
$$sin a cos b = frac12 left(sin(a-b) + sin(a+b)right)$$
And we have
$$sin a + sin b = 2 sinfrac12(a+b)cosfrac12(a-b)$$
From there, we seem to be missing how to get to the right-hand side of the equation.

We first expand
$$sin a + sin b = 2 sinfrac12(a+b) cosfrac12(a-b)$$
Then we add $sin(a+b)$, which is $sin a cos b + cos a sin b$.
We now have:

$$2 sinfrac12(a+b) cosfrac12(a-b) + frac12 left(sin(a-b) + sin (a+b)right) + frac12 left( sin(b-a) + sin (a+b)right)$$

From there, we can't see how to obtain the right-hand side of the equation which is
$$4 sinfrac12(a+b) cosfrac12a cosfrac12b$$

Following Thomas Shelby and Dr. Sonnhard Graubner, I provide you a full proof:
begin{align}sin(a)+sin(b)+sin(a+b)&=2cosleft(frac{a-b}{2}right)sinleft(frac{a+b}{2}right)+2sinleft(frac{a+b}{2}right)cosleft(frac{a+b}{2}right)\&=2sinleft(frac{a+b}{2}right)left[cosleft(frac{a-b}{2}right)+cosleft(frac{a+b}{2}right)right]end{align}
Since
begin{align}cosleft(a-bright)+cosleft(a+bright)=2cos(a)cos(b)end{align}
[see: last page of https://services.math.duke.edu/~leili/teaching/uwm/math222s11/problems/quizzes/trig.pdf or search for An elementary proof of two formulas in trigonometry on google] then,
begin{align}cosleft(frac{a-b}{2}right)+cosleft(frac{a+b}{2}right)=2cosleft(frac{a}{2}right)cosleft(frac{b}{2}right)end{align}
Hence,
begin{align}2sinleft(frac{a+b}{2}right)left[cosleft(frac{a-b}{2}right)+cosleft(frac{a+b}{2}right)right]=&2sinleft(frac{a+b}{2}right)times 2cosleft(frac{a}{2}right)cosleft(frac{b}{2}right)\=&4sinleft(frac{a+b}{2}right)cosleft(frac{a}{2}right)cosleft(frac{b}{2}right)end{align}

Use that $$sin(A)+sin(B)=2cosleft(frac{A-B}{2}right)sinleft(frac{A+B}{2}right)$$ and $$sin(A+B)=2sinleft(frac{A+B}{2}right)cosleft(frac{A+B}{2}right)$$

It may help setting $A=a/2$ and $B=b/2$, so the left-hand side becomes
$$
sin2A+sin2B+sin2(A+B)
$$
The right-hand side screams “sum-to-product”! OK, let's apply the formula
$$
sin2A+sin2B=2sin(A+B)cos(A-B)
$$
so the left-hand side becomes
begin{align}
sin2A+sin2B+sin2(A+B)
&=2sin(A+B)cos(A-B)+2sin(A+B)cos(A+B)\
&=2sin(A+B)bigl(cos(A-B)+cos(A+B)bigr) \
&=2sin(A+B)(cos Acos B+sin Asin B+cos Acos B-sin Asin B)\
&=4sin(A+B)cos Acos B
end{align}

begin{align}
sin a + sin b + sin (a+b)&=2 sinleft(frac{a+b}2right) cosleft(frac{a-b}2right)+2 sinleft(frac{a+b}2right) cosleft(frac{a+b}2right)\
&=2 sinleft(frac{a+b}2right)left(cosleft(frac{a-b}2right)+cosleft(frac{a+b}2right)right)
end{align}
Now use the identity $$cos A +cos B=2cos left(frac{A+B}2right)cos left(frac{A-B}2right)$$.

An alternative strategy uses $1+cos x=2cos^2tfrac{x}{2}$ to obtain $$sin a+sin b+sin (a+b)=2left(sin acos^2tfrac{b}{2}+sin bcos^2tfrac{a}{2}right).$$To finish, use $sin x=2sintfrac{x}{2}costfrac{x}{2}$ to obtain$$sin a+sin b+sin (a+b)=4costfrac{a}{2}costfrac{b}{2}left(sintfrac{a}{2}costfrac{b}{2}+sintfrac{b}{2}costfrac{a}{2}right)=4costfrac{a}{2} costfrac{b}{2}sintfrac{a+b}{2}.$$