Question

If x a third quadrant angle and y an acute angle such that cos x=-(3/5) and sec y= (3/2) ,find the exact value of sin(x-y) , exact value of tan(x-y) and quadrant containing (x-y)?

Answer 1

Find sin (x - y), knowing cos x = -3/5 and sec y = 3/2

Explanation:

Apply the trig identity:
sin (x - y) = sin x.cos y - sin y.cos x
`cos x = -3/5` --> `sin^2 x = = 1 - 9/25 = 16 /25` --> `sin x = -4/5`
`cos y = 2/3` -->`sin^2 y = 1 - 4/9` --> `sin^2 y = 5/9`--> `sin y = sqrt5/3`

`sin (x - y) = (-4/5)(2/3) - (sqrt5/3)(-3/5) = -8/15 + 3sqrt5/15 = (3sqrt5 - 8)/15`
Apply trig identity:
`cos (x - y) = cos x.cos y + sin x.sin y =`
`= (-3/5)(2/3) + (-4/5)(sqrt5)/3) = - 6/15 - (4sqrt5)/15 =`
`= -(4sqrt5 + 6)/15`
`tan (x - y) = (sin)/(cos) = -(3sqrt5 - 8)/(4sqrt5 + 6)`
The arc (x - y) is located in Quadrant III since its cos and its sin are both negative.
Check by calculator:
`cos x = - 3/5` --> x = 360 - 126.86 = 233.14 deg (Quadrant III) ; `cos y = 2/3` --> y = 48.19 deg
Arc (x - y) = 233.14 - 48.19 = 184.95 deg --> sin (x - y) = -0.09
`(3sqrt5 - 8)/15 = -0.09` OK