Trig Identity Problem: Finding Theta When Tan(theta) = Cos(theta)

[TW Cantor]

Homework Statement

the problem is:
tan(theta) = cos(theta)
find theta: -pi < theta < pi

Homework Equations

tan(theta)=sin(theta)/cos(theta)
sin^2(theta)+cos^2(theta)=1

The Attempt at a Solution

 

[Mark44]

Homework Statement

the problem is:
tan(theta) = cos(theta)
find theta: -pi < theta < pi

Homework Equations

tan(theta)=sin(theta)/cos(theta)
sin^2(theta)+cos^2(theta)=1

The Attempt at a Solution

You will need to use both identities you show in the Relevant equations.

 

[TW Cantor]

i have tried using both equations but it just ends up getting more complicated. could you maybe tell me which one to use first? :-)

 

[Mark44]

Show me what you've done and we can go from there.

 

[TW Cantor]

i have basicly tried rearranging the equation to try and eliminate one of the functions from it.
so I've tried going from:
tan(θ) = cos(θ)
to:
sin(θ)/cos(θ) = cos(θ)

sin(θ) = cos²(θ)

sin²(θ) = cos²(θ)*sin(θ)

is this the right thing to do or am i going wrong?

 

[Mark44]

i have basicly tried rearranging the equation to try and eliminate one of the functions from it.
so I've tried going from:
tan(θ) = cos(θ)
to:
sin(θ)/cos(θ) = cos(θ)

sin(θ) = cos²(θ)

Here (above) is where you use the other identity: sin²(θ) + cos²(θ) = 1. You want an equation in terms of sin(θ).

sin²(θ) = cos²(θ)*sin(θ)

is this the right thing to do or am i going wrong?

 

[TW Cantor]

so you get:
sin²(θ) + sin(θ) -1 = 0
and then just solve as a quadratic?

 

[TW Cantor]

so I've got one result: 0.666 radians
but i drew out the cosine graph and the tan graph and they meet at two separate points. how do i calculate the second value?

 

[Mark44]

so you get:
sin²(θ) + sin(θ) -1 = 0
and then just solve as a quadratic?

Yes.

so I've got one result: 0.666 radians
but i drew out the cosine graph and the tan graph and they meet at two separate points. how do i calculate the second value?

When you solved the equation above you should have gotten two values for sin(θ).

 

[TW Cantor]

i did get two values for sin(θ). one was 0.6180339 which gave a value for θ of 0.66624
the other value of sin(θ) was -1.6180339 so when i try sin^-1 it gets a math error

 

[Mark44]

There should be another value in the interval [0, pi] for which sin(θ) = .6180339. Hint: the graph of y = sin(x) is symmetrical about the line x = pi/2.

 

[TW Cantor]

ahhhh yes. didnt think of putting it on a sine graph, i was still using the tan and cosine graphs. i got θ = 0.666, 2.475 :-) thanks a lot for your help :-)

 

[HallsofIvy]

Homework Statement

the problem is:
tan(theta) = cos(theta)
find theta: -pi < theta < pi

Homework Equations

tan(theta)=sin(theta)/cos(theta)
sin^2(theta)+cos^2(theta)=1

The Attempt at a Solution

Something missing here, isn't there? Yes, use tan(theta)= sin(theta)/cos(theta) to write the equation in terms of sine and cosine and then use sin^2(theta)+ cos^2(theta)= 1 to reduce to only cosine (or only sine).