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Question: How do you evaluate \(\sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right)\) ?...

How do you evaluate sin⁡((3π)2)\sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) ?

Explanation

Solution

First, we need to analyze the given information so that we can able to solve the problem.Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.Here we are asked to calculate sin⁡((3π)2)\sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right)
We need to apply the appropriate trigonometric identities to obtain the required answer.

Formula to be used:
The trigonometric identity that is used to solve the given problem is as follows.
sin⁡(270+θ)=−cos⁡θ\sin \left( {270 + \theta } \right) = - \cos \theta

Complete step by step answer:
Here we are asked to calculate sin⁡((3π)2)\sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right)
Now we are asked to calculate sin⁡((3π)2)\sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right)
We all know that the value of π\pi is 180∘{180^\circ }
Thus, we have
sin⁡((3π)2)=sin⁡((3×180∘)2)\sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = \sin \left( {\dfrac{{\left( {3 \times {{180}^\circ }} \right)}}{2}} \right)
⇒sin⁡((3π)2)=sin⁡(3×90∘)\Rightarrow \sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = \sin \left( {3 \times {{90}^\circ }} \right)
⇒sin⁡((3π)2)=sin⁡(270∘)\Rightarrow \sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = \sin \left( {{{270}^\circ }} \right)
⇒sin⁡((3π)2)=sin⁡(270∘+0)\Rightarrow \sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = \sin \left( {{{270}^\circ } + 0} \right)
The trigonometric identity that is used to solve the given problem is as follows.
sin⁡(270+θ)=−cos⁡θ\sin \left( {270 + \theta } \right) = - \cos \theta
⇒sin⁡((3π)2)=sin⁡(270∘+0)\Rightarrow \sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = \sin \left({{ {270}^\circ } + 0} \right)
⇒sin⁡((3π)2)=−cos⁡0\Rightarrow \sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = - \cos 0
⇒sin⁡((3π)2)=−1\Rightarrow \sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = - 1
Hence, sin⁡((3π)2)=−1\sin \left( {\dfrac{{\left( {3\pi} \right)}}{2}} \right) = - 1

Note:

Here in the first quadrant, all ratios are positive. In the second quadrant, only sine is positive, and in the third quadrant, only the tangent ratio is positive. And in the fourth quadrant, only the cosine ratio is positive.
When we are given sin⁡(270+θ)\sin \left( {270 + \theta } \right) , we need to understand sin⁡270\sin 270 is in between the third and fourth quadrant and sin⁡(270+θ)\sin \left( {270 + \theta } \right) is in the fourth quadrant. In the fourth quadrant, only the cos ratio is positive and sin is negative. That’s why we got minus sign in this trigonometric identity sin⁡(270+θ)=−cos⁡θ\sin \left( {270 + \theta } \right) = - \cos \theta