Question

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

Answer 1

Hint: We will be using the concept of trigonometric functions to solve the problem. We will first find the value of LHS by substituting the value of A and B. Then we will find the RHS by substituting the value of A and B and prove both the values to be equal.

Complete step-by-step answer:

Now, we have been given that$A=60{}^\circ \ and\ B=30{}^\circ $.

Now, we will first take LHS of the equation and substitute the value of A and B in it. So, we have,

$\begin{align}

  & \cos \left( A+B \right)=\cos \left( 60{}^\circ +30{}^\circ \right) \\

 & =\cos \left( 90{}^\circ \right) \\

\end{align}$

Now, we know that the value of $\cos \left( 90{}^\circ \right)=0$. Therefore, we have that,
$\cos \left( A+B \right)=0.............\left( 1 \right)$

Now, we will take RHS of the equation and substitute the value of A and B in it. So, we have,
$\cos A\cos B-\sin A\sin B=\cos 60{}^\circ \cos 30{}^\circ -\sin 60{}^\circ \sin 30{}^\circ $
Now, we know that the value of,

$\begin{align}

  & \sin 30{}^\circ =\cos 60{}^\circ =\dfrac{1}{2} \\

 & \sin 60{}^\circ =\cos 30{}^\circ =\dfrac{\sqrt{3}}{2} \\

\end{align}$

So, we have the value as,

$\begin{align}

  & \cos A\cos B-\sin A\sin B=\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2} \\

 & =\dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4} \\

 & =\dfrac{4}{4} \\

 & \cos A\cos B-\sin A\sin B=0..............\left( 2 \right) \\

\end{align}$

Now, from (1) and (2) we have that, if $A=60{}^\circ \ and\ B=30{}^\circ $ then$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$.


Note: To solve these type of question it is important to note that we have used a trigonometric properties like,

$\begin{align}

  & \sin 30{}^\circ =\cos 60{}^\circ =\dfrac{1}{2} \\

 & \cos 30{}^\circ =\sin 60{}^\circ =\dfrac{\sqrt{3}}{2} \\

\end{align}$

Also, it has to be noted that to verify the given equation, we have simply substituted the values of A and B and find its value.