Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there any way to solve the following sum of trigonometric functions for theta without using a solver? $$25\sin(\theta)-1.5\cos(\theta)=20$$

share|cite|improve this question
up vote 4 down vote accepted

$$ 25\sin\theta-1.5\cos\theta = \sqrt{25^2+1.5^2}\left( \frac{25}{\sqrt{25^2+1.5^2}}\sin\theta - \frac{1.5}{\sqrt{25^2+1.5^2}}\cos\theta \right) $$ $$ = \sqrt{25^2+1.5^2}(\cos\varphi\sin\theta-\sin\varphi\cos\theta) = \sqrt{25^2+1.5^2} \sin(\varphi-\theta). $$ So you want $$ \sin(\varphi-\theta)=\frac{20}{\sqrt{25^2+1.5^2}}. $$ Take arcsines.

share|cite|improve this answer
+1. I just fixed the plus in the first set of parentheses. – Ayman Hourieh Sep 9 '12 at 10:21

If you're okay with a numeric solution, define a function $$ f\left(\theta\right) = 25 \sin \theta - 1.5 \cos \theta - 20 = 0 $$ and use a root finding algorithm.

share|cite|improve this answer
I think OP was looking for a solution without using a solver – Daryl Sep 9 '12 at 2:41
Eric, but the closed form is readily available. Besides, it shows the multitude of solutions. This answer might attract down-votes, I am afraid. – Sasha Sep 9 '12 at 2:55
Then the OP should've asked for an analytic solution. To me, a "solver" is something like Mathematica. Besides, I qualified my answer with the phrase "if you're okay with a numeric solution." Finally, it's pretty clear that $f\left(\theta^*+2 \pi n\right) = 0$, where $\theta^*$ is a solution and $n$ is an integer. – Eric Angle Sep 9 '12 at 14:15

The solution to $A\sin(t) + B\cos(t) = C$ is $t = 2\left(\pi n + \arctan{(\frac{A \pm \sqrt{A^2+B^2-C^2}}{B+C})}\right)$

share|cite|improve this answer
Nice. But it'd be more helpful for OP if you show some derivation steps, or a name of a reference. – user2468 Sep 9 '12 at 3:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.