Rewrite the following as a product of trigonometric functions.
-Find $\sin ( \mathrm { s } - \mathrm { t } )$ given that $\cos \mathrm { s } = \frac { 1 } { 3 }$, with $\mathrm { s }$ in quadrant $\mathrm { I }$, and $\sin \mathrm { t } = \frac { 1 } { 4 }$, with $\mathrm { t }$ in quadrant II.
A) $\frac { - 12 \sqrt { 30 } - 1 } { 2 }$
B) $\frac { - 30 \sqrt { 2 } - 1 } { 12 }$
C) $\frac { - 2 \sqrt { 30 } - 1 } { 12 }$
D) $\frac { - 12 \sqrt { 2 } - 1 } { 30 }$

Question

Quizplus · Accepted Answer

To find $\sin(s - t)$, we use the identity $\sin(s - t) = \sin s \cos t - \cos s \sin t$. Given $\cos s = \frac{1}{3}$ and $\sin t = \frac{1}{4}$, we first find $\sin s$ and $\cos t$. Since $s$ is in quadrant I, where sine is positive, and $t$ is in quadrant II, where cosine is negative, we use the Pythagorean identities $\sin^2 s + \cos^2 s = 1$ and $\sin^2 t + \cos^2 t = 1$ to find $\sin s = \sqrt{1 - \cos^2 s} = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$ and $\cos t = -\sqrt{1 - \sin^2 t} = -\sqrt{1 - \left(\frac{1}{4}\right)^2} = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4}$. Then, $\sin(s - t) = \left(\frac{2\sqrt{2}}{3}\right)\left(-\frac{\sqrt{15}}{4}\right) - \left(\frac{1}{3}\right)\left(\frac{1}{4}\right) = -\frac{\sqrt{30}}{6} - \frac{1}{12} = \frac{-2\sqrt{30} - 1}{12}$.

Rewrite the Following as a Product of Trigonometric Functions sin⁡(s−t)\sin ( \mathrm { s } - \mathrm { t } )sin(s−t)

Rewrite the Following as a Product of Trigonometric Functions $\sin ( \mathrm { s } - \mathrm { t } )$