Question

If $\sin (A+B)=\sin A\cos B+\cos A\sin B$ and $\cos (A-B)=\cos A\cos B+\sin A\sin B$, find the values of $(i)\sin {{75}^{\circ }}$ and $(ii)\cos {{15}^{\circ }}$.

Answer 1

Hint: Assume angle A and B such that their sum is 75 degrees and their difference is 15 degrees. Now, form two linear equations using this information and determine the values of A and B. Now, use the relation provided in the question: $\sin (A+B)=\sin A\cos B+\cos A\sin B$ to determine the value of $\sin {{75}^{\circ }}$ and $\cos (A-B)=\cos A\cos B+\sin A\sin B$ to determine the value of $\cos {{15}^{\circ }}$.

Complete step-by-step answer:


We have to find the value of $\sin {{75}^{\circ }}$ and $\cos {{15}^{\circ }}$ by using the identities: $\sin (A+B)=\sin A\cos B+\cos A\sin B$ and $\cos (A-B)=\cos A\cos B+\sin A\sin B$ respectively.

On comparing the above equations, we get,

$\begin{align}

  & A+B={{75}^{\circ }}.....................(i) \\

 & A-B={{15}^{\circ }}......................(ii) \\

\end{align}$

Let us solve these two equations.

Adding these two equations, we get,

$\begin{align}

  & 2A={{90}^{\circ }} \\

 & \Rightarrow A={{45}^{\circ }} \\

\end{align}$

Substituting, $A={{45}^{\circ }}$ in equation (ii), we get,

$\begin{align}

  & {{45}^{\circ }}+B={{75}^{\circ }} \\

 & \Rightarrow B={{75}^{\circ }}-{{45}^{\circ }} \\

 & \Rightarrow B={{30}^{\circ }} \\

\end{align}$

Now, let us substitute these values in the given identities.

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

$\begin{align}

  & \Rightarrow \sin ({{45}^{\circ }}+{{30}^{\circ }})=\sin {{45}^{\circ }}\cos {{30}^{\circ }}+\cos

{{45}^{\circ }}\sin {{30}^{\circ }} \\

 & \Rightarrow \sin ({{75}^{\circ }})=\sin {{45}^{\circ }}\cos {{30}^{\circ }}+\cos {{45}^{\circ

}}\sin {{30}^{\circ }} \\

\end{align}$

We know that, $\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\sin {{30}^{\circ

}}=\dfrac{1}{2}\text{ and }\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$. Substituting these in the

above relation, we get,

\[\begin{align}

  & \sin ({{75}^{\circ }})=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times

 \dfrac{1}{2} \\

 & \Rightarrow \sin ({{75}^{\circ }})=\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}} \\

 & \Rightarrow \sin ({{75}^{\circ }})=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\

\end{align}\]

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

$\begin{align}

  & \Rightarrow \cos ({{45}^{\circ }}-{{30}^{\circ }})=\cos {{45}^{\circ }}\cos {{30}^{\circ }}+\sin {{45}^{\circ }}\sin {{30}^{\circ }} \\

 & \Rightarrow \sin ({{15}^{\circ }})=\cos {{45}^{\circ }}\cos {{30}^{\circ }}+\cos {{45}^{\circ }}\sin {{30}^{\circ }} \\

 & \Rightarrow \sin ({{15}^{\circ }})=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2} \\

 & \Rightarrow \sin ({{15}^{\circ }})=\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{1}{2\sqrt{2}} \\

 & \Rightarrow \sin ({{15}^{\circ }})=\dfrac{\sqrt{3}+1}{2\sqrt{2}} \\

\end{align}$


Note: One may note that we are getting the same values of the two expressions. This is because 15 degrees and 75 degrees are complement of each other, that is, their sum is 90 degrees. Therefore, $\sin {{75}^{\circ }}=\cos \left( {{90}^{\circ }}-{{75}^{\circ }} \right)=\cos {{15}^{\circ }}$. So, basically we do not need to calculate the value of expression (ii) using the formula. But here, it is said that we have to use the given relation, so we have not applied the complementary angle rule.