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To derive trig identities (something useful in the exam), we will use Euler relation as starting point, which is $e^{ix}=\cos x+i\sin x$.

1 $\cos (A+B)$ and $\sin (A+B)$

$$\begin{array}{}\text{(1)}& e^{i(A+B)}=\cos (A+B)+i\sin (A+B)\end{array}$$ But $e^{i(A+B)}=e^{iA}{e}^{iB}$ therefore$$\begin{array}{rl}{e}^{iA}{e}^{iB}& =(\cos A+i\sin A)(\cos B+i\sin B)\\ & =\cos A\cos B+i\cos A\sin B+i\sin A\cos B-\sin A\sin B\\ \text{(2)}& & =(\cos A\cos B-\sin A\sin B)+i(\cos A\sin B+\sin A\cos B)\end{array}$$

Now (1) is the same as (2). Hence the real part and the imaginary parts must be the same. Therefore $$\begin{array}{}\text{(3)}& \cos (A+B)& =\cos A\cos B-\sin A\sin B\text{(4)}& \sin (A+B)& =\cos A\sin B+\sin A\cos B\end{array}$$

2 $\cos (A-B)$ and $\sin (A-B)$

This can be derived in similar way to the above using $e^{i(A-B)}=\cos (A-B)+i\sin (A-B)$ 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 $\sin (-B)$ to $-\sin B$ and leaving $\cos B$ the same since $\cos (-B)=\cos B$. This is because $\cos$ is even and $\sin$ is odd, then (3) becomes$$\begin{array}{}\text{(3A)}& \cos (A-B)& =\cos A\cos B+\sin A\sin B\text{(4A)}& \sin (A-B)& =-\cos A\sin B+\sin A\cos B\end{array}$$

So we really just need to find (3) to find the 4 formulas for addition and subtractions of angles.

3 $\cos \left(2A\right)$ and $\sin \left(2A\right)$

These also can be found from (3,4). By replacing $B$ with $A$ resulting in$$\begin{array}{rl}\cos (A+A)& =\cos A\cos A-\sin A\sin A\\ \sin (A+A)& =\cos A\sin A+\sin A\cos A\end{array}$$

Therefore$$\begin{array}{}\text{(3C)}& \cos \left(2A\right)& ={\cos }^{2}A-{\sin }^{2}A\text{(4C)}& \sin \left(2A\right)& =2\cos A\sin A\end{array}$$

Or we could use Euler formula, but the above is simpler. To use Euler formula, we write$$\begin{array}{}\text{(5)}& e^{i\left(2A\right)}=\cos \left(2A\right)+i\sin \left(2A\right)\end{array}$$ But $e^{i\left(2A\right)}=e^{iA}{e}^{iA}$ therefore$$\begin{array}{rl}{e}^{iA}{e}^{iA}& =(\cos A+i\sin A)(\cos A+i\sin A)\\ & ={\cos }^{2}A+2i\cos A\sin A-{\sin }^{2}A\\ \text{(6)}& & =({\cos }^{2}A-{\sin }^{2}A)+i(2\cos A\sin A)\end{array}$$

Comparing (5,6) shows that$$\begin{array}{rl}\cos \left(2A\right)& ={\cos }^{2}A-{\sin }^{2}A\\ \sin \left(2A\right)& =2\cos A\sin A\end{array}$$

Which is the same as (3C,4C) above.

4 $\cos \left(\frac{x}{2}\right)$

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$$\begin{array}{rl}\cos \left(2A\right)& ={\cos }^{2}A-(1-{\cos }^{2}A)\\ & =2{\cos }^{2}A-1\end{array}$$

Hence$${\cos }^{2}A=\frac{\cos \left(2A\right)+1}{2}$$ Let $A=\frac{x}{2}$ then the above becomes$$\begin{array}{rl}{\cos }^2\left(\frac{x}{2}\right)& =\frac{\cos \left(x\right)+1}{2}\\ \cos \left(\frac{x}{2}\right)& =\pm \sqrt{\frac{\cos \left(x\right)+1}{2}}\end{array}$$

The sign depends on the quadrant of $\frac{x}{2}$.

5 $\sin \left(\frac{x}{2}\right)$

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$$\begin{array}{rl}\cos \left(2A\right)& =1-{\sin }^{2}A-{\sin }^{2}A\\ & =1-2{\sin }^{2}A\end{array}$$

Hence$${\sin }^{2}A=\frac{1-\cos \left(2A\right)}{2}$$ Let $A=\frac{x}{2}$ then the above becomes$$\begin{array}{rl}{\sin }^2\left(\frac{x}{2}\right)& =\frac{1-\cos \left(x\right)}{2}\\ \sin \left(\frac{x}{2}\right)& =\pm \sqrt{\frac{1-\cos \left(x\right)}{2}}\end{array}$$

The sign depends on the quadrant of $\frac{x}{2}$.

6 $\sin \left(\alpha \right)+\sin \left(\beta \right)$

This can be found by adding (4) and (4A). Let $$\begin{array}{rl}\text{(7)}& A+B& =\alpha A-B& =\beta \end{array}$$

Then (4)+(4A) now becomes$$\begin{array}{rl}\sin \left(\alpha \right)+\sin \left(\beta \right)& =(\cos A\sin B+\sin A\cos B)-\cos A\sin B+\sin A\cos B\\ \text{(8)}& & =2\sin A\cos B\end{array}$$

Now we solve for $A,B$ from (7). Which gives$$\begin{array}{rl}A& =\frac{\alpha +\beta }{2}\\ B& =\frac{\alpha -\beta }{2}\end{array}$$

Substituting the above in (8) gives$$\sin \left(\alpha \right)+\sin \left(\beta \right)=2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$$

7 $\cos \left(\alpha \right)+\cos \left(\beta \right)$

This can be found by adding (3) and (3A). Let $$\begin{array}{rl}\text{(7)}& A+B& =\alpha A-B& =\beta \end{array}$$

Then (3)+(3A) now becomes$$\begin{array}{rl}\cos \left(\alpha \right)+\cos \left(\beta \right)& =(\cos A\cos B-\sin A\sin B)+(\cos A\cos B+\sin A\sin B)\\ \text{(9)}& & =2\cos A\cos B\end{array}$$

Now we solve for $A,B$ from (7). Which gives$$\begin{array}{rl}A& =\frac{\alpha +\beta }{2}\\ B& =\frac{\alpha -\beta }{2}\end{array}$$

Substituting the above in (9) gives$$\cos \left(\alpha \right)+\cos \left(\beta \right)=2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)$$

8 $\sin \left(\alpha \right)-\sin \left(\beta \right)$

This can be found from (4)-(4A). Let $$\begin{array}{rl}\text{(7)}& A+B& =\alpha A-B& =\beta \end{array}$$

Then (4)-(4A) now becomes$$\begin{array}{rl}\sin \left(\alpha \right)-\sin \left(\beta \right)& =(\cos A\sin B+\sin A\cos B)+\cos A\sin B-\sin A\cos B\\ \text{(10)}& & =2\cos A\sin B\end{array}$$

Now we solve for $A,B$ from (7). Which gives $$\begin{array}{rl}A& =\frac{\alpha +\beta }{2}\\ B& =\frac{\alpha -\beta }{2}\end{array}$$

Substituting the above in (10) gives$$\sin \left(\alpha \right)-\sin \left(\beta \right)=2\cos \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)$$

9 $\cos \left(\alpha \right)-\cos \left(\beta \right)$

This can be found from (3)-(3A). Let $$\begin{array}{rl}\text{(7)}& A+B& =\alpha A-B& =\beta \end{array}$$

Then (3)-(3A) now becomes$$\begin{array}{rl}\cos \left(\alpha \right)-\cos \left(\beta \right)& =(\cos A\cos B-\sin A\sin B)-(\cos A\cos B+\sin A\sin B)\\ \text{(11)}& & =-2\sin A\sin B\end{array}$$

Now we solve for $A,B$ from (7). Which gives$$\begin{array}{rl}A& =\frac{\alpha +\beta }{2}\\ B& =\frac{\alpha -\beta }{2}\end{array}$$

Substituting the above in (11) gives$$\cos \left(\alpha \right)-\cos \left(\beta \right)=-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)$$

10 $\cos \left(A\right)\cos \left(B\right)$

Adding (3)+(3A) gives$$\begin{array}{rl}\cos (A+B)& =\cos A\cos B-\sin A\sin B\\ \cos (A-B)& =\cos A\cos B+\sin A\sin B\\ \cos (A+B)+\cos (A-B)& =2\cos A\cos B\end{array}$$

Hence$$\cos A\cos B=\frac{1}{2}(\cos (A+B)+\cos (A-B))$$

11 $\sin \left(A\right)\cos \left(B\right)$

Adding (4)+(4A) gives$$\begin{array}{rl}\sin (A+B)& =\cos A\sin B+\sin A\cos B\\ \sin (A-B)& =-\cos A\sin B+\sin A\cos B\\ \sin (A+B)+\sin (A-B)& =2\sin A\cos B\end{array}$$

Hence$$\sin A\cos B=\frac{1}{2}(\sin (A+B)+\sin (A-B))$$

12 $\sin \left(A\right)\sin \left(B\right)$

(3)-(3A) gives$$\begin{array}{rl}\cos (A+B)& =\cos A\cos B-\sin A\sin B\\ \cos (A-B)& =\cos A\cos B+\sin A\sin B\\ \cos (A+B)-\cos (A-B)& =-2\sin A\sin B\end{array}$$

Hence$$\sin A\sin B=\frac{1}{2}(\cos (A-B)-\cos (A+B))$$