Question

How do you simplify $\sin (x - \dfrac{\pi }{2})$ ?

Answer 1

Hint: In the above question you were asked to simplify $\sin (x - \dfrac{\pi }{2})$ . For simplifying this you can use the formula of trigonometric compound angles which is $\sin (A - B) = \sin A\cos B - \cos A\sin B$ where A would be x and B would be $\dfrac{\pi }{2}$. So let us see how we can solve this problem.

Complete Step by Step Solution:
In the given question we have to simplify $\sin (x - \dfrac{\pi }{2})$ . Here we will use $\sin (A - B) = \sin A\cos B - \cos A\sin B$ the formula from trigonometric compound angles.
So, according to our problem A is x and B is $\dfrac{\pi }{2}$
 $\Rightarrow \sin (x - \dfrac{\pi }{2}) = \sin x\cos (\dfrac{\pi }{2}) - \cos x\sin (\dfrac{\pi }{2})$
We know that, $\cos (\dfrac{\pi }{2}) = 0$ and $\sin (\dfrac{\pi }{2}) = 1$ . By putting the values of $\cos (\dfrac{\pi }{2})$ and $\sin (\dfrac{\pi }{2})$ in the above expression we get,
 $\Rightarrow \sin (x - \dfrac{\pi }{2}) = \sin x(0) - \cos x(1)$
 $\Rightarrow \sin (x - \dfrac{\pi }{2}) = - \cos x$

Therefore, $\sin (x - \dfrac{\pi }{2})$ is $- \cos x$

Additional Information:
Here we see a formula of compound angle but there are more formulas. sin(A+B), cos(A+B), cos(A-B), tan(A+B), tan(A-B). All these trigonometric compound angles have different formulas which we will study in detail in later classes.

Note:
In the above solution, we have used the formula of trigonometric compound angles. Also, we used the values of $\cos (\dfrac{\pi }{2})$ and $\sin (\dfrac{\pi }{2})$. As we know that the value of sin increases with the increase in angle and the value of cos decreases with the increase of angle, so we concluded the value as 1 and 0 respectively for sin and cos.