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Should not Sina.Cosb and Cosa.Sinb be same if we exchange a and b? Why there is an extra negative sign for Cosa.Sinb?

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    $\begingroup$ Because $\sin(b-a)=-\sin(a-b)$ $\endgroup$
    – J. W. Tanner
    Jun 2 at 20:34
  • $\begingroup$ Thank you. I understand now. $\endgroup$
    – Dinesh Katoch
    Jun 2 at 20:36
  • $\begingroup$ Recal that $\sin a\cos b+\cos a \sin b=\sin (a+b) $ and $\sin a\cos b-\cos a \sin b=\sin (a-b) $. $\endgroup$
    – user
    Jun 2 at 20:36
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    $\begingroup$ I think it is rather silly to have both formulas in the same table of formulas though, isn't it? After all, if you have $\cos x \sin y$ you can just write it as $\sin y \cos x$ and then apply the first formula. $\endgroup$
    – David K
    Jun 3 at 5:00

1 Answer 1

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Should not $\sin a\cos b$ and $\cos a \sin b$ be same if we exchange $a$ and $b$? Why there is an extra negative sign for $\cos a\sin b$?

Because if you exchange $a$ and $b$, you get the term $$\sin(b-a),$$

which is the same as $$\color{blue}-\sin(a-b).$$

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