Question

How do you find the 6 trigonometric functions for $-\dfrac{\pi }{2}$?

Answer 1

Hint: In trigonometry, the six basic trigonometric functions that are widely used are $\sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\sec \theta ,\cos ec\theta $. We need to substitute the value of $-\dfrac{\pi }{2}$ in place in $\theta $ to get the required result.

Complete step by step answer:
The six trigonometric functions are $\sin \theta ,\cos \theta ,\tan \theta ,\cot \theta ,\sec \theta ,\cos ec\theta $.
For the $\sin \theta $function,
sine is the trigonometric function of any specified angle that is used in the context of a right angle.
It is usually defined as the ratio of the length of the side opposite to an angle $\theta $ to the length of the hypotenuse of the right-angle triangle denoted by$\sin \theta $.
According to the question,
$\Rightarrow \theta =-\dfrac{\pi }{2}$
So, we need to find the value of $\sin \left( -\dfrac{\pi }{2} \right)$
We know that $\sin \left( -\theta \right)=-\sin \theta $
Substituting the same,
We get,
$\Rightarrow \sin \left( -\dfrac{\pi }{2} \right)=-\sin \left( \dfrac{\pi }{2} \right)$
$\Rightarrow -\sin \left( \dfrac{\pi }{2} \right)=-1$
For the $\cos \theta $ function,
cosine is the trigonometric function of any specified angle that is used in the context of a right angle.
It is usually defined as the ratio of the length of the side adjacent to an angle $\theta $ to the length of the hypotenuse of the right-angle triangle denoted by $\cos \theta $.
According to the question,
$\Rightarrow \theta =-\dfrac{\pi }{2}$
So, we need to find the value of $\cos \left( -\dfrac{\pi }{2} \right)$
We know that $\cos \left( -\theta \right)=\cos \theta $
Substituting the same,
We get,
$\Rightarrow \cos \left( -\dfrac{\pi }{2} \right)=\cos \left( \dfrac{\pi }{2} \right)$
$\Rightarrow \cos \left( \dfrac{\pi }{2} \right)=0$
For the $\tan \theta $ function,
Tangent is the trigonometric function of any specified angle that is used in the context of a right angle.
It is usually defined as the ratio of the length of the side opposite to an angle $\theta $ to the length of the side adjacent to an angle $\theta $of the right-angle triangle denoted by$\tan \theta $.
According to the question,
$\Rightarrow \theta =-\dfrac{\pi }{2}$
So, we need to find the value of $\tan \left( -\dfrac{\pi }{2} \right)$
We know that $\tan \left( -\theta \right)=-\tan \theta $
Substituting the same,
We get,
$\Rightarrow \tan \left( -\dfrac{\pi }{2} \right)=-\tan \left( \dfrac{\pi }{2} \right)$
$\Rightarrow -\tan \left( \dfrac{\pi }{2} \right)=-\infty $
For the $\cot \theta $function,
The reciprocal of tangent function is cotangent function.
$\Rightarrow \cot \theta =\dfrac{1}{\tan \theta }$
According to the question,
$\Rightarrow \theta =-\dfrac{\pi }{2}$
$\Rightarrow \cot \left( -\dfrac{\pi }{2} \right)=\dfrac{1}{\tan \left( -\dfrac{\pi }{2} \right)}$
From the above,
$\Rightarrow \tan \left( -\dfrac{\pi }{2} \right)=-\infty $
Substituting the same,
We get,
$\Rightarrow \cot \left( -\dfrac{\pi }{2} \right)=\dfrac{1}{-\infty }$
$\Rightarrow \cot \left( -\dfrac{\pi }{2} \right)=0$
For the $\cos ec\theta $ function,
The reciprocal of sine function is the cosecant function.
$\Rightarrow \cos ec\theta =\dfrac{1}{\sin \theta }$
According to the question,
$\Rightarrow \theta =-\dfrac{\pi }{2}$
$\Rightarrow \cos ec\left( -\dfrac{\pi }{2} \right)=\dfrac{1}{\sin \left( -\dfrac{\pi }{2} \right)}$
From the above,
$\Rightarrow \sin \left( -\dfrac{\pi }{2} \right)=-1$
Substituting the same,
We get,
$\Rightarrow \cos ec\left( -\dfrac{\pi }{2} \right)=\dfrac{1}{-1}$
$\Rightarrow \cos ec\left( -\dfrac{\pi }{2} \right)=-1$
For the $\sec \theta $ function,
The reciprocal of cosine function is the secant function.
$\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }$
According to the question,
$\Rightarrow \theta =-\dfrac{\pi }{2}$
$\Rightarrow \sec \left( -\dfrac{\pi }{2} \right)=\dfrac{1}{\cos \left( -\dfrac{\pi }{2} \right)}$
From the above,
$\Rightarrow \cos \left( -\dfrac{\pi }{2} \right)=0$
Substituting the same,
We get,
$\Rightarrow \sec \left( -\dfrac{\pi }{2} \right)=\dfrac{1}{0}$
$\Rightarrow \sec \left( -\dfrac{\pi }{2} \right)=\infty $

Note: We need to know the trigonometric ratios of $\left( -\theta \right)$to solve the question easily. The relation between the trigonometric functions such as cotangent, tangent and sine, cosecant helps us to solve the problem in less time.