Question

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

Answer 1

Hint: First analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. This can be done by using the suitable trigonometric identities (formulae) and simplify any one side of the equation.

Complete step by step solution:
Sine, cosine and tangent of an angle are trigonometric ratios. These ratios are also called trigonometric functions. When we plot a graph of the trigonometric ratios with respect to all the real values of an angle, we get a graph that has a periodic property. This means that the graph repeats itself after equal intervals of the angle.Other than the trigonometric ratios sine, cosine and tangent we have other trigonometric ratios called cosecant, secant and cotangent.

All the above six trigonometric ratios (functions) are dependent on each other. There are different properties and identities that relate the trigonometric ratios.The equation that has to be verified is,
$\cos \left( x-\dfrac{\pi }{2} \right)=-\sin x$
Let us first analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. The left hand side of the equation is,
$\cos \left( x-\dfrac{\pi }{2} \right)$ ….. (i)
We can write the above expression as,
$\cos \left( x-\dfrac{\pi }{2} \right)=\cos \left\{ -\left( \dfrac{\pi }{2}-x \right) \right\}$

Now, we shall use the identity $\cos \left( -\theta \right)=\cos \theta $.
Then this gives us that,
$\cos \left( x-\dfrac{\pi }{2} \right)=\cos \left( \dfrac{\pi }{2}-x \right)$ …. (ii)
We know that $\cos \left( \dfrac{\pi }{2}-x \right)=\sin x$
Substitute this value in (ii).
$\therefore\cos \left( x-\dfrac{\pi }{2} \right)=\sin x$

However, the right hand side of the given equation is -sin(x). And $\sin x\ne -\sin x$.Therefore, the given equation is incorrect.

Note:It is not compulsory to verify a given equation only by simplifying the left hand side of the equation. You can also simply the right hand side and check whether it results as the same as the left hand side. Or you can also simplify both sides of the equation and check whether they gave the same result.It depends on which side of the equation is complex.