Question

find the exact values of the six trigonometric functions of the given angle. do not use a calculator. 300°. select the correct choice below and fill in a answer boxes within your choice. sin 300°=-√3/2 (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.) a. cos 300°= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.) b. the function value is undefined. select the correct choice below and fill in any answer boxes within your choice. a. cos 300°= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.) b. the function value is undefined.

Explanation:

Step1: Rewrite the angle

$300^{\circ}=360^{\circ}- 60^{\circ}$.

Step2: Use trigonometric identities

For $\sin300^{\circ}$, since $\sin(A - B)=\sin A\cos B-\cos A\sin B$, and for $A = 360^{\circ}$, $\sin360^{\circ}=0$, $\cos360^{\circ}=1$, $B = 60^{\circ}$, $\sin60^{\circ}=\frac{\sqrt{3}}{2}$, $\cos60^{\circ}=\frac{1}{2}$, then $\sin300^{\circ}=\sin(360^{\circ}-60^{\circ})=\sin360^{\circ}\cos60^{\circ}-\cos360^{\circ}\sin60^{\circ}=0\times\frac{1}{2}-1\times\frac{\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}$.
For $\cos300^{\circ}$, using $\cos(A - B)=\cos A\cos B+\sin A\sin B$, $\cos300^{\circ}=\cos(360^{\circ}-60^{\circ})=\cos360^{\circ}\cos60^{\circ}+\sin360^{\circ}\sin60^{\circ}=1\times\frac{1}{2}+0\times\frac{\sqrt{3}}{2}=\frac{1}{2}$.

Answer:

$\sin300^{\circ}=-\frac{\sqrt{3}}{2}$, $\cos300^{\circ}=\frac{1}{2}$