Question

How do you prove $\cos \left( {x - \dfrac{{3\pi }}{2}} \right) = - \sin x$?

Answer 1

Hint: In this question we have to simplify the expression which is a trigonometric angle, to do this we will make use of the sum of angle identities in the trigonometry which is given by $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ and trigonometric ratios, i.e., $\cos \dfrac{{3\pi }}{2} = 0$and $\sin \dfrac{{3\pi }}{2} = - 1$, then substitute the values given in the identity, and then simplify the expression to get the required simplified result.

Complete step by step solution:
Given function is $\cos \left( {x - \dfrac{{3\pi }}{2}} \right) = - \sin x$,
Now take left hand side of the equation i.e.,
$ \Rightarrow \cos \left( {x - \dfrac{{3\pi }}{2}} \right)$,
Now using the sum angle identity which is given by $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$,
Here $A = x$ and $B = \dfrac{{3\pi }}{2}$, by substituting the values in the identity we get,
$ \Rightarrow \cos \left( {x - \dfrac{{3\pi }}{2}} \right) = \cos x\cos \dfrac{{3\pi }}{2} + \sin x\sin \dfrac{{3\pi }}{2}$,
Now we know that $\cos \dfrac{{3\pi }}{2} = 0$and $\sin \dfrac{{3\pi }}{2} = - 1$, and by substituting the values in the above expression we get,
$ \Rightarrow \cos \left( {x - \dfrac{{3\pi }}{2}} \right) = \cos x\left( 0 \right) + \sin x\left( { - 1} \right)$,
Now simplifying the expression we get,
$ \Rightarrow \cos \left( {x - \dfrac{{3\pi }}{2}} \right) = 0 - \sin x$,
Now simplifying we get,
$ \Rightarrow \cos \left( {x - \dfrac{{3\pi }}{2}} \right) = - \sin x$.
So, the left hand side is equal to right hand side,
Hence proved.

Final Answer:
$\therefore $The given expression $\cos \left( {x - \dfrac{{3\pi }}{2}} \right)$ will be equal to $ - \sin x$.

Note:
Angle sum identities and angle difference identities can be used to find the function values of any angles however, the most practical use is to find exact values of an angle that can be written as a sum or difference using the familiar values for the sine, cosine and tangent of the ${30^o}$, ${45^o}$, ${60^o}$ and ${90^o}$ angles and their multiples. Some important formulas are:
$\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$,
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$,
$\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$,
$\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$,
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$,
$\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$,
To find a cosecant, secant or cotangent function you change the expression to one of the three basic functions, do the necessary calculations. Then use the reciprocal identity again to change the answer back to the original identity.