Solve the problem.
-If $f ( \theta ) = \sin \theta$ and $f ( a ) = - \frac { 1 } { 6 }$, find the exact value of $f ( a ) + f ( a + 2 \pi ) + f ( a + 4 \pi )$
A) $- \frac { 1 } { 2 }$
B) $- \frac { 1 } { 2 } + 6 \pi$
C) $- \frac { 1 } { 6 }$
D) $\frac { 3 } { 2 }$

Question

Quizplus · Accepted Answer

The period of the sine function is $2 \pi$, so $f ( a + 2 \pi ) = f ( a )$ and $f ( a + 4 \pi ) = f ( a )$. Therefore, $f ( a ) + f ( a + 2 \pi ) + f ( a + 4 \pi ) = 3f ( a ) = 3 \left( - \frac { 1 } { 6 } \right) = - \frac { 1 } { 2 }$.

Solve the Problem f(θ)=sin⁡θf ( \theta ) = \sin \thetaf(θ)=sinθ And f(a)=−16f ( a ) = - \frac { 1 } { 6 }f(a)=−61​

Solve the Problem $f ( \theta ) = \sin \theta$ And $f ( a ) = - \frac { 1 } { 6 }$