# Trig - Addition Formula Question

1. Dec 4, 2011

### studentxlol

1. The problem statement, all variables and given/known data

Solve the equation cos(x-60)=sinx

2. Relevant equations

cosAcosB+sinAsinB

3. The attempt at a solution

cos(x-60)=sinx
cosxcos60+sinxsin60=sinx
1/2cosx+(√3)/2sinx=sinx

How do I then solve to find x for0<x<360

2. Dec 4, 2011

### SammyS

Staff Emeritus
Subtract sin(x) from both sides.

Divide by cos(x).

3. Dec 4, 2011

### Curious3141

Suggestion: don't even do it this way.

You know that sin x = cos (90-x)

So re-express RHS like that:

cos (x-60) = cos (90-x)

x-60 = 90-x + 360n (where n is an integer)

2x = 150 + 360n

x = 75 + 180n

So x = 75 or 255 for n = 0 and 1 respectively. Those are the only two solutions in the required range.

4. Dec 4, 2011

### SammyS

Staff Emeritus
Did you try what I suggested ?

What did you get ?

(1/2)cos(x)+((√3)/2-1)sin(x)=0

Now, divide by cos(x) .

$\displaystyle \frac{\sin(x)}{\cos(x)}=\tan(x)$ ---- Right?