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The question is to prove the compound angle identity $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ starting from the $\sin$ compound angle identity. Before this, the task wants me to show that $\sin(\frac \pi 2 - x) = \cos(x)$ and I did not have any problems there. I guess I have to use this fact somehow so thats what I've tried:

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$ $$\sin\left(\frac \pi 2 - (a+b)\right)=\sin\left(\frac \pi 2\right)\cos(a+b)-\cos\left(\frac \pi 2\right)\sin(a+b)$$ $$\cos(a+b)=1\cos(a+b)-0\sin(a+b)$$ $$\cos(a+b)=\cos(a+b)$$ And that brings me nowhere. How can I prove this identity in the required way?

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    $\begingroup$ Try expanding something like $\sin \left( \left(\frac {\pi}2-a\right)-b\right)$ $\endgroup$
    – lulu
    Feb 12, 2021 at 19:06
  • $\begingroup$ thanks! this helped a lot $\endgroup$
    – David A.
    Feb 12, 2021 at 19:20
  • $\begingroup$ You may as well put your work in as a self-answer so the problem has an accepted answer. $\endgroup$
    – user694818
    Feb 12, 2021 at 19:59
  • $\begingroup$ allright, I already wondered how I could mark this question as solved (but I still have to wait two days until I can accept my own answer) $\endgroup$
    – David A.
    Feb 12, 2021 at 20:29

2 Answers 2

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Direct calculation:$$\begin{align}\cos(a+b)&=\sin(\tfrac{\pi}{2}-a-b)\\&=\sin(\tfrac{\pi}{2}-a)\cos(-b)+\cos(\tfrac{\pi}{2}-a)\sin(-b)\\&=\cos a\cos b-\sin a\sin b.\end{align}$$

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Solution to the problem: $$\sin((\frac \pi 2-a)-b)=\sin(\frac \pi 2 -a)\cos(b)-\cos(\frac \pi 2-a)\sin(b)$$ $$\sin((\frac \pi 2-a)-b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$

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