Question

Evaluate the value of $\dfrac{{\sin 18^\circ }}{{\cos 72^\circ }}$ .

Answer 1

Here, we are asked to find the value of the trigonometric fraction $\dfrac{{\sin 18^\circ }}{{\cos 72^\circ }}$ .
Now, use the property $\sin x = \cos \left( {90^\circ - x} \right)$ and find the value of $\sin 18^\circ$ in the terms of cosine function.
Thus, to get the required answer, substitute the value of $\sin 18^\circ$ in terms of cosine function in the given trigonometric equation.

Complete step-by-step answer:
Here, we are asked to find the value of the trigonometric fraction $\dfrac{{\sin 18^\circ }}{{\cos 72^\circ }}$ .
We know the property that, $\sin x$ can also be written as $\cos \left( {90^\circ - x} \right)$ i.e. $\sin x = \cos \left( {90^\circ - x} \right)$ .
So, using the above property, we can write $\sin 18^\circ$ as $\cos \left( {90^\circ - 18^\circ } \right)$
$\therefore \sin 18^\circ = \cos \left( {90^\circ - 18^\circ } \right) = \cos 72^\circ$ .
Now, we will substitute the value of $\sin 18^\circ$ as $\cos 72^\circ$ in the given trigonometric fraction.
$\therefore \dfrac{{\sin 18^\circ }}{{\cos 72^\circ }} = \dfrac{{\cos 72^\circ }}{{\cos 72^\circ }} = 1$
Thus, we get the required value of the given trigonometric fraction $\dfrac{{\sin 18^\circ }}{{\cos 72^\circ }}$ as 1.

Note: Alternatively, we can also write $\cos 72^\circ$ in the terms of sine function by using the property $\cos y = \cos \left( {90^\circ - y} \right)$ . Thus, by substituting the value of $\cos 72^\circ$ in terms of sine function in the given trigonometric fraction, we get the required answer.
Some angle properties of trigonometric functions:
(i) $\sin x = \cos \left( {90^\circ - x} \right)$
(ii) $\cos x = \sin \left( {90^\circ - x} \right)$
(iii) $\tan x = \cot \left( {90^\circ - x} \right)$
(iv) $\sin x = \sin \left( {360^\circ + x} \right)$
(v) $\cos x = \cos \left( {360^\circ + x} \right)$
(vi) $\tan x = \tan \left( {360^\circ + x} \right)$