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

  • $\begingroup$ 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. $\endgroup$ – Doug M Mar 31 '16 at 0:07
  • $\begingroup$ One can change a sine to a cos or vice-versa. For example $\sin(\theta+65)=\cos(90-(\theta+65))=\cos(25-\theta)$. $\endgroup$ – André Nicolas Mar 31 '16 at 0:08
  • $\begingroup$ $\cos(65)=\sin(25)$ so the 2nd and 3rd terms just cancel out. $\endgroup$ – Nikunj Mar 31 '16 at 0:15
  • $\begingroup$ 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 $\endgroup$ – Jalejo Mar 31 '16 at 0:17

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


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

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



∴$\theta≈55.6°, 326.5°$


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

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

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

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

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

$\theta$ ~ 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:



Returning to the original problem,



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.


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