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=(8)/(17)`,
`\alpha` is in Quadrant IV,
`\sin\beta=-(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=(8)/(17)`.

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


`\sin^2\alpha+((8)/(17))^2=1 \\ \sin^2\alpha+(64)/(289)=1 \\ \sin^2\alpha=(225)/(289) \\ \\\sin\alpha=\pm(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=-(15)/(17)`.

As similar process is then done with
`\sin\beta=-(24)/(25)`.


`(-(24)/(25))^2+\cos^2\beta=1 \\ (576)/(625)+\cos^2\beta=1 \\ \cos^2\beta=(49)/(625) \\ \\\cos\beta=\pm(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=-(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)=(-(15)/(17)*-(7)/(25))-((8)/(17)*-(24)/(25)) \\\\\sin(\alpha-\beta)=(105)/(425)-(-(192)/(425)) \\\\ \boxed{\sin(\alpha-\beta)=(297)/(425)}`