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Question: How do you evaluate \(\sin \left( {\dfrac{{7\pi }}{4}} \right)\)?...

How do you evaluate sin⁡(7π4)\sin \left( {\dfrac{{7\pi }}{4}} \right)?

Explanation

Solution

Here we need to know that whenever we need to calculate such values we need to notice the quadrant in which it is lying. So when we calculate we come to know that the angle is in the fourth quadrant and here the value of sin⁡\sin is negative. Hence we can write sin⁡(7π4)\sin \left( {\dfrac{{7\pi }}{4}} \right) in the form ofsin⁡(2π−θ)\sin \left( {2\pi - \theta } \right) and then apply sin⁡(2π−θ)=−sin⁡θ\sin \left( {2\pi - \theta } \right) = - \sin \theta .

Complete step by step solution:
Here we are given to find the value of sin⁡(7π4)\sin \left( {\dfrac{{7\pi }}{4}} \right)
We must know that we have four quadrants in which:

  1. The 1st1{\text{st}} quadrant lies between 0∘ and 90∘0^\circ {\text{ and 90}}^\circ
  2. The 2nd{\text{2nd}} quadrant lies between 90∘ and 180∘90^\circ {\text{ and 180}}^\circ
  3. The 3rd{\text{3rd}} quadrant lies between 180∘ and 270∘180^\circ {\text{ and 270}}^\circ
  4. The 4th4{\text{th}} quadrant lies between 270∘ and 360∘270^\circ {\text{ and 360}}^\circ
    We also know that π radians=180∘\pi {\text{ radians}} = 180^\circ
    Now if we will calculate the given angle which is given in radian in degrees we will get:
    π radians=180∘\pi {\text{ radians}} = 180^\circ
    (7π4)radian=180π×7π4=315∘\left( {\dfrac{{7\pi }}{4}} \right){\text{radian}} = \dfrac{{180}}{\pi } \times \dfrac{{7\pi }}{4} = 315^\circ
    So we come to know that this angle lies in the fourth quadrant; the value of the trigonometric function sin⁡\sin is negative in the fourth quadrant. Hence our answer will also be negative.
    We must know that sin⁡ and csc\sin {\text{ and csc}} are positive only in the first and second quadrant. In others they are negative.
    Now we can write (7π4)=2π−π4\left( {\dfrac{{7\pi }}{4}} \right) = 2\pi - \dfrac{\pi }{4}.
    We know that we have the formula:
    sin⁡(2π−θ)=−sin⁡θ\sin \left( {2\pi - \theta } \right) = - \sin \theta
    Now we can compare (7π4)=2π−π4\left( {\dfrac{{7\pi }}{4}} \right) = 2\pi - \dfrac{\pi }{4} with (2π−θ)\left( {2\pi - \theta } \right) we will get θ=π4\theta = \dfrac{\pi }{4} and we will get:
    sin⁡(7π4)=sin⁡(2π−π4)\sin \left( {\dfrac{{7\pi }}{4}} \right) = \sin \left( {2\pi - \dfrac{\pi }{4}} \right)
    Now applying the above formulasin⁡(2π−θ)=−sin⁡θ\sin \left( {2\pi - \theta } \right) = - \sin \theta , we will get:

sin⁡(7π4)=sin⁡(2π−π4)=−sin⁡(π4)=−12\sin \left( {\dfrac{{7\pi }}{4}} \right) = \sin \left( {2\pi - \dfrac{\pi }{4}} \right) = - \sin \left( {\dfrac{\pi }{4}} \right) = - \dfrac{1}{{\sqrt 2 }}.

Note: Here the student must know the trigonometric functions that are positive or negative in each quadrant. For example: If we would have been told to find cos⁡(7π4)\cos \left( {\dfrac{{7\pi }}{4}} \right) then we must know that cosine function is positive in the fourth quadrant. Hence its value will come out to be positive. All the properties must be known.