Solving a trigonometric equation for the angle

[QuarkDecay]

Homework Statement

The equation;
0.966= -0.354sin(φ+60) + 0.935cos(φ+60) and we're trying to find φ.

Homework Equations

3. The Attempt at a Solution [/B]
(edited)
I tried doing ^2;
0.933= 0.125sin²(φ+60) -2*0.331sin(φ+60)cos(φ+60) + 0.874cos²(φ+60)

x=φ+60

0.933= 0.125sin²x - 0.331sin(2x) + 0.874cos²x⇒
0.933= 0.125(1-cos²x) - 0.331sin(2x) + 0.874cos²x ⇒
0.933= 0.125 - 0.331sin(2x) + 0.749cos²x

 

[mjc123]

So (a+b)² = a² + b²?

 

[QuarkDecay]

Yes you're right. I was just about to edit that

 

[Ray Vickson]

Homework Statement

The equation;
0.966= -0.354sin(φ+60) + 0.935cos(φ+60) and we're trying to find φ.

Homework Equations

3. The Attempt at a Solution [/B]
(edited)
I tried doing ^2;
0.933= 0.125sin²(φ+60) -2*0.331sin(φ+60)cos(φ+60) + 0.874cos²(φ+60)

x=φ+60

0.933= 0.125sin²x - 0.331sin(2x) + 0.874cos²x⇒
0.933= 0.125(1-cos²x) - 0.331sin(2x) + 0.874cos²x ⇒
0.933= 0.125 - 0.331sin(2x) + 0.749cos²x

There is a standard method for converting an expression of the form $a \cos(A) + b \sin(A)$ into the form $c \sin(A+r_1)$ or $d \cos(A+r_2)$, where $c, d, r_1, r_2$ are constants that can be computed in terms of $a, b.$ That is something that every engineer and physicist should know. The method uses the trigonometric identities $\cos(A+r) = \cos(r) \cos(A) - \sin(r) \sin(A)$ and $\sin(A+r) = \sin(r) \cos(A) + \cos(r) \sin(A).$

See, eg., https://www.myphysicslab.com/springs/trig-identity-en.html