Question

Why does $\cos \left( {90 - x} \right) = \sin \left( x \right)$ and $\sin \left( {90 - x} \right) = \cos \left( x \right)$ ?

Answer 1

First, we need to analyze the given information so that we are able to solve the problem. Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Here we are asked to prove that $\cos \left( {90 - x} \right) = \sin \left( x \right)$ and$\sin \left( {90 - x} \right) = \cos \left( x \right)$
We need to apply the appropriate trigonometric identities to obtain the required answer.
Formula to be used:
The trigonometric identities that are used to solve the given problem are as follows.
a)$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
b)$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$

Complete step by step answer:
Here we are asked to prove that $\cos \left( {90 - x} \right) = \sin \left( x \right)$ and $\sin \left( {90 - x} \right) = \cos \left( x \right)$
We need to apply the appropriate trigonometric identities to obtain the required answer.
a) Let us consider $\cos \left( {90 - x} \right)$
To prove: $\cos \left( {90 - x} \right) = \sin \left( x \right)$
We shall apply the trigonometric identity given below.
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
Thus $\cos \left( {90 - x} \right) = \cos 90\cos x + \sin 90\sin x$
We know that $\cos 90 = 0$ and $\sin 90 = 1$
$\Rightarrow \cos \left( {90 - x} \right) = 0 \times \cos x + 1 \times \sin x$
$\Rightarrow \cos \left( {90 - x} \right) = \sin x$
Hence we proved $\cos \left( {90 - x} \right) = \sin \left( x \right)$
b) Now, let us consider $\sin \left( {90 - x} \right)$
To prove: $\sin \left( {90 - x} \right) = \cos \left( x \right)$
We shall apply the trigonometric identity given below.
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
Thus $\sin \left( {90 - x} \right) = \sin 90\cos x - \cos 90\sin x$
We know that $\cos 90 = 0$ and $\sin 90 = 1$
$\Rightarrow \sin \left( {90 - x} \right) = 1 \times \cos x - 0 \times \sin x$
$\Rightarrow \sin \left( {90 - x} \right) = \cos x$
Hence we proved $\sin \left( {90 - x} \right) = \cos \left( x \right)$

Note: Here we are asked to prove that $\cos \left( {90 - x} \right) = \sin \left( x \right)$ and $\sin \left( {90 - x} \right) = \cos \left( x \right)$
We can prove the given trigonometric identities using the concept of quadrants. Considering the quadrants of a graph, we are able to solve this problem. There are many methods to prove the given identities. We have proved this by just applying the angle formula.