If A = $30^{o}$ and B = $60^{o}$ ,verify that

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

Open in App

Solution

$left parenthesis i right parenthesis space L H S space equals space sin left parenthesis 30 plus 60 right parenthesis space equals space sin 90 space equals 1 R H S space equals space sin 30 cos 60 plus cos 30 sin 60 space equals space 1 half.1 half plus fraction numerator square root of 3 over denominator 2 end fraction. fraction numerator square root of 3 over denominator 2 end fraction space equals space 1 L H S thin space equals thin space R H S space h e n c e space v e r i f i e d.$

(ii) $L H S = c o s (30^{o} + 60^{o}) = c o s 90^{o} = 0$

$R H S = c o s 30^{o} c o s 60^{o} - s i n 30^{o} s i n 60^{o}$

$= \frac{\sqrt{3}}{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{\sqrt{3}}{2} = 0 = L H S$

Hence, verified.

Suggest Corrections

Similar questions

A = B = 60^{\circ}

cos (A - B) = cos A cos B + sin A sin B

sin (A - B) = sin A cos B - cos A sin B

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

If sin(A - B) = sinA cos B - cosA sinB and cos(A - B)= cos A cos B + sin Asin B then find
sin15 and cos 15 .

If cos (A-B) = $\frac{3}{5}$ and tanAtanB = 2,then

Join BYJU'S Learning Program