Prove that $(sin 32^{o} cos 58^{o} + cos 32^{o} sin 58^{o})$ .

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Solution

Now,
$(sin 32^{o} cos 58^{o} + cos 32^{o} sin 58^{o})$
$= sin (32^{o} + 58^{o})$ [ since $sin (A + B) = sin A . cos B + cos A . sin B$ ]
$= sin 90^{o}$
$= 1$ .

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