Problem 47 - Precalculus: A Concise Course

Question

Use the value of the trigonometric function to evaluate the indicated functions. \(\sin t=\frac{4}{5}\) (a) \(\sin (\pi-t)\) (b) \(\sin (t+\pi)\)

Step-by-Step Solution

Answer

(a) The value of \( \sin(\pi-t) \) is \( \frac{4}{5} \). (b) The value of \( \sin(t+\pi) \) is \( -\frac{4}{5} \).

Step 1: Identify the identities

For \( \sin t = \frac{4}{5} \), we know that sine is positive in the first and second quadrants. With the given value we cannot directly find which quadrant we are in. Therefore, instead we will use the co-function identity for sine, which is \( \sin(\pi - t) = \sin t \) and the periodic property, \( \sin(t + \pi) = -\sin t \). This does not depend on which quadrant we are in.

Step 2: Evaluate \(\sin(\pi-t)\)

Substitute \( t \) in the equation \( \sin(\pi-t) = \sin t \). Because we know that \( \sin t = \frac{4}{5} \), we replace \( \sin t \) with \( \frac{4}{5} \). Therefore, \( \sin(\pi-t) = \frac{4}{5} \).

Step 3: Evaluate \(\sin(t+\pi)\)

For the second part, substitute \( t \) in the equation \( \sin(t+\pi) = -\sin t \). Because we know that \( \sin t = \frac{4}{5} \), we replace \( \sin t \) with \( \frac{4}{5} \). Therefore, \( \sin(t+\pi) = - \frac{4}{5} \).