Show Trigonometric Identities from Complex indentity

Question

So the exercise says to show

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

and

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

By using the following identity:

$e^{i(a+b)}=e^{ia}e^{ib}$

How do I show it? I suspect it should be fairly trivial since the exercise gives only 1 point.

Answer 1

Do you know that $$e^{xi}=\cos x +i\sin x?$$

Notice that $e^{i(a+b)}= \cos(a+b) + i \cdot \sin(a+b)$ and

$e^{ia} \cdot e^{ib} = \left(\cos{a}+i\cdot \sin(a)\right) \times \left(\cos{b}+i\cdot \sin(b)\right)=(\cos a \cos b-\sin a\sin b)+i(\sin a\cos b+\cos a\sin b)$

So $$\cos(a+b)=\cos a\cos b-\sin a\sin b$$ and $$\sin(a+b)=\sin a\cos b+\cos a\sin b$$

Answer 2

• $e^{i(a+b)}= \cos(a+b) + i \cdot \sin(a+b)$

• $e^{ia} \cdot e^{ib} = \left(\cos{a}+i\cdot \sin(a)\right) \times \left(\cos{b}+i\cdot \sin(b)\right)$

• Multiply and look at the imaginary part which is $i\cdot \sin(a)\cos(b) + i\cdot \cos(a)\cdot\sin(b)$

Answer 3

You should use the fact that $e^{ix}=\cos(x)+i\sin(x)$ and then look at the product $e^{i(a+b)}=e^{ia}e^{ib}$. If you replaced the left and right hand sides with $e^{ix}$ in terms of trigonometric functions, you get $$\cos(a+b)+i\sin(a+b)=(\cos(a)+i\sin(a))\times(\cos(b)+i\sin(b))$$ Expanding the left hand side and equating the real and imaginary parts of this equation separately will give the answer.