Question

If $A={{30}^{{}^\circ }}$ and $B={{60}^{{}^\circ }}$, verify that $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$.

Answer 1

Hint:We know that the given relation is $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$. We will put the given values as $A={{30}^{{}^\circ }}$ and $B={{60}^{{}^\circ }}$, then we will solve LHS and RHS individually and if we get $LHS=RHS$, then the above relation is verified. We will also use the values $\sin {{90}^{{}^\circ }}=1$, $\sin {{30}^{{}^\circ }}=\dfrac{1}{2}$, $\sin {{60}^{{}^\circ }}=\dfrac{\sqrt{3}}{2}$, $\cos {{60}^{{}^\circ }}=\dfrac{1}{2}$ and $cos{{30}^{{}^\circ }}=\dfrac{\sqrt{3}}{2}$.

Complete step-by-step answer:
It is given in the question that $A={{30}^{{}^\circ }}$ and $B={{60}^{{}^\circ }}$,and we have to verify that $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$. We will solve LHS and RHS individually by putting the value of $A={{30}^{{}^\circ }}$ and $B={{60}^{{}^\circ }}$ both sides. If both the sides get the same value then the given relation will be verified.
To proceed with the question, let us first consider the below table. It shows the values of the trigonometric ratios at standard angles.

In LHS, we have $\sin \left( A+B \right)$, on putting the given values as $A={{30}^{{}^\circ }}$ and $B={{60}^{{}^\circ }}$, we get
$=\sin \left( {{30}^{{}^\circ }}+{{60}^{{}^\circ }} \right)$ that is
$=\sin \left( {{90}^{{}^\circ }} \right)$, we know that $\sin {{90}^{{}^\circ }}=1$, therefore we get
$=1$.
Thus \[LHS=1\].
Now, in RHS, we have $\sin A\cos B+\cos A\sin B$, on putting the given values as $A={{30}^{{}^\circ }}$ and $B={{60}^{{}^\circ }}$, we get
\[=\sin {{30}^{{}^\circ }}\cos {{60}^{{}^\circ }}+\cos {{30}^{{}^\circ }}\sin {{60}^{{}^\circ }}\], using the values of $\sin {{30}^{{}^\circ }}=\dfrac{1}{2}$, $\sin {{60}^{{}^\circ }}=\dfrac{\sqrt{3}}{2}$, $\cos {{60}^{{}^\circ }}=\dfrac{1}{2}$ and $cos{{30}^{{}^\circ }}=\dfrac{\sqrt{3}}{2}$in the last step, we get
\[=\dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}\], further simplifying we get
\[=\dfrac{1}{4}+\dfrac{3}{4}\], performing the addition in last step, we get
$=1$.
Thus $RHS=1$.
On comparing LHS and RHS, we see that $LHS=RHS$ each equal to 1.
Thus we have proved the given relation $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ holds true.

Note: Usually students make mistakes in hurry while substituting the value of angle and this leads to the formation of wrong answers. Many time student take the values as $cos{{30}^{{}^\circ }}=\dfrac{1}{2}$ and $cos{{60}^{{}^\circ }}=\dfrac{\sqrt{3}}{2}$ which is not correct at all. Thus, it is recommended to put all the values correct in order to get the correct result.Students should remember the standard trigonometric angles and formulas for solving these types of questions.