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Question Number 189533 by mustafazaheen last updated on 18/Mar/23
when  a+b=60^°      find  ((cos^2 a−sin^2 b)/(cos(a−b)))=?
$${when}\:\:{a}+{b}=\mathrm{60}^{°} \:\:\:\:\:{find}\:\:\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)}=? \\ $$
Answered by som(math1967) last updated on 18/Mar/23
((cos^2 a−sin^2 b)/(cos(a−b)))  =((cos(a+b)cos(a−b))/(cos(a−b)))  =cos(a+b)=cos60=(1/2)  if (a−b)≠90
$$\frac{{cos}^{\mathrm{2}} {a}−{sin}^{\mathrm{2}} {b}}{{cos}\left({a}−{b}\right)} \\ $$$$=\frac{{cos}\left({a}+{b}\right){cos}\left({a}−{b}\right)}{{cos}\left({a}−{b}\right)} \\ $$$$={cos}\left({a}+{b}\right)={cos}\mathrm{60}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${if}\:\left({a}−{b}\right)\neq\mathrm{90} \\ $$

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