Given [tex]\cos\alpha=\frac{8}{17}[/tex], [tex]\alpha[/tex] is in Quadrant IV, [tex]\sin\beta=-\frac{24}{25}[/tex], and [tex]\beta[/tex] is in Quadrant III, find [tex]\sin(\alpha-\beta)[/tex]
We can use the angle subtraction formula of sine to answer this question.
[tex]\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta[/tex]
We already know that [tex]\cos\alpha=\frac{8}{17}[/tex].
We can use the Pythagorean identity [tex]\sin^2\theta+\cos^2\theta=1[/tex] to find [tex]\sin\alpha[/tex].
[tex]\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}[/tex]
Since [tex]\alpha[/tex] is in Quadrant IV, and sine is represented as y value on the unit circle, we must assume the negative value [tex]\sin\alpha=-\frac{15}{17}[/tex].
As similar process is then done with [tex]\sin\beta=-\frac{24}{25}[/tex].
[tex](-\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}[/tex]
And since [tex]\beta[/tex] is in Quadrant III, and cosine in represented as x value on the unit cercle, we must assume the negative value [tex]\cos\beta=-\frac{7}{25}[/tex].
Now we can fill in our angle subtraction formula!
[tex]\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}}[/tex]
