Question

How do you find the value of the six trigonometric functions \[t=\dfrac{7\pi }{4}\] ?

Answer 1

Hint: The six trigonometric functions are \[\sin t\],\[\cos t\] ,\[\tan t\] ,\[\text{cosec}t\] ,\[\sec t\] and \[\cot t\] . In the given problem, the value of \[t\] is given to be as \[\dfrac{7\pi }{4}\] , which when expressed in terms of degrees, is equivalent to \[\dfrac{7\times 180}{4}=315\] degrees. Thus, we need to find the values of \[\sin {{315}^{\circ }}\],\[\cos {{315}^{\circ }}\] ,\[\tan {{315}^{\circ }}\] ,\[\text{cosec}{{315}^{\circ }}\] ,\[\sec {{315}^{\circ }}\] and \[\cot {{315}^{\circ }}\] . To solve these types of questions smoothly, we need to keep in mind all the relations and formulae. The equations are described as:
\[\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
 & \cot \theta =\dfrac{1}{\tan \theta } \\
\end{align}\]
We also need to remember some of the basic defined relations, which are:
\[\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
 & \text{cose}{{\text{c}}^{2}}\theta -{{\cot }^{2}}\theta =1 \\
\end{align}\]

Complete step-by-step answer:
Now, we start working with the solution: First, we need to change the value of \[\dfrac{7\pi }{4}\] to some other form. We know that on dividing 7 by 4, we get the remainder as 3. So rewriting \[\dfrac{7\pi }{4}\] as \[2\pi -\dfrac{\pi }{4}\] , we now try evaluating the values of all the trigonometric functions. We know, that in the first quadrant all the trigonometric functions yield positive values, in the second quadrant only \[\sin x\] and \[\text{cosec}x\] are positive, in the third quadrant only \[\tan x\] and \[\text{cot}x\] are positive, in the fourth quadrant only \[\cos x\] and \[\sec x\] are positive. We see that \[2\pi -\dfrac{\pi }{4}\] forms an angle which lies in the fourth quadrant, hence only the values of \[\cos x\] and \[\sec x\] will be positive, all the other trigonometric functions will yield a negative value as the angle do not fall in their respective quadrants.
We write,
\[\sin \dfrac{7\pi }{4}=\sin \left( 2\pi -\dfrac{\pi }{4} \right)\] , rewriting it with a negative sign,
\[\Rightarrow \sin \dfrac{7\pi }{4}=-\sin \left( \dfrac{\pi }{4} \right)\]
Writing the value in degrees, we get,
\[\Rightarrow \sin \dfrac{7\pi }{4}=-\sin \left( {{45}^{\circ }} \right)\]
We write it’s value as, \[\sin \dfrac{7\pi }{4}=-\dfrac{1}{\sqrt{2}}\] . Now, we can easily find the value of \[\text{cosec}\dfrac{7\pi }{4}\] . We know \[\text{cosec}\theta \text{=}\dfrac{1}{\sin \theta }\] , Therefore,
\[\text{cosec}\dfrac{7\pi }{4}=\dfrac{1}{\sin \dfrac{7\pi }{4}}=-\sqrt{2}\]
We write,
\[\cos \dfrac{7\pi }{4}=\cos \left( 2\pi -\dfrac{\pi }{4} \right)\] , rewriting it,
\[\Rightarrow \cos \dfrac{7\pi }{4}=\cos \left( \dfrac{\pi }{4} \right)\]
Writing the value in degrees, we get,
\[\Rightarrow \cos \dfrac{7\pi }{4}=\cos \left( {{45}^{\circ }} \right)\]
We write it’s value as, \[\cos \dfrac{7\pi }{4}=\dfrac{1}{\sqrt{2}}\] . Now, we can easily find the value of \[\sec \dfrac{7\pi }{4}\] . We know \[\text{sec}\theta \text{=}\dfrac{1}{\cos \theta }\] , Therefore,
\[\text{sec}\dfrac{7\pi }{4}=\dfrac{1}{\cos \dfrac{7\pi }{4}}=\sqrt{2}\]
We write,
\[\tan \dfrac{7\pi }{4}=\tan \left( 2\pi -\dfrac{\pi }{4} \right)\] , rewriting it with a negative sign,
\[\Rightarrow \tan \dfrac{7\pi }{4}=-\tan \left( \dfrac{\pi }{4} \right)\]
Writing the value in degrees, we get,
\[\Rightarrow \tan \dfrac{7\pi }{4}=-\tan \left( {{45}^{\circ }} \right)\]
We write it’s value as, \[\tan \dfrac{7\pi }{4}=-1\] . Now, we can easily find the value of \[\text{cot}\dfrac{7\pi }{4}\] . We know \[\text{cot}\theta \text{=}\dfrac{1}{\tan \theta }\] , Therefore,
\[\text{cot}\dfrac{7\pi }{4}=\dfrac{1}{\tan \dfrac{7\pi }{4}}=-1\]

Note: We should remember all the standard values of the trigonometric function angles or else solving the problem may be difficult. The conversion from radian to degree should be done carefully or we may end up into a different answer. The partial fraction should also be done cautiously and must be rechecked.