If \theta = \arctan\left(-\frac{4}{3}\right), what is the exact value of \sin(2\theta)? — ACT

If $\theta = \arctan\left(-\frac{4}{3}\right)$ , what is the exact value of $\sin(2\theta)$ ?

The exact value of

\sin(2\theta)

-\frac{24}{25}

The angle

\theta = \arctan\left(-\frac{4}{3}\right)

lies in Quadrant IV (

-\frac{\pi}{2} < \theta < 0

) because the range of

\arctan(x)

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

. Using a right triangle in Quadrant IV with opposite side

-4

and adjacent side

3

, the hypotenuse is

5

. Hence,

\sin(\theta) = -\frac{4}{5}

and

\cos(\theta) = \frac{3}{5}

. Using the double-angle identity

\sin(2\theta) = 2\sin(\theta)\cos(\theta)

, we get

2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{24}{25}

Determine the quadrant and trigonometric ratios for

\theta

Since

\theta = \arctan\left(-\frac{4}{3}\right)

, the angle

\theta

is in Quadrant IV where

-\frac{\pi}{2} < \theta < 0

. In this quadrant, the opposite side is

-4

, the adjacent side is

3

, and the hypotenuse is

\sqrt{3^2 + (-4)^2} = 5

. Therefore,

\sin(\theta) = -\frac{4}{5}

and

\cos(\theta) = \frac{3}{5}

The range of the principal arctangent function is

\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

, so a negative input places the angle in Quadrant IV.

Apply the double-angle formula for sine

\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2\left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{24}{25}

The double-angle identity for sine expresses

\sin(2\theta)

in terms of

\sin(\theta)

and

\cos(\theta)

Evaluating trigonometric functions of double angles involving inverse trigonometric functions

Estimated Time:1m 15s

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