Question

What is the value of given trigonometric expression $\sin {{15}^{\circ }}$?
(a) $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
(b) $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
(c) $\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$
(d) $\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$

Answer 1

Hint: Use the trigonometric identity $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$. Substitute $x={{45}^{\circ }},y={{30}^{\circ }}$ in the above formula and use the values $\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\sin {{30}^{\circ }}=\dfrac{1}{2}$ to calculate the value of $\sin {{15}^{\circ }}$.

Complete step-by-step answer:
We have to calculate the value of $\sin {{15}^{\circ }}$.
We will use the trigonometric identity $\sin \left( x-y \right)=\sin x\cos y-\cos x\sin y$.
Substituting $x={{45}^{\circ }},y={{30}^{\circ }}$ in the above equation, we have $\sin \left( {{45}^{\circ }}-{{30}^{\circ }} \right)=\sin {{15}^{\circ }}=\sin {{45}^{\circ }}\cos {{30}^{\circ }}-\cos {{45}^{\circ }}\sin {{30}^{\circ }}$. We can also substitute $x={{60}^{\circ }},y={{45}^{\circ }}$ in the trigonometric equation. In fact, we can substitute all those x and y whose values are known to us and $x-y={{15}^{\circ }}$.
We know the values $\sin {{45}^{\circ }}=\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\sin {{30}^{\circ }}=\dfrac{1}{2}$.
Thus, we have $\sin {{15}^{\circ }}=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2}$.
Rearranging the terms, we have $\sin \left( {{15}^{\circ }} \right)=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$.
Hence, the value of $\sin \left( {{15}^{\circ }} \right)$ is $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$, which is option A.
Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

Note: We can also solve this question by using the trigonometric identity $\cos 2x=1-2{{\sin }^{2}}x$. Substitute $x={{15}^{\circ }}$ in the above equation and use $\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ to simplify the above equation. To find the value of $\sin \left( {{15}^{\circ }} \right)$, we must also use the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$.