Question

How to prove that $\sin (\dfrac{\pi }{2} - \theta ) = \cos \theta $ ?

Answer 1

Hint: In this question, we have to prove that the left-hand side of the equation is equal to the right-hand side. For that, we will take one of the sides and solve it so that it becomes equal to the other side. On the left-hand side, we are given the sine of the difference of two angles. We know by an identity that the sine of the difference of two angles is equal to the product of the sine of the first angle and the cosine of the other angle minus the product of the cosine of the first angle and the sine of the second angle, that is, $\sin (A - B) = \sin A\cos B - \cos A\sin B$ .

Complete step-by-step solution:
Let us start solving the left-hand side –
$\sin (\dfrac{\pi }{2} - \theta ) = \sin \dfrac{\pi }{2}\cos \theta - \cos \dfrac{\pi }{2}\sin \theta $
We know that $\sin \dfrac{\pi }{2} = 1$ and $\cos \dfrac{\pi }{2} = 0$ , so we get –
$
  \sin (\dfrac{\pi }{2} - \theta ) = 1.\cos \theta - 0.\sin \theta \\
   \Rightarrow \sin (\dfrac{\pi }{2} - \theta ) = \cos \theta \\
 $
Now, the left-hand side has become equal to the right-hand side.
Hence proved that $\sin (\dfrac{\pi }{2} - \theta ) = \cos \theta $ .

Note: Trigonometry tells us the relation between the two sides of a right-angled triangle and one of its angles except for the right angle, trigonometry involves six trigonometric functions namely sine, cosine, tangent, cosecant, secant and cotangent function.

All the trigonometric functions are related to each other by some identities like the one we have used in the above solution. For solving any trigonometry question, we must remember the trigonometric value of some basic trigonometric functions. The result obtained signifies the periodic property of the trigonometric functions and this identity is used very commonly for solving other trigonometry related questions.