# Thread: Trig Identity/Pre-Calc help

1. ## Trig Identity/Pre-Calc help

I have a problem as follows:

cos((π/3) - X)

I then used the addition identity of cos(x-y) = cos x cos y + sin x sin y and got:

cos(π/3)*cosX + sin(π/3)*sinX

Then i used the unit circle to get

(1/2)*cosX + √(3)*sinX

And am now stuck. I'm supposed to solve for 'x' but i'm not sure how to continue.
Help please?
-I'm supposed to use trig identities, of course.

I posted this in both urgent hw help and pre-calc because its both urgent and in pre-calc . I'll remove one as soon as either get a reply.

2. Dear Dust,

what is your problem?? You wrote a simple expression not an equation or else.

Bye the way, once again how many sin(pi/3) ?

3. I have to solve for x. Thats the problem - I don't know how to further simplify to the point where i can solve X.

And .. just one pi/three? I don't understand your question; sorry.

Its just cos of (pi/3) minus the variable x

4. Originally Posted by Dust

cos(π/3)*cosX + sin(π/3)*sinX

(1/2)*cosX + √(3)*sinX
There is an error above

it should be
$\frac{1}{2}cos(x) + \frac{\sqrt{3}}{2}Sin(x) = \frac{cos(x) +\sqrt{3}Sin(x)}{2}$

maybe that's why you got stuck

5. solve, solve, but what??
You wrote an expression: cos(pi/3 - x) without sign '='.
Man can usually solve an equation, e. g. 2x = 10
but only 2x is meaningless to solve.

My question: sin(pi/3) = ?

6. Skalkaz - the point of this problem is to solve for x. so the problem equals whatever x is. But you don't know it till you solve it so it doesn't really matter.

Thank you, Fonso.

7. How do you continue solving this problem?

8. No problem.

9. Originally Posted by Dust
Skalkaz - the point of this problem is to solve for x. so the problem equals whatever x is. But you don't know it till you solve it so it doesn't really matter.

Thank you, Fonso.
Skalkaz is exactly right, you can't solve an equation for x without an equation.

You have given the left hand side

$\cos{(\frac{\pi}{3} - x)}$

Now you need an equals sign and a right hand side.

Then we can solve.