To derive trig identities (something useful in the exam), we will use Euler relation as starting point, which is \(e^{ix}=\cos x+i\sin x\).
\begin {equation} e^{i\left ( A+B\right ) }=\cos \left ( A+B\right ) +i\sin \left ( A+B\right ) \tag {1} \end {equation} But \(e^{i\left ( A+B\right ) }=e^{iA}e^{iB}\) therefore\begin {align} e^{iA}e^{iB} & =\left ( \cos A+i\sin A\right ) \left ( \cos B+i\sin B\right ) \nonumber \\ & =\cos A\cos B+i\cos A\sin B+i\sin A\cos B-\sin A\sin B\nonumber \\ & =\left ( \cos A\cos B-\sin A\sin B\right ) +i\left ( \cos A\sin B+\sin A\cos B\right ) \tag {2} \end {align}
Now (1) is the same as (2). Hence the real part and the imaginary parts must be the same. Therefore \begin {align} \cos \left ( A+B\right ) & =\cos A\cos B-\sin A\sin B\tag {3}\\ \sin \left ( A+B\right ) & =\cos A\sin B+\sin A\cos B\tag {4} \end {align}
This can be derived in similar way to the above using \(e^{i\left ( A-B\right ) }=\cos \left ( A-B\right ) +i\sin \left ( A-B\right ) \) and so on. But more easily, it can be derived from (3,4) directly by just changing replacing \(B\) by \(-B\) everywhere and then changing \(\sin \left ( -B\right ) \) to \(-\sin B\) and leaving \(\cos B\) the same since \(\cos \left ( -B\right ) =\cos B\). This is because \(\cos \) is even and \(\sin \) is odd, then (3) becomes\begin {align} \cos \left ( A-B\right ) & =\cos A\cos B+\sin A\sin B\tag {3A}\\ \sin \left ( A-B\right ) & =-\cos A\sin B+\sin A\cos B\tag {4A} \end {align}
So we really just need to find (3) to find the 4 formulas for addition and subtractions of angles.
These also can be found from (3,4). By replacing \(B\) with \(A\) resulting in\begin {align*} \cos \left ( A+A\right ) & =\cos A\cos A-\sin A\sin A\\ \sin \left ( A+A\right ) & =\cos A\sin A+\sin A\cos A \end {align*}
Therefore\begin {align} \cos \left ( 2A\right ) & =\cos ^{2}A-\sin ^{2}A\tag {3C}\\ \sin \left ( 2A\right ) & =2\cos A\sin A\tag {4C} \end {align}
Or we could use Euler formula, but the above is simpler. To use Euler formula, we write\begin {equation} e^{i\left ( 2A\right ) }=\cos \left ( 2A\right ) +i\sin \left ( 2A\right ) \tag {5} \end {equation} But \(e^{i\left ( 2A\right ) }=e^{iA}e^{iA}\) therefore\begin {align} e^{iA}e^{iA} & =\left ( \cos A+i\sin A\right ) \left ( \cos A+i\sin A\right ) \nonumber \\ & =\cos ^{2}A+2i\cos A\sin A-\sin ^{2}A\nonumber \\ & =\left ( \cos ^{2}A-\sin ^{2}A\right ) +i\left ( 2\cos A\sin A\right ) \tag {6} \end {align}
Comparing (5,6) shows that\begin {align*} \cos \left ( 2A\right ) & =\cos ^{2}A-\sin ^{2}A\\ \sin \left ( 2A\right ) & =2\cos A\sin A \end {align*}
Which is the same as (3C,4C) above.
From the double angle formula (3C)\[ \cos \left ( 2A\right ) =\cos ^{2}A-\sin ^{2}A \] But \(\cos ^{2}A+\sin ^{2}A=1\) then \(\sin ^{2}A=1-\cos ^{2}A\) and the above becomes\begin {align*} \cos \left ( 2A\right ) & =\cos ^{2}A-\left ( 1-\cos ^{2}A\right ) \\ & =2\cos ^{2}A-1 \end {align*}
Hence\[ \cos ^{2}A=\frac {\cos \left ( 2A\right ) +1}{2}\] Let \(A=\frac {x}{2}\) then the above becomes\begin {align*} \cos ^{2}\left ( \frac {x}{2}\right ) & =\frac {\cos \left ( x\right ) +1}{2}\\ \cos \left ( \frac {x}{2}\right ) & =\pm \sqrt {\frac {\cos \left ( x\right ) +1}{2}} \end {align*}
The sign depends on the quadrant of \(\frac {x}{2}\).
From the double angle formula (3C)\[ \cos \left ( 2A\right ) =\cos ^{2}A-\sin ^{2}A \] But \(\cos ^{2}A+\sin ^{2}A=1\) then \(\cos ^{2}A=1-\sin ^{2}A\) and the above becomes\begin {align*} \cos \left ( 2A\right ) & =1-\sin ^{2}A-\sin ^{2}A\\ & =1-2\sin ^{2}A \end {align*}
Hence\[ \sin ^{2}A=\frac {1-\cos \left ( 2A\right ) }{2}\] Let \(A=\frac {x}{2}\) then the above becomes\begin {align*} \sin ^{2}\left ( \frac {x}{2}\right ) & =\frac {1-\cos \left ( x\right ) }{2}\\ \sin \left ( \frac {x}{2}\right ) & =\pm \sqrt {\frac {1-\cos \left ( x\right ) }{2}} \end {align*}
The sign depends on the quadrant of \(\frac {x}{2}\).
This can be found by adding (4) and (4A). Let \begin {align} A+B & =\alpha \tag {7}\\ A-B & =\beta \nonumber \end {align}
Then (4)+(4A) now becomes\begin {align} \sin \left ( \alpha \right ) +\sin \left ( \beta \right ) & =\left ( \cos A\sin B+\sin A\cos B\right ) -\cos A\sin B+\sin A\cos B\nonumber \\ & =2\sin A\cos B\tag {8} \end {align}
Now we solve for \(A,B\) from (7). Which gives\begin {align*} A & =\frac {\alpha +\beta }{2}\\ B & =\frac {\alpha -\beta }{2} \end {align*}
Substituting the above in (8) gives\[ \sin \left ( \alpha \right ) +\sin \left ( \beta \right ) =2\sin \left ( \frac {\alpha +\beta }{2}\right ) \cos \left ( \frac {\alpha -\beta }{2}\right ) \]
This can be found by adding (3) and (3A). Let \begin {align} A+B & =\alpha \tag {7}\\ A-B & =\beta \nonumber \end {align}
Then (3)+(3A) now becomes\begin {align} \cos \left ( \alpha \right ) +\cos \left ( \beta \right ) & =\left ( \cos A\cos B-\sin A\sin B\right ) +\left ( \cos A\cos B+\sin A\sin B\right ) \nonumber \\ & =2\cos A\cos B\tag {9} \end {align}
Now we solve for \(A,B\) from (7). Which gives\begin {align*} A & =\frac {\alpha +\beta }{2}\\ B & =\frac {\alpha -\beta }{2} \end {align*}
Substituting the above in (9) gives\[ \cos \left ( \alpha \right ) +\cos \left ( \beta \right ) =2\cos \left ( \frac {\alpha +\beta }{2}\right ) \cos \left ( \frac {\alpha -\beta }{2}\right ) \]
This can be found from (4)-(4A). Let \begin {align} A+B & =\alpha \tag {7}\\ A-B & =\beta \nonumber \end {align}
Then (4)-(4A) now becomes\begin {align} \sin \left ( \alpha \right ) -\sin \left ( \beta \right ) & =\left ( \cos A\sin B+\sin A\cos B\right ) +\cos A\sin B-\sin A\cos B\nonumber \\ & =2\cos A\sin B\tag {10} \end {align}
Now we solve for \(A,B\) from (7). Which gives \begin {align*} A & =\frac {\alpha +\beta }{2}\\ B & =\frac {\alpha -\beta }{2} \end {align*}
Substituting the above in (10) gives\[ \sin \left ( \alpha \right ) -\sin \left ( \beta \right ) =2\cos \left ( \frac {\alpha +\beta }{2}\right ) \sin \left ( \frac {\alpha -\beta }{2}\right ) \]
This can be found from (3)-(3A). Let \begin {align} A+B & =\alpha \tag {7}\\ A-B & =\beta \nonumber \end {align}
Then (3)-(3A) now becomes\begin {align} \cos \left ( \alpha \right ) -\cos \left ( \beta \right ) & =\left ( \cos A\cos B-\sin A\sin B\right ) -\left ( \cos A\cos B+\sin A\sin B\right ) \nonumber \\ & =-2\sin A\sin B\tag {11} \end {align}
Now we solve for \(A,B\) from (7). Which gives\begin {align*} A & =\frac {\alpha +\beta }{2}\\ B & =\frac {\alpha -\beta }{2} \end {align*}
Substituting the above in (11) gives\[ \cos \left ( \alpha \right ) -\cos \left ( \beta \right ) =-2\sin \left ( \frac {\alpha +\beta }{2}\right ) \sin \left ( \frac {\alpha -\beta }{2}\right ) \]
Adding (3)+(3A) gives\begin {align*} \cos \left ( A+B\right ) & =\cos A\cos B-\sin A\sin B\\ \cos \left ( A-B\right ) & =\cos A\cos B+\sin A\sin B\\ \cos \left ( A+B\right ) +\cos \left ( A-B\right ) & =2\cos A\cos B \end {align*}
Hence\[ \cos A\cos B=\frac {1}{2}\left ( \cos \left ( A+B\right ) +\cos \left ( A-B\right ) \right ) \]
Adding (4)+(4A) gives\begin {align*} \sin \left ( A+B\right ) & =\cos A\sin B+\sin A\cos B\\ \sin \left ( A-B\right ) & =-\cos A\sin B+\sin A\cos B\\ \sin \left ( A+B\right ) +\sin \left ( A-B\right ) & =2\sin A\cos B \end {align*}
Hence\[ \sin A\cos B=\frac {1}{2}\left ( \sin \left ( A+B\right ) +\sin \left ( A-B\right ) \right ) \]
(3)-(3A) gives\begin {align*} \cos \left ( A+B\right ) & =\cos A\cos B-\sin A\sin B\\ \cos \left ( A-B\right ) & =\cos A\cos B+\sin A\sin B\\ \cos \left ( A+B\right ) -\cos \left ( A-B\right ) & =-2\sin A\sin B \end {align*}
Hence\[ \sin A\sin B=\frac {1}{2}\left ( \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ) \]