1. ## Using trig identities

Hi, I've attempted and completed this problem and all of it's parts, but I want to know if I've followed all the right steps and if my answers are correct. I've put the problems in blue and all of my work and my answers in red. Please tell me if this is not clear. I did my best to make it as clear as possible. Thank you in advance for your help.

1. Suppose that cosθ = −1/3, where π <θ < 3π/2 , and sinα = −4/5, where 3π/2 <α < 2π

Before starting the problems, I drew pictures of the angles on a graph and determined the following:

For θ:

cosθ = -1/3, sinθ = (-2√2)/3

For α:

cosα = 3/5, sinα = -4/5

The next part of the problem is to find and simplify the following.

a) tanθ

tanθ = ((-2√2)/3)/(-1/3) = ((-6√2)/-3) = 2√2

b) cos θ/2

Using the half angle formula for cos θ/2 which is +/- (√(1+cosθ)/2), I did this:

-(√(1+(-1/3))/2) = -(√(2/3)/2) = -√(1/3) = -1/√3

c) cos2α

I used the formula for lowering powers for cos2α which is 1-sin^2α and then tried to solve it:

1-2(-4/5)^2 = -7/25

d) cos(θ−α)

I used the addition angle identity for cosine:

cos(θ−α) = (cosθ)(cosα) + (sinθ)(sinα)

= (-1/3)(3/5)+((-2√2)/3)(-4/5)

= (-3/15)+((8√2)/15)

= (5√2)/15

e) sinα/2

I used the half angle formula:

sinα/2 = +/- (√(1-cosα)/2)

= (√(1-(3/5))/2)

= (√(2/5)/2)

= √(1/5)

= 1/(√5)

f) sin(θ+α)

I used the additional angle identity for sine:

sin(θ+α) = (sinθ)(cosα)+(cosθ)(sinα)

= ((-2√2)/3)(3/5) + (-1/3)(-4/5)

= ((-6√2)/15) + (4/15)

= -(2√2)/15

2. ## Re: Using trig identities

Originally Posted by FatimaA
Hi, I've attempted and completed this problem and all of it's parts, but I want to know if I've followed all the right steps and if my answers are correct. I've put the problems in blue and all of my work and my answers in red. Please tell me if this is not clear. I did my best to make it as clear as possible. Thank you in advance for your help.

1. Suppose that cosθ = −1/3, where π <θ < 3π/2 , and sinα = −4/5, where 3π/2 <α < 2π

Before starting the problems, I drew pictures of the angles on a graph and determined the following:

For θ:

cosθ = -1/3, sinθ = (-2√2)/3

Yes, since $\displaystyle \theta$ is in the third quadrant both sine and cosine are negative and, of course, [tex](-1/3)^2+ (-2\sqrt{2}/3)^2= 1/9+ 8/9= 1.

For α:

cosα = 3/5, sinα = -4/5
Yes, $\displaystyle \alpha$ is in the fourth quadrant so cosine is positive and sine is negative. Also $\displaystyle (3/5)^2+ (-4/5)^2= 9/25+ 16/25= 1$.

The next part of the problem is to find and simplify the following.

a) tanθ

tanθ = ((-2√2)/3)/(-1/3) = ((-6√2)/-3) = 2√2
Yes, that is correct.

b) cos θ/2

Using the half angle formula for cos θ/2 which is +/- (√(1+cosθ)/2), I did this:

-(√(1+(-1/3))/2) = -(√(2/3)/2) = -√(1/3) = -1/√3
Okay, since $\displaystyle \pi\le \theta\le 3\pi/2$, $\displaystyle \pi/2\le \theta\le 3\pi/4$, in the second quadrant, cosine is negative and you use "-".

c) cos2α

I used the formula for lowering powers for cos2α which is 1-sin^2α and then tried to solve it:

You mean $\displaystyle cos(2\alpha)= 1- 2 cos^2(\alpha)$
I would, rather, use [tex]cos(2\alpha)= cos^2(\alpha)- sin^2(\alpha)= (3/5)^2- (-4/5)^2= (9- 16)/25= -7/25

1-2(-4/5)^2 = -7/25

d) cos(θ−α)

I used the addition angle identity for cosine:

cos(θ−α) = (cosθ)(cosα) + (sinθ)(sinα)

= (-1/3)(3/5)+((-2√2)/3)(-4/5)

= (-3/15)+((8√2)/15)

= (5√2)/15
Yes, that is correct.

e) sinα/2

I used the half angle formula:

sinα/2 = +/- (√(1-cosα)/2)

= (√(1-(3/5))/2)

= (√(2/5)/2)

= √(1/5)

= 1/(√5)

f) sin(θ+α)

I used the additional angle identity for sine:

sin(θ+α) = (sinθ)(cosα)+(cosθ)(sinα)

= ((-2√2)/3)(3/5) + (-1/3)(-4/5)

= ((-6√2)/15) + (4/15)

= -(2√2)/15

3. ## Re: Using trig identities

Thanks for taking the time to look at this. Are the last two, e and f, correct?