Question

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

Answer 1

Hint: 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 $\cos {150^0}$ and $\sin {150^0}$. We need to find its exact values.
Let us start with ${150^0} = {180^0} - {30^0}$ and then we will replace into the given values $\cos {150^0} = \cos ({180^0} - {30^0})$
Since using the trigonometric identities, the value of the $\cos ({180^0} - \theta ) = - \cos \theta $
Thus $\cos {150^0} = \cos ({180^0} - {30^0})$ can be expressed as in the form of $\cos {150^0} = \cos ({180^0} - {30^0}) = - \cos {30^0}$
Once using the trigonometry values and the table we know that the $\cos {30^0} = \dfrac{{\sqrt 3 }}{2}$

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
\[\cos \]	\[1\]	\[\dfrac{{\sqrt 3 }}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	\[0\]

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

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
\[\sin \]	\[0\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	\[1\]

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

Note: Both values of the $\sin {150^0} = \sin {30^0}$ are the same because the reference angle for the $150$ is equal to the $30$ 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.