Question

How do you evaluate sine, cosine, tangent of $- \dfrac{\pi }{2}$ without using a calculator?

Answer 1

Hint : We have to find the value of the given three trigonometric ratios of a negative angle. We can use the trigonometric identities of sine and cosine to convert the given angle into an angle whose value of trigonometric ratios is trivially known. To find the tangent of the given angle we can first find sine and cosine and then use the ratio $\tan A = \dfrac{{\sin A}}{{\cos A}}$

Complete step by step solution:
We have to evaluate the value of three trigonometric ratios, sine, cosine and tangent, of the given angle $- \dfrac{\pi }{2}$. We have to solve this without using a calculator.
We can try to convert the given negative angle into some angle whose value of trigonometric values is trivially known.
We can use trigonometric identities for this purpose.
First we try to find $\sin \left( { - \dfrac{\pi }{2}} \right)$
For sine of difference of two angles, we have an identity,
$\sin (A - B) = \sin A\cos B - \cos A\sin B$
Putting $A = 0$ and $B = \dfrac{\pi }{2}$ , we get,
$\sin (0 - \dfrac{\pi }{2}) = \sin 0\cos \dfrac{\pi }{2} - \cos 0\sin \dfrac{\pi }{2}$
We know that $\sin 0 = 0$ , $\cos 0 = 1$ and $\sin \dfrac{\pi }{2} = 1$ . Thus we get,
$\sin \left( { - \dfrac{\pi }{2}} \right) = - \sin \dfrac{\pi }{2} = - 1$
Now to find cosine ratio we can use the following trigonometric identity,
$\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
Again putting $A = 0$ and $B = \dfrac{\pi }{2}$ , we get,
$\cos \left( {0 - \dfrac{\pi }{2}} \right) = \cos 0\cos \dfrac{\pi }{2} + \sin 0\sin \dfrac{\pi }{2}$
We know that $\sin 0 = 0$ , $\cos 0 = 1$ and $\cos \dfrac{\pi }{2} = 0$ . Thus we get,
$\cos \left( { - \dfrac{\pi }{2}} \right) = \cos \dfrac{\pi }{2} = 0$
Now to find tangent of $- \dfrac{\pi }{2}$ we will use the ratio $\tan A = \dfrac{{\sin A}}{{\cos A}}$
Thus, $\tan \left( { - \dfrac{\pi }{2}} \right) = \dfrac{{\sin \left( { - \dfrac{\pi }{2}} \right)}}{{\cos \left( { - \dfrac{\pi }{2}} \right)}} = \dfrac{{ - \sin \left( {\dfrac{\pi }{2}} \right)}}{{\cos \left( {\dfrac{\pi }{2}} \right)}} = - \tan \left( {\dfrac{\pi }{2}} \right) \to - \infty$
Hence, we get the values as,
$\sin \left( { - \dfrac{\pi }{2}} \right) = - 1 \\\ \cos \left( { - \dfrac{\pi }{2}} \right) = 0 \\\ \tan \left( { - \dfrac{\pi }{2}} \right) \to - \infty \;$

Note : We have used trigonometric identities to find the value of trigonometric ratios of an angle by converting that negative angle into angles whose trigonometric ratios are known. We have converted $- \dfrac{\pi }{2}$ into a difference of $0$ and $\dfrac{\pi }{2}$. We could have used any other combination to arrive at the same result.