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Question: How do you evaluate \(\sin \left( {11\dfrac{\pi }{2}} \right)\)?...

How do you evaluate sin⁡(11π2)\sin \left( {11\dfrac{\pi }{2}} \right)?

Explanation

Solution

This problem deals with evaluating the value of the given trigonometric function. In order to solve this some basic trigonometric identities and properties of trigonometric angles are used. Such as the values of the trigonometric angles in different quadrants. Formulas which are used here are:
⇒sin⁡(2π−θ)=−sin⁡θ\Rightarrow \sin \left( {2\pi - \theta } \right) = - \sin \theta
⇒sin⁡(π2)=1\Rightarrow \sin \left( {\dfrac{\pi }{2}} \right) = 1

Complete step-by-step solution:
Given a trigonometric function of an angle.
The given angle is equal to 11π2\dfrac{{11\pi }}{2}.
The trigonometric ratio is sine trigonometric function.
We know that the value of the function sin⁡(2π−θ)\sin \left( {2\pi - \theta } \right) is a negative value of the sine trigonometric ratio of the given value of the angle, which is equal to sin⁡(−θ)\sin \left( { - \theta } \right) which is eventually equal to −sin⁡(θ) - \sin \left( \theta \right).
The given value of the trigonometric function is sin⁡(11π2)\sin \left( {11\dfrac{\pi }{2}} \right), consider it as shown below:
⇒sin⁡(11π2)\Rightarrow \sin \left( {11\dfrac{\pi }{2}} \right)
The above value of angle of the sine trigonometric function can be rewritten as shown below:
⇒sin⁡(11π2)=sin⁡(2π−π2)\Rightarrow \sin \left( {11\dfrac{\pi }{2}} \right) = \sin \left( {2\pi - \dfrac{\pi }{2}} \right)
We know that the value of sin⁡(2π−θ)\sin \left( {2\pi - \theta } \right) is equal to −sin⁡(θ) - \sin \left( \theta \right), hence applying this property to the above expression, as shown below:
⇒sin⁡(2π−π2)=sin⁡(−π2)\Rightarrow \sin \left( {2\pi - \dfrac{\pi }{2}} \right) = \sin \left( { - \dfrac{\pi }{2}} \right)
We know that the value of sin⁡(−θ)=−sin⁡(θ)\sin \left( { - \theta } \right) = - \sin \left( \theta \right), as shown below:
⇒sin⁡(2π−π2)=−sin⁡(π2)\Rightarrow \sin \left( {2\pi - \dfrac{\pi }{2}} \right) = - \sin \left( {\dfrac{\pi }{2}} \right)
We know that value of sin⁡(π2)=1\sin \left( {\dfrac{\pi }{2}} \right) = 1, so substituting this in the above expression a shown below:
⇒sin⁡(2π−π2)=−1\Rightarrow \sin \left( {2\pi - \dfrac{\pi }{2}} \right) = - 1
So the value of the given expression which is sin⁡(11π2)\sin \left( {11\dfrac{\pi }{2}} \right) is equal to sin⁡(2π−π2)\sin \left( {2\pi - \dfrac{\pi }{2}} \right) which is eventually equal to -1.
∴sin⁡(11π2)=−1\therefore \sin \left( {11\dfrac{\pi }{2}} \right) = - 1
The value of sin⁡(11π2)=−1\sin \left( {11\dfrac{\pi }{2}} \right) = - 1.

Note: Please note that the above problem is solved with the help of some basic trigonometric angle properties and some trigonometric identities. Here all the values of sine trigonometric angles of all the four quadrants which are present in the coordinate system are given below:
⇒sin⁡(π2−θ)=sin⁡θ\Rightarrow \sin \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta
⇒sin⁡(π2+θ)=sin⁡θ\Rightarrow \sin \left( {\dfrac{\pi }{2} + \theta } \right) = \sin \theta
⇒sin⁡(π−θ)=sin⁡θ\Rightarrow \sin \left( {\pi - \theta } \right) = \sin \theta
⇒sin⁡(π+θ)=−sin⁡θ\Rightarrow \sin \left( {\pi + \theta } \right) = - \sin \theta
⇒sin⁡(3π2−θ)=−sin⁡θ\Rightarrow \sin \left( {\dfrac{{3\pi }}{2} - \theta } \right) = - \sin \theta
⇒sin⁡(2π−θ)=−sin⁡θ\Rightarrow \sin \left( {2\pi - \theta } \right) = - \sin \theta