Question

How do you express $\sin \left( {\dfrac{\pi }{4}} \right) \cdot \sin \left( {\dfrac{{4\pi }}{3}} \right)$ without using products of trigonometric functions?

Answer 1

In this question, we first find the value of $\sin \left( {\dfrac{\pi }{4}} \right)$ by making use of Pythagoras theorem.
Then we find the value of $\sin \left( {\dfrac{{4\pi }}{3}} \right)$ by using the trigonometric formula given by
$\sin (A + B) = \sin A\cos B + \cos A\sin B$.
Then after getting the values of $\sin \left( {\dfrac{\pi }{4}} \right)$ and $\sin \left( {\dfrac{{4\pi }}{3}} \right)$ we just multiply them to get the solution.

Complete step by step solution:
We need to find $\sin \left( {\dfrac{\pi }{4}} \right) \cdot \sin \left( {\dfrac{{4\pi }}{3}} \right)$
First we find $\sin \left( {\dfrac{\pi }{4}} \right)$.
In the trigonometric circle $\dfrac{\pi }{4}$ is the bisectrix between 0 and $\dfrac{\pi }{2}$, where $x = y.$
By the Pythagoras theorem we know that ${x^2} + {y^2} = 1.$ (This is true only if $x = y.$)
Hence we get,
${x^2} + {y^2} = 1$
$\Rightarrow {y^2} + {y^2} = 1\,$ $\left( {\because x = y} \right)$
$\Rightarrow 2{y^2} = 1$
Now taking 2 to the right hand side we get,
$\Rightarrow {y^2} = \dfrac{1}{2}$
Now taking square root on both sides we get,
$\Rightarrow y = \sqrt {\dfrac{1}{2}}$
$\Rightarrow y = \dfrac{1}{{\sqrt 2 }}$.
We can write 2 as $\sqrt 2 \times \sqrt 2$.
Hence, $\dfrac{{\sqrt 2 }}{2} = \dfrac{{\sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }}$
Cancelling $\sqrt 2$ from numerator and denominator we get,
$\dfrac{{\sqrt 2 }}{2} = \dfrac{1}{{\sqrt 2 }}$.
Therefore $y = \dfrac{{\sqrt 2 }}{2}$.
Since here hypotenuse is 1, to get $\sin \left( {\dfrac{\pi }{4}} \right)$ we divide y by 1 .
Hence $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{y}{1}$
$\Rightarrow \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{{\dfrac{2}{1}}}$
$\Rightarrow \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}$.
Now we calculate the value of $\sin \left( {\dfrac{{4\pi }}{3}} \right)$.
We can write $\dfrac{{4\pi }}{3} = \pi + \dfrac{\pi }{3}$
So, $\sin \left( {\dfrac{{4\pi }}{3}} \right) = \sin \left( {\pi + \dfrac{\pi }{3}} \right)$ ……(1)
We now use the trigonometric formula of sine.
$\sin (A + B) = \sin A\cos B + \cos A\sin B$ ……(2)
Here $A = \pi$and $B = \dfrac{\pi }{3}$.
Substituting the values of A and B in the equation (2), we get,
$\therefore \sin \left( {\pi + \dfrac{\pi }{3}} \right) = \sin \pi \cdot \cos \dfrac{\pi }{3} + \cos \pi \cdot \sin \dfrac{\pi }{3}$ …..(3)
We know that,
$\sin \pi = 0$
$\cos \dfrac{\pi }{3} = \dfrac{1}{2}$
$\cos \pi = - 1$
$\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
Now substituting all these values in the equation (3), we obtain the value of $\sin \left( {\dfrac{{4\pi }}{3}} \right)$.
Now by (3), we have,
$\Rightarrow \sin \left( {\pi + \dfrac{\pi }{3}} \right) = 0 \cdot \dfrac{1}{2} + ( - 1) \cdot \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow \sin \left( {\pi + \dfrac{\pi }{3}} \right) = 0 - \dfrac{{\sqrt 3 }}{2}$
$\Rightarrow \sin \left( {\pi + \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 3 }}{2}$
Since by the equation (1), we have, $\sin \left( {\dfrac{{4\pi }}{3}} \right) = \sin \left( {\pi + \dfrac{\pi }{3}} \right)$
Hence, $\sin \left( {\dfrac{{4\pi }}{3}} \right) = - \dfrac{{\sqrt 3 }}{2}$.
Therefore substituting the values of $\sin \left( {\dfrac{\pi }{4}} \right)$ and $\sin \left( {\dfrac{{4\pi }}{3}} \right)$ in the given problem we get,

$\sin \left( {\dfrac{\pi }{4}} \right) \cdot \sin \left( {\dfrac{{4\pi }}{3}} \right)$ $= \dfrac{{\sqrt 2 }}{2} \cdot \left( { - \dfrac{{\sqrt 3 }}{2}} \right)$
$\Rightarrow \sin \left( {\dfrac{\pi }{4}} \right) \cdot \sin \left( {\dfrac{{4\pi }}{3}} \right) = - \dfrac{{\sqrt 6 }}{4}$
Therefore we get $\sin \left( {\dfrac{\pi }{4}} \right) \cdot \sin \left( {\dfrac{{4\pi }}{3}} \right) = - \dfrac{{\sqrt 6 }}{4}$.

Note:
Alternative method :
We know from trigonometric table the value of $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
Where $\dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}$
Hence $\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}$
Now we find $\sin \left( {\dfrac{{4\pi }}{3}} \right)$.
Also we know that $\sin \left( {\pi - a} \right) = \sin a$
Substitute $a = \dfrac{{4\pi }}{3}$ in the above formula, we get,
$\sin \left( {\pi - \dfrac{{4\pi }}{3}} \right) = \sin \dfrac{{4\pi }}{3}$
$\Rightarrow \sin \left( {\dfrac{{3\pi - 4\pi }}{3}} \right) = \sin \dfrac{{4\pi }}{3}$
$\Rightarrow \sin \left( { - \dfrac{\pi }{3}} \right) = \sin \dfrac{{4\pi }}{3}$
Hence we get $\sin \dfrac{{4\pi }}{3} = \sin \left( { - \dfrac{\pi }{3}} \right)$
We know that the value of $\sin \dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}$
Hence, $\sin \left( { - \dfrac{\pi }{3}} \right) = - \dfrac{{\sqrt 3 }}{2}$
Therefore we get,
$\sin \left( {\dfrac{\pi }{4}} \right) \cdot \sin \left( {\dfrac{{4\pi }}{3}} \right)$ $= \dfrac{{\sqrt 2 }}{2} \cdot \left( { - \dfrac{{\sqrt 3 }}{2}} \right)$
$\Rightarrow \sin \left( {\dfrac{\pi }{4}} \right) \cdot \sin \left( {\dfrac{{4\pi }}{3}} \right) = - \dfrac{{\sqrt 6 }}{4}$
One must remember that trigonometric ratios are positive in the particular quadrant.
Also we must know the trigonometric values of sine and cosine. So it is important to learn the trigonometric table to solve the given problem without much difficulty.