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Question

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

How do you simplify sin(xπ2)\sin (x - \dfrac{\pi }{2}) ?

Explanation

Solution

In the above question you were asked to simplify sin(xπ2)\sin (x - \dfrac{\pi }{2}) . For simplifying this you can use the formula of trigonometric compound angles which is sin(AB)=sinAcosBcosAsinB\sin (A - B) = \sin A\cos B - \cos A\sin B where A would be x and B would be π2\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(xπ2)\sin (x - \dfrac{\pi }{2}) . Here we will use sin(AB)=sinAcosBcosAsinB\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 π2\dfrac{\pi }{2}
sin(xπ2)=sinxcos(π2)cosxsin(π2)\Rightarrow \sin (x - \dfrac{\pi }{2}) = \sin x\cos (\dfrac{\pi }{2}) - \cos x\sin (\dfrac{\pi }{2})
We know that, cos(π2)=0\cos (\dfrac{\pi }{2}) = 0 and sin(π2)=1\sin (\dfrac{\pi }{2}) = 1 . By putting the values of cos(π2)\cos (\dfrac{\pi }{2}) and sin(π2)\sin (\dfrac{\pi }{2}) in the above expression we get,
sin(xπ2)=sinx(0)cosx(1)\Rightarrow \sin (x - \dfrac{\pi }{2}) = \sin x(0) - \cos x(1)
sin(xπ2)=cosx\Rightarrow \sin (x - \dfrac{\pi }{2}) = - \cos x

Therefore, sin(xπ2)\sin (x - \dfrac{\pi }{2}) is cosx- \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(π2)\cos (\dfrac{\pi }{2}) and sin(π2)\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.