# Find the exact values.

• Jun 13th 2013, 08:28 AM
FatimaA
Find the exact values.
Let sin x = -3/5, where 3π/2x , and cos y = -5/13, where π ≤ x ≤ 3π/2. Find the exact value of

a) cos2x

b) cos(y/2)

c) cos(x + y)

d) sin (x - y)

My work:

a) cos2x = 1 - 2sin^2x

for sin x = -3/5 -----> 1-2(-3/5)^2 = 7/25

for cosy = -5/13 -----> 1-2(-12/13)^2 = -119/169

b) cos(y/2) =
√(1+ cosy)/2

for cosy = -5/3 -----> √(1+(-5/13))/2 = 2/√13

for siny = -3/5 -----> √(1+(4/5))/2 = 3/(2√5)

c) cos(x + y) = (cosx)(cosy) - (sinx)(siny)

(4/5)(-5/13) - (-3/5)(-12/13)

(-20/65) - (36/65) = -56/65

d) sin(x - y) = (sinx)(cosy) + (cosx)(siny)

(-3/5)(-5/13) + (4/5)(-12/13)

= (-15/65) + (-48/65) = -63/65

• Jun 13th 2013, 11:20 AM
ebaines
Re: Find the exact values.
Quote:

Originally Posted by FatimaA
My work:

a) cos2x = 1 - 2sin^2x

for sin x = -3/5 -----> 1-2(-3/5)^2 = 7/25

Right

Quote:

Originally Posted by FatimaA
for cosy = -5/13 -----> 1-2(-12/13)^2 = -119/169

You have found the value for cos(2y), but that wasn't asked for.

Quote:

Originally Posted by FatimaA
b) cos(y/2) = √(1+ cosy)/2

for cosy = -5/3 -----> √(1+(-5/13))/2 = 2/√13

Good.

Quote:

Originally Posted by FatimaA
for siny = -3/5 -----> √(1+(4/5))/2 = 3/(2√5)

Looks like you're trying to find the value of cos(x/2), which wasn't asked for. If so, the 2 belongs inside the square root sign, so it should be 3/sqrt(10)

Quote:

Originally Posted by FatimaA
c) cos(x + y) = (cosx)(cosy) - (sinx)(siny)

(4/5)(-5/13) - (-3/5)(-12/13)

(-20/65) - (36/65) = -56/65

You have siny = -12/13, but are you sure it isn't actually +12/13? Your post is a bit garbled regarding what quadrant y is in, so I can't really tell whether this is correct or not.

Quote:

Originally Posted by FatimaA
d) sin(x - y) = (sinx)(cosy) + (cosx)(siny)

No, sin(x-y) = sinx cosy - cos x siny
• Jun 13th 2013, 04:08 PM
HallsofIvy
Re: Find the exact values.
You understand, do you not, that your "$\displaystyle 3\pi/2= x= 2\pi$" and "$\displaystyle \pi= y= 3\pi/2$" are nonsense? x cannot be equal to two different numbers (and neither $\displaystyle 3\pi/2$ nor $\displaystyle 2\pi$ has sine equal to -3/5). I think you meant $\displaystyle 3\pi/2\le x\le 2\pi$ and $\displaystyle \pi\le y\le 3\pi/2$. That is that x is in the fourth quadrant and y is in the third quadrant.