# Solve: $\cos(\theta + 25^\circ) + \sin(\theta +65^\circ) = 1$

This afternoon I've been studying the pythagorean identities & compound angles. I've got a problem with a question working with 2 sets of compound angles:

Solve, in the interval $0^\circ \leq \theta \leq 360^\circ$, $$\cos(\theta + 25^\circ) + \sin(\theta +65^\circ) = 1$$

I've attempted expanding but reach a point with no common factors & see how to manipulate the trig ratios to move on; is there a solution without expanding?

$$\cos\theta\cos 25^\circ-\sin\theta \sin 25^\circ+\sin\theta\cos 65^\circ+\sin 65^\circ\cos\theta=1$$

$$\cos \theta\;\left(\cos 25^\circ+\sin 65^\circ\right) +\sin\theta\;\left(\cos 65^\circ-\sin 25^\circ\right)=1$$

Could you tell me if I've made a mistake or how I could continue; thanks

coffee is wearing out

• 65 and 25 are difficult angles to work with. Maybe there is some number in between that is nicer. You can carry that difference all the way to the end. Commented Mar 31, 2016 at 0:07
• One can change a sine to a cos or vice-versa. For example $\sin(\theta+65)=\cos(90-(\theta+65))=\cos(25-\theta)$. Commented Mar 31, 2016 at 0:08
• $\cos(65)=\sin(25)$ so the 2nd and 3rd terms just cancel out. Commented Mar 31, 2016 at 0:15
• Yeah, changing into a single trig ratio worked; I reached $2\cos(\theta)\cos(25)=1$, then isolating $\cos(\theta)$ gives $\theta$ = 56.5. Thank you Commented Mar 31, 2016 at 0:17

"Solve, in the interval 0≤θ≤360, cos(θ+25)+sin(θ+65)=1"

$\cos(\theta+25)+\cos(25-\theta)=1$

∵ $\sin(\theta+65)=\cos(90-(\theta+65))=\cos(25-\theta)$

$\cos(\theta)\cos(25)- \sin(\theta)\sin(25)+\cos(25)\cos(\theta)+\sin(\theta)\sin(25)=1$

$2\cos(\theta)\cos(25)=1$

$\cos(\theta)=\frac{1}{2\cos(25)}$

∴$\theta≈55.6°, 326.5°$

$\cos(\theta + 25) + \sin(\theta+65) = 1$

$\cos(\theta + 25) + \cos(25-\theta) = 1$

$\cos(\theta + 25) + \cos(\theta-25) = 1$

$2\cos\theta \cos(25) = 1$

$\cos \theta = (1/2) \sec 25$

$\theta$ ~ 57 degrees, 303 degrees

Since you've reduced it down to $X\sin(\theta)+Y\cos(\theta)=Z$, the solution may be found with some trigonometric identities:

$$\sin(\theta+\alpha)=\sin(\theta)\cos(\alpha)+\cos(\theta)\sin(\alpha)$$

$$\frac{\sin(\theta+\alpha)}{\cos(\alpha)}=\sin(\theta)+\tan(\alpha)\cos(\theta)$$

Returning to the original problem,

$$X\sin(\theta)+Y\cos(\theta)=Z$$

$$\sin(\theta)+\frac{Y}X\cos(\theta)=\frac{Z}X$$

Looking at this, we want $\frac{Y}X=\tan(\alpha)$, which would allow this to reduce down to a single trig function that we can take the inverse of.