Question

If $\sin \theta =\dfrac{1}{4}$, $\theta $ being in quadrant II, how do you find exact value of $\sin \left( \theta -\dfrac{\pi }{3} \right)$ ?

Answer 1

Hint:We explain the function $\sin \theta =\dfrac{1}{4}$ and the quadrant value for the angle $\theta $. We express the identity functions of other ratio of cos with ratio of sin. It’s given that $\sin \theta =\dfrac{1}{4}$. Thereafter we put the value to find the value of each of the remaining trigonometric functions. We also use the associative angle formula of $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.

Complete step by step answer:
It’s given that $\sin \theta =\dfrac{1}{4}$, $\theta $ being in quadrant II. In that quadrant only $\sin \theta $ and $\csc \theta $ are positive whereas all the other ratios are negative.
We know the sum of the square law of ${{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1$.
We can put the value of $\sin \theta =\dfrac{1}{4}$ in the equation of ${{\left( \sin x \right)}^{2}}+{{\left( \cos x \right)}^{2}}=1$.
Putting the value of $\sin \theta $, we get ${{\left( \dfrac{1}{4} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1$.
Now we perform the binary operations.
${{\left( \dfrac{1}{4} \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1 \\
\Rightarrow {{\left( \cos \theta \right)}^{2}}=1-\dfrac{1}{16}=\dfrac{15}{16} \\
\Rightarrow \left( \cos \theta \right)=\pm \dfrac{\sqrt{15}}{4} \\ $
The value of $\cos \theta $ in quadrant II will be negative and that’s why $\left( \cos \theta \right)=-\dfrac{\sqrt{15}}{4}$.
Now we use the identity formula of $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ for $\sin \left( \theta -\dfrac{\pi }{3} \right)$.
So, $\sin \left( \theta -\dfrac{\pi }{3} \right)=\sin \theta \cos \dfrac{\pi }{3}-\cos \theta \sin \dfrac{\pi }{3}$.
Putting the values, we get,
$\sin \left( \theta -\dfrac{\pi }{3} \right)=\dfrac{1}{4}\times \dfrac{1}{2}-\left( -\dfrac{\sqrt{15}}{4} \right)\times \dfrac{\sqrt{3}}{2}\\
\Rightarrow\sin \left( \theta -\dfrac{\pi }{3} \right) =\dfrac{1}{8}+\dfrac{3\sqrt{5}}{8}\\
\therefore\sin \left( \theta -\dfrac{\pi }{3} \right) =\dfrac{1+3\sqrt{5}}{8}$

Therefore, the exact value of $\sin \left( \theta -\dfrac{\pi }{3} \right)$ is $\dfrac{1+3\sqrt{5}}{8}$.

Note: In addition to the reciprocal relationships of certain trigonometric functions, two other types of relationships exist. These relationships, known as trigonometric identities, include functional relationships and Pythagorean relationships.