Question 36

The maximum value of the function $y = 4 \sin x \sin(x - 60°)$ is achieved when the value of $x$ = ....

$x = 30^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 60^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 90^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 120^\circ + k \cdot 180^\circ$ , $k$ is an integer

$x = 150^\circ + k \cdot 180^\circ$ , $k$ is an integer

Explanation

Determine the derivative of the function

y = 4 \sin x \sin(x - 60°)

y' = 4[\cos x \cdot \sin(x - 60°) + \sin x \cdot \cos(x - 60°)]

Use trigonometric identities $\sin A \cos B + \cos A \sin B = \sin(A + B)$ and $\cos A \cos B - \sin A \sin B = \cos(A + B)$

y' = 4[\sin(x - 60°) \cos x + \cos(x - 60°) \sin x]

y' = 4 \sin[(x - 60°) + x]

y' = 4 \sin(2x - 60°)

Or using the identity $\sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)]$

y = 4 \sin x \sin(x - 60°)

y = 4 \cdot \frac{1}{2}[\cos(x - (x - 60°)) - \cos(x + (x - 60°))]

y = 2[\cos(60°) - \cos(2x - 60°)]

y = 2 \cos 60° - 2 \cos(2x - 60°)

y' = -2 \cdot (-\sin(2x - 60°)) \cdot 2

y' = 4 \sin(2x - 60°)

The maximum value occurs when $y' = 0$ or when $\sin(2x - 60°)$ reaches its maximum value, which is $\sin(2x - 60°) = 1$ .

However, since $y = 2 \cos 60° - 2 \cos(2x - 60°) = 1 - 2 \cos(2x - 60°)$ , the maximum value occurs when $\cos(2x - 60°)$ is minimum, which is $\cos(2x - 60°) = -1$ .

\cos(2x - 60°) = -1

2x - 60° = 180° + k \cdot 360°

2x = 240° + k \cdot 360°

x = 120° + k \cdot 180°

With $k$ an integer.

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Number 36

Explanation