Question

Given cos alpha = 8/17, alpha in quadrant IV, and sin beta = -24/25, beta in quadrant III, find sin(alpha-beta)

Answer 1

Given $\cos\alpha=\frac{8}{17}$, $\alpha$ is in Quadrant IV,  $\sin\beta=-\frac{24}{25}$, and $\beta$ is in Quadrant III, find $\sin(\alpha-\beta)$

We can use the angle subtraction formula of sine to answer this question.

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

We already know that $\cos\alpha=\frac{8}{17}$.

We can use the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ to find $\sin\alpha$.

$\sin^2\alpha+(\frac{8}{17})^2=1 \\ \sin^2\alpha+\frac{64}{289}=1 \\ \sin^2\alpha=\frac{225}{289} \\ \\\sin\alpha=\pm\frac{15}{17}$

Since $\alpha$ is in Quadrant IV, and sine is represented as y value on the unit circle, we must assume the negative value $\sin\alpha=-\frac{15}{17}$.

As similar process is then done with  $\sin\beta=-\frac{24}{25}$.

$(-\frac{24}{25})^2+\cos^2\beta=1 \\ \frac{576}{625}+\cos^2\beta=1 \\ \cos^2\beta=\frac{49}{625} \\ \\\cos\beta=\pm\frac{7}{25}$

And since $\beta$ is in Quadrant III, and cosine in represented as x value on the unit cercle, we must assume the negative value $\cos\beta=-\frac{7}{25}$.

Now we can fill in our angle subtraction formula!

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta \\\\ \sin(\alpha-\beta)=(-\frac{15}{17}\times-\frac{7}{25})-(\frac{8}{17}\times-\frac{24}{25}) \\\\\sin(\alpha-\beta)=\frac{105}{425}-(-\frac{192}{425}) \\\\ \boxed{\sin(\alpha-\beta)=\frac{297}{425}}$