If A = 30° and B = 60°, verify that

Question:

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

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

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

Solution:

(i) Given

$A=30^{\circ}$ and $B=60^{\circ} \ldots \ldots$ (1)

To verify:

$\sin (A+B)=\sin A \cos B+\cos A \sin B$.....(2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

$\sin (A+B)=\sin (30+60)$

$=\sin 90$

$=1$

Now consider RHS of the expression to be verified in equation (2)

Therefore;

$\sin A \cos B+\cos A \sin B=\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ}$

$=\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}$

$=\frac{1+3}{4}$

$=1$

Hence it is verified that,

$\sin (A+B)=\sin A \cos B+\cos A \sin B$

(ii) Given:

$A=30^{\circ}$ and $B=60^{\circ} \ldots \ldots$ (1)

To verify:

$\cos (A+B)=\cos A \cos B-\sin A \sin B$.....(2)

Now consider LHS of the expression to be verified in equation (2)

Therefore,

$\cos (30+60)=\cos 90$

$=0$

Now consider RHS of the expression to be verified in equation (2)

Therefore,

$\cos A \cos B-\sin A \sin B=\cos 30 \cos 60-\sin 30 \sin 60$

$=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}$

$=0$

Hence it is verified that,

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

Leave a comment

None
Free Study Material