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