Question

Prove the following:
$\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}=\dfrac{1}{\sqrt{2}}$

Answer 1

Hint: If we can see the L.H.S of the given expression, we will find that it is the expansion of the trigonometric identity $\sin \left( {{36}^{\circ }}+{{9}^{\circ }} \right)=\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}$so we can write L.H.S as sin 45°. Now, we know that the value of sin 45° is$\dfrac{1}{\sqrt{2}}$.

Complete step-by-step answer:
The equation that we have to prove is:
$\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}=\dfrac{1}{\sqrt{2}}$
We are going to solve the L.H.S of the above equation.
$\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}$
The above expression is in the form of the trigonometric identity$\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$where A = 36° and B = 9°. So, we can write the above expression as:
$\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}=\sin \left( {{36}^{\circ }}+{{9}^{\circ }} \right)=\sin {{45}^{\circ }}$
And we know that the value of sin45° is equal to$\dfrac{1}{\sqrt{2}}$. So, the value of L.H.S of the given equation is:
$\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$
R.H.S of the given equation is equal to$\dfrac{1}{\sqrt{2}}$.
As we have got the L.H.S of the given expression is equal to$\dfrac{1}{\sqrt{2}}$ so L.H.S = R.H.S of the given equation.
Hence, we have proved the L.H.S = R.H.S of the given equation$\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}=\dfrac{1}{\sqrt{2}}$.

Note: The alternative way of proving the above problem is shown below. $\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}=\dfrac{1}{\sqrt{2}}$
The way is to write the R.H.S of the given equation in the form of L.H.S as follows:
As you can see that L.H.S contains the angles of sine and cosine so we can write$\dfrac{1}{\sqrt{2}}$as sin 45°.
Now, try to write sin 45° in terms of 36° and 9°. As sum of 36° and 9° gives 45° so we can write$\sin {{45}^{\circ }}=\sin \left( {{36}^{\circ }}+{{9}^{\circ }} \right)$.
Now, we are going to apply the trigonometric identity of sin (A + B) on sin (36° + 9°).
$\sin \left( {{36}^{\circ }}+{{9}^{\circ }} \right)=\sin {{36}^{\circ }}\cos {{9}^{\circ }}+\cos {{36}^{\circ }}\sin {{9}^{\circ }}$
The expression we got is equal to L.H.S of the given equation.
Hence, we have proved L.H.S = R.H.S of the given equation.