Question

How do you find the value of $\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$ ?

Answer 1

Hint : Here we are given with the trigonometric function with the two angles. Here we will use the sum and difference formulas to get the question solved. We know that trigonometric functions are having various identities as well as formulas to solve the given problem.
$\dfrac{\pi }{3} = {60^ \circ }\& \dfrac{\pi }{4} = {45^ \circ }$
Formula used:
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$

Complete step-by-step answer :
We are given the trigonometric difference problem.
We will use the formula mentioned above to solve it.
Given that $\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$
Comparing it with $\sin \left( {A - B} \right)$ we get
$\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{3}\cos \dfrac{\pi }{4} - \cos \dfrac{\pi }{3}\sin \dfrac{\pi }{4}$
We know that
$\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$ , $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
$\cos \dfrac{\pi }{3} = \dfrac{1}{2}$ , $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
Putting these values in the formula above we get,
$\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2} \times \dfrac{1}{{\sqrt 2 }}$
On multiplying we get,
$\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} - \dfrac{1}{{2\sqrt 2 }}$
Since the denominator is same we can directly perform the mathematical operations on the terms in numerator,
$\sin \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$
If the value of $\sqrt 3$ is given to us we will substitute it and then we can proceed otherwise this is our answer.
So, the correct answer is “$\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$”.

Note : Note that this problem is simply based on the formula. Nothing else is to be added or to be removed. But note that the angles are in radians (that is in terms of $\pi$ ); we either can convert them in degrees or if we are known with the radian values we can directly solve like the way above. If we are also known the value of $\sin \left( {{{60}^ \circ } - {{45}^ \circ }} \right) = \sin {15^ \circ }$ we can directly tick the correct option from the multiple choices if given. But the habit of using formulas is good because we need not to remember so many values!