Question

Using the trigonometric identity for the sine of sum of two angles, prove that:
$\sin {{75}^{\circ }}=\dfrac{\left( \sqrt{6}+\sqrt{2} \right)}{4}$.

Answer 1

Hint: In this given question, we can use the formula for $\sin \left( A+B \right)$ which is equal to $\sin A\cos B+\cos A\sin B$,where we can write $\sin {{75}^{\circ }}$ in the form of $\sin \left( A+B \right)$, and substitute A and B with two angles whose sum is equal to ${{75}^{\circ }}$. Then, we may expand this formula to get our proof done by multiplying both the numerator and denominator by $\sqrt{2}$.

Complete step-by-step answer:
In this given question, we are asked to prove that $\sin {{75}^{\circ }}=\dfrac{\left( \sqrt{6}+\sqrt{2} \right)}{4}$.
In proving the above equation by the satisfying the condition of Left Hand Side(LHS) equal to Right Hand Side (RHS), we are going to use the formula for $\sin \left( A+B \right)$ which is equal to $\sin A\cos B+\cos A\sin B$ that is $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$…..……(1.1).
Now, substituting the values of A and B with ${{30}^{\circ }}$ and ${{45}^{\circ }}$ respectively, we get:
$LHS=\sin {{75}^{\circ }}=\sin \left( {{30}^{\circ }}+{{45}^{\circ }} \right)$
$=\sin {{30}^{\circ }}\cos {{45}^{\circ }}+\cos {{30}^{\circ }}\sin {{45}^{\circ }}$(using equation 1.1) …………………(1.2)
Now, we know that the values of $\sin {{30}^{\circ }}=\dfrac{1}{2}$, $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$, $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$.
So, putting these values in equation 1.2, we get:
\[LHS=\dfrac{1}{2}\times \dfrac{1}{\sqrt{2}}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{\sqrt{2}}\]
$=\dfrac{1}{2\sqrt{2}}+\dfrac{\sqrt{3}}{2\sqrt{2}}$
$=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
Now, multiplying both the numerator and denominator by $\sqrt{2}$, we get:
\[LHS=\dfrac{\sqrt{2}\left( \sqrt{3}+1 \right)}{\sqrt{2}\left( 2\sqrt{2} \right)}\]
$=\dfrac{\left( \sqrt{6}+\sqrt{2} \right)}{4}$
$=RHS$
Hence, we have proved that the LHS is equal to the RHS.
Therefore, $\sin {{75}^{\circ }}=\dfrac{\left( \sqrt{6}+\sqrt{2} \right)}{4}$.

Note: Here, we are using the transformation formula for the sine of the difference of two angles, that is:
$\sin A\cos B+\cos A\sin B=\sin \left( A+B \right)$.
But in some other questions we may also have to use the transformation formula for the sine of the sum of two angles, that is:
$\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$.