Solving Trigonometric Equations: TanΘ=4/3

[CrossFit415]

Homework Statement

TanΘ = 4/3

pi < Θ < 3pi/2

Find...
a.) Sin(2Θ)

b.) cos(2Θ)

c.) Sin Θ/2

d.) Cos Θ/s

The Attempt at a Solution

So...

The line will be in quadrant III.

Would sin = 4 and cos = 3?
What I used for a.) is ...
a.) Sin(2Θ) = 2sinΘcosΘ
= 2(4)(3)
=24
But the book got Sin(2Θ) = 27/25

What formula should I use? Thanks

 

[eumyang]

Homework Statement

TanΘ = 4/3

pi < Θ < 3pi/2

Find...
a.) Sin(2Θ)

b.) cos(2Θ)

c.) Sin Θ/2

d.) Cos Θ/s

The Attempt at a Solution

So...

The line will be in quadrant III.

Would sin = 4 and cos = 3?

No. While tan θ = y/x and it is positive in Q III, don't forget that x and y both have to be negative. So x = -3 and y = -4.

Also, sin θ = y/r and cos θ = x/r. You forgot to find r.

What I used for a.) is ...
a.) Sin(2Θ) = 2sinΘcosΘ
= 2(4)(3)
=24
But the book got Sin(2Θ) = 27/25

You sure that's the book's answer? Because I am getting 24/25.

 

[CrossFit415]

Ahh I meant 24/25, you're right. Thank you. I forgot to add the "r" for sin and cos. But then how would I find r ?.. I know that sin θ = 4/r and cos θ= 3/r ?

 

[Mark44]

sin θ = -[/color]4/r and cos θ= -[/color]3/r

Use the identity sin²(θ) + cos²(θ) = 1 to solve for r.

 

[Mark44]

a.) Sin(2Θ)

The line will be in quadrant III.

Would sin = 4 and cos = 3?

Neither sin(θ) nor cos(θ) can ever be larger than 1 or smaller than -1.

What I used for a.) is ...
a.) Sin(2Θ) = 2sinΘcosΘ
= 2(4)(3)
=24

This answer is too large, for the reason given above.

But the book got Sin(2Θ) = 27/25

What formula should I use? Thanks

 

[CrossFit415]

sin θ = -[/color]4/r and cos θ= -[/color]3/r

Use the identity sin²(θ) + cos²(θ) = 1 to solve for r.

So that would equal to 25 = 1 ?

 

[Mark44]

So that would equal to 25 = 1 ?

I don't understand what you're asking. Things aren't equal to an equation, and what you show as an equation is not true (i.e. 25 \neq 1).

 

[CrossFit415]

But then how would I use that identity to find "r"? Thanks

 

[Mark44]

Substitute -4/r for sinθ and -3/4 for cosθ to get an equation that involves r, then solve for r.

 

[eumyang]

So that would equal to 25 = 1 ?

But then how would I use that identity to find "r"? Thanks

Substitute -4/r for sinθ and -3/r for cosθ to get an equation that involves r, then solve for r.

Did you do what Mark44 said?

\cos^2 \theta + \sin^2 \theta = 1
\left( \frac{-3}{r} \right)^2 + \left( \frac{-4}{r} \right)^2 = 1

Solve the above equation for r.

 

[CrossFit415]

Did you do what Mark44 said?

\cos^2 \theta + \sin^2 \theta = 1
\left( \frac{-3}{r} \right)^2 + \left( \frac{-4}{r} \right)^2 = 1

Solve the above equation for r.

Yes, I got 25/r

 

[vela]

That's not correct. The r should be squared, and that expression should be equal to something. You can't start with an equation and end up with just an expression. When you "solve for r," you should end up with the equation r=something.

 

[CrossFit415]

That's not correct. The r should be squared, and that expression should be equal to something. You can't start with an equation and end up with just an expression. When you "solve for r," you should end up with the equation r=something.

r2 = 25
r = sqrt 25
r = 5?

I'm even getting my algebra incorrect lol.

 

[Mark44]

Yes. So what you have now is sin(θ) = -4/5 and cos(θ) = -3/5. Now you can evaluate sin(2θ) = 2sinθ cosθ.

Do something similar to evaluate cos(2θ).

For sin(θ/2) and cos(θ/2) you're going to need to use the half-angle identities.

 

[CrossFit415]

Ahh I see now. Thank you my friends.