If $A=60{}^\circ \ and\ B=30{}^\circ $, verify that: $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.
Answer
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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,
$\begin{align}
& \sin \left( A-B \right)=\sin \left( 60{}^\circ -30{}^\circ \right) \\
& =\sin \left( 30{}^\circ \right) \\
\end{align}$
Now, we know that the value of $\sin \left( 30{}^\circ \right)=\dfrac{1}{2}$. Therefore, we have that,
$\sin \left( A-B \right)=\dfrac{1}{2}.............\left( 1 \right)$
Now, we will take RHS of the equation and substitute the value of A and B in it. So, we have,
$\sin A\cos B-\cos A\sin B=\sin 60{}^\circ \cos 30{}^\circ -\cos 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,
$\begin{align}
& \sin A\cos B-\cos A\sin B=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\times \dfrac{1}{2} \\
& =\dfrac{3}{4}-\dfrac{1}{4} \\
& =\dfrac{2}{4} \\
& \sin A\cos B-\cos A\sin B=\dfrac{1}{2}..............\left( 2 \right) \\
\end{align}$
Now, from (1) and (2) we have that, if $A=60{}^\circ \ and\ B=30{}^\circ $ then$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.
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.
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,
$\begin{align}
& \sin \left( A-B \right)=\sin \left( 60{}^\circ -30{}^\circ \right) \\
& =\sin \left( 30{}^\circ \right) \\
\end{align}$
Now, we know that the value of $\sin \left( 30{}^\circ \right)=\dfrac{1}{2}$. Therefore, we have that,
$\sin \left( A-B \right)=\dfrac{1}{2}.............\left( 1 \right)$
Now, we will take RHS of the equation and substitute the value of A and B in it. So, we have,
$\sin A\cos B-\cos A\sin B=\sin 60{}^\circ \cos 30{}^\circ -\cos 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,
$\begin{align}
& \sin A\cos B-\cos A\sin B=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\times \dfrac{1}{2} \\
& =\dfrac{3}{4}-\dfrac{1}{4} \\
& =\dfrac{2}{4} \\
& \sin A\cos B-\cos A\sin B=\dfrac{1}{2}..............\left( 2 \right) \\
\end{align}$
Now, from (1) and (2) we have that, if $A=60{}^\circ \ and\ B=30{}^\circ $ then$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.
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.
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