Precalculus problem

Question

Given that sin(x) = 4/5 with x in quadrant 2, and cos(y) =-sqrt5/5 with y in quadrant 3, find the exact value of: sin(x-y) and cos(x-y)

Answer 1

x = 53.13 degrees in quadrant 1. in quadrant 2 that's 180-53.13 = 126.87 degrees

y = -45 degrees in quadrant 4. in quadrant 3 it's 180+45 = 225 degrees

x-y= 126.87-225 = -98.13 or 126.87 +45 = 171.87 degrees

sin(171.87) = 0.141491587...or

sin(x-y) =sin(-98.13) = -0.989949746

cos(x-y) = -0.141419587 or

cos(171.87) = -989949746

cosine and sine are about -1 and sqr2, or vice versa

but you want the "exact" and that's just rounded to about 10 decimal places

y is the base of a right triangle/hypotenuse, with sides 5, sqr5, sqr5, y=exactly 45 degrees or pi/4 radians

x is the opposite side/hypotenuse of a 3,4,5 triangle sinx=4/5 and x = sin^-1(4/5)

sin(x-y) = sinxcosy+cosxsiny and

cos(x-y)= cosxcosy+sinxsiny

sin(x-y) = (4/5)(-sqr5/5)-(-sqr5/5)(-3/5) = (-1/5)(sqr5/5) = -0.08944

cos(x-y)=(3/5)(-sqr5/5)+(4/5)(-sqr5/5) = (-7/5)(sqr5/5) = -0.626099...

the first method with "inexact" answers should be a check on whatever you come up with trying to get "exact" answers. They should match much better.

while the quadrant narrows down the possible values of x and y, it opens up ambiguities as to the infinite angles that could still match the cosine and sine values, Depending on which you choose, you get different answers.

Answer 2

First: draw a diagram with your two angles: x in second quadrant and y in third quadrant. Next draw a right triangle in each two quadrants using given information. Use Pythagorean theorem to find missing sides of the two triangles you formed.

Based on given information you can find that:

sin x=4/5 so cos x=-3/5

and

cos y=-√5 /5 and cos y=-2√5 /5

Next you use sum and difference formulas to find sin(x-y) and cos(x-y)

sin(x-y)=sin x cos y - sin y cos x

You just have to plug in and do all math.

Do the same for cos(x-y)= cos x cos y+ sin x sin y.

Hope it helps.