Question

Find the value of the following. $\cos {72}^{\circ }$

Answer 1

Hint: Since it is not possible to find the angles of the trigonometric function individually, try to bring those angles in the form of the angles which are easily available to us or which we can find out easily.

Complete step-by-step answer:
We have to find the value of a trigonometric function of the angle ${72}^{\circ }$ . Since we directly do not know the value of the given angle. We will try to use some already known trigonometric identities to find the answer of the question given.
We know that $\cos \left({\displaystyle \frac{\pi }{2}}-x\right)=\sin x$ . So, we will use this identity to represent $\cos \left({72}^{\circ }\right)$ in terms of $\sin$ . We observe that, $\cos {72}^{\circ }=\cos \left({90}^{\circ }-{18}^{\circ }\right)$
On using the identity $\cos \left({\displaystyle \frac{\pi }{2}}-x\right)=\sin x$ we get,
 $\Rightarrow \cos {72}^{\circ }=\sin {18}^{\circ }$
So, we will try to find the value of $\sin {18}^{\circ }$ . Let us take $A$ to be ${18}^{\circ }$ , that is $A={18}^{\circ }$ .
Then we can see that $5A={90}^{\circ }$ .
 $\Rightarrow 2A+3A={90}^{\circ }$
 $\Rightarrow 2A={90}^{\circ }-3A$
Now on applying sine function to the angles on both sides of the equation we get,
 $\sin 2A=\sin \left({90}^{\circ }-3A\right)$
Now again by using the identity $\cos \left({\displaystyle \frac{\pi }{2}}-x\right)=\sin x$ , we get
 $\sin 2A=\cos 3A$
Now we will use the identities for multiple angles to get,
 $\Rightarrow 2\sin A\cos A=4{\cos }^{3}A-3\cos A$ [Using $\sin 2x=2\sin x\cos x$ and $\cos 3x=4{\cos }^{3}x-3\cos x$ ]
 $\Rightarrow 2\sin A\cos A-4{\cos }^{3}A+3\cos A=0$
 $\Rightarrow \cos A(2\sin A-4{\cos }^{2}A+3)=0$
Since $\cos A=\cos {18}^{\circ }\ne 0$ , we have
 $2\sin A-4{\cos }^{2}A+3=0$
 $\Rightarrow 2\sin A-4(1-{\sin }^{2}A)+3=0$ [Using ${\cos }^{2}x+{\sin }^{2}x=1$ ]
 $\Rightarrow 2\sin A-4+4{\sin }^{2}A+3=0$
 $\Rightarrow 2\sin A+4{\sin }^{2}A-1=0$ , which is a quadratic equation in $\sin A$ . So, we use the quadratic formula $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ given for the equation $a{x}^2+bx+c=0$ .
Therefore, $\sin A={\displaystyle \frac{-2\pm \sqrt{4-4(4)(-1)}}{2(4)}}={\displaystyle \frac{-2\pm \sqrt{4+16}}{8}}$
 $\Rightarrow \sin A={\displaystyle \frac{-2\pm 2\sqrt{5}}{8}}$
 $\Rightarrow \sin A={\displaystyle \frac{-1\pm \sqrt{5}}{4}}$
Since ${18}^{\circ }$ belongs to the first quadrant, we know that $\sin {18}^{\circ }$ is positive.
Therefore, $\sin {18}^{\circ }={\displaystyle \frac{\sqrt{5}-1}{4}}$ .
So, we have $\cos {72}^{\circ }=\sin {18}^{\circ }={\displaystyle \frac{\sqrt{5}-1}{4}}$ .
Hence the value of $\cos {72}^{\circ }$ is ${\displaystyle \frac{\sqrt{5}-1}{4}}$ .
So, the correct answer is “ ${\displaystyle \frac{\sqrt{5}-1}{4}}$ ”.

Note: While solving this kind of problem, one should remember that three are many ways of getting to the answer using various trigonometric identities. But we have to be clear with what we need to find and use a relatively easier method. Remember to mention the identities used wherever required. Such as $\sin 2x=2\sin x\cos x$ etc.