# Thread: Finding cos and sin from tan?

1. ## Finding cos and sin from tan?

Give the expression for one of the three functions sin θ, cos θ, or tan θ, with θ in the first quadrant. Find expressions for the other two functions. Your answer will be algebraic expressions in terms of x.

a) x = 2 cos θ
b) x = 9 tan θ

What are we suppose to do? I don't get what the x in front of the equation and the variables such as "2" and "9" in front of the cos and tan mean. And how would we solve it for the other two trig functions? Please help me with this and show your work. Thank you so much!

2. Originally Posted by florx
Give the expression for one of the three functions sin θ, cos θ, or tan θ, with θ in the first quadrant. Find expressions for the other two functions. Your answer will be algebraic expressions in terms of x.

a) x = 2 cos θ
b) x = 9 tan θ

What are we suppose to do? I don't get what the x in front of the equation and the variables such as "2" and "9" in front of the cos and tan mean. And how would we solve it for the other two trig functions? Please help me with this and show your work. Thank you so much!
You make use of the following identities:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

and

$\sin^2{\theta} + \cos^2{\theta} = 1$.

3. Originally Posted by Prove It
You make use of the following identities:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

and

$\sin^2{\theta} + \cos^2{\theta} = 1$.
I already know how to use them already. It's just that I don't know what to do with the x and about the variable in front of the cos and tan.

4. ## suggestion

Originally Posted by Prove It
You make use of the following identities:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

and

$\sin^2{\theta} + \cos^2{\theta} = 1$.

5. Originally Posted by florx
Give the expression for one of the three functions sin θ, cos θ, or tan θ, with θ in the first quadrant. Find expressions for the other two functions. Your answer will be algebraic expressions in terms of x.

a) x = 2 cos θ
So $cos(\theta)= x/2$. You can think of $\theta$ as an angle in a right triangle with "near side" x and hypotenuse of length 2. By the Pythagorean formula, the "opposite side" has length $\sqrt{4- x^2}$. You can get all trig functions from that.

b) x = 9 tan θ
Now $tan(\theta)= x/9$. A right triangle with opposite side of length x and near side of length 9 has hypotenuse of length $\sqrt{x^2+ 9}$. Again, you can get all trig functions from that.

What are we suppose to do? I don't get what the x in front of the equation and the variables such as "2" and "9" in front of the cos and tan mean.
Seriously? the "2" and the "9" multiply the trig functions. If you don't know that you should not be taking trigonometry!

And how would we solve it for the other two trig functions? Please help me with this and show your work. Thank you so much!