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Question: What are the exact values of \(\cos {150^0}\) and \(\sin {150^0}\) ?...

What are the exact values of cos⁡1500\cos {150^0} and sin⁡1500\sin {150^0} ?

Explanation

Solution

First the given values are in the form of the trigonometric values because it contains the trigonometric identities of sine and cosine.
Thus, we start with the position of the problem and to check its quadrant. We will conclude the signs of the values and also with the help of trigonometry identity tables we get the required answer.

Complete step-by-step solution:
Since from the given, we have the two values of the sine and cosine as cos⁡1500\cos {150^0} and sin⁡1500\sin {150^0}. We need to find its exact values.
Let us start with 1500=1800−300{150^0} = {180^0} - {30^0} and then we will replace into the given values cos⁡1500=cos⁡(1800−300)\cos {150^0} = \cos ({180^0} - {30^0})
Since using the trigonometric identities, the value of the cos⁡(1800−θ)=−cos⁡θ\cos ({180^0} - \theta ) = - \cos \theta
Thus cos⁡1500=cos⁡(1800−300)\cos {150^0} = \cos ({180^0} - {30^0}) can be expressed as in the form of cos⁡1500=cos⁡(1800−300)=−cos⁡300\cos {150^0} = \cos ({180^0} - {30^0}) = - \cos {30^0}
Once using the trigonometry values and the table we know that the cos⁡300=32\cos {30^0} = \dfrac{{\sqrt 3 }}{2}

Angle in degrees0∘0^\circ 30∘30^\circ 45∘45^\circ 60∘60^\circ 90∘90^\circ
cos⁡\cos 1132\dfrac{{\sqrt 3 }}{2}12\dfrac{1}{{\sqrt 2 }}12\dfrac{1}{2}00

Hence, we get the value of −cos⁡300=−32 - \cos {30^0} = \dfrac{{ - \sqrt 3 }}{2}
Now the same method follows for the sine, with 1500=1800−300{150^0} = {180^0} - {30^0} and then we will replace into the given values sin⁡1500=sin⁡(1800−300)\sin {150^0} = \sin ({180^0} - {30^0})
Since using the trigonometric identities, the value of the sin⁡(1800−θ)=sin⁡θ\sin ({180^0} - \theta ) = \sin \theta
Thus sin⁡1500=sin⁡(1800−300)\sin {150^0} = \sin ({180^0} - {30^0}) can be expressed as in the form of sin⁡1500=sin⁡(1800−300)=sin⁡300\sin {150^0} = \sin ({180^0} - {30^0}) = \sin {30^0}
Once using the trigonometry values and the table we know that the sin⁡300=12\sin {30^0} = \dfrac{1}{2}

Angle in degrees0∘0^\circ 30∘30^\circ 45∘45^\circ 60∘60^\circ 90∘90^\circ
sin⁡\sin 0012\dfrac{1}{2}12\dfrac{1}{{\sqrt 2 }}32\dfrac{{\sqrt 3 }}{2}11

Hence, we get the value of sin⁡300=12\sin {30^0} = \dfrac{1}{2}
Therefore, the exact values of the sine and cosine for cos⁡1500\cos {150^0} and sin⁡1500\sin {150^0} are cos⁡1500=−32\cos {150^0} = - \dfrac{{\sqrt 3 }}{2} and sin⁡1500=12\sin {150^0} = \dfrac{1}{2}.

Note: Both values of the sin⁡1500=sin⁡300\sin {150^0} = \sin {30^0} are the same because the reference angle for the 150150 is equal to the 3030 triangle formed in the unit circle.
The angle is referenced in the form when the perpendicular is dropped from the unit circle to the x-axis, which forms a right triangle.