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Use the Addition Formulas To Calculate cos(π3)\cos \left( \frac { \pi } { 3 } \right)

Question 61

Multiple Choice

Use the addition formulas: sin(x+y) =sinxcosy+cosxsinysin(xy) =sinxcosycosxsinycos(x+y) =cosxcosysinxsinycos(xy) =cosxcosy+sinxsiny\begin{array} { l } \sin ( x + y ) = \sin x \cdot \cos y + \cos x \cdot \sin y \\\sin ( x - y ) = \sin x \cdot \cos y - \cos x \cdot \sin y \\\cos ( x + y ) = \cos x \cdot \cos y - \sin x \cdot \sin y \\\cos ( x - y ) = \cos x \cdot \cos y + \sin x \cdot \sin y\end{array}
To calculate cos(π3) \cos \left( \frac { \pi } { 3 } \right) , given that sin(π6) =12\sin \left( \frac { \pi } { 6 } \right) = \frac { 1 } { 2 } and cos(π6) =32\cos \left( \frac { \pi } { 6 } \right) = \frac { \sqrt { 3 } } { 2 } .


A) cos(π3) =0\cos \left( \frac { \pi } { 3 } \right) = 0
B) cos(π3) =32\cos \left( \frac { \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 2 }
C) cos(π3) =12\cos \left( \frac { \pi } { 3 } \right) = \frac { 1 } { 2 }
D) cos(π3) =32\cos \left( \frac { \pi } { 3 } \right) = - \frac { \sqrt { 3 } } { 2 }
E) cos(π3) =12\cos \left( \frac { \pi } { 3 } \right) = - \frac { 1 } { 2 }

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