# Is there a trig identity to help solve this equation? $A \cdot \cos\theta = B+ \sin\theta$

I'm trying to solve for $\theta$ in a simple equation:

$A \cdot \cos(\theta) = B+ \sin(\theta)$

($A$ and $B$ are constants)

But all the trig identities I've tried just make the equation worse.

What am I missing? How do you approach this problem?

You have $$A\cos\theta-\sin\theta=B$$

Now write the left hand side as $$R\cos(\theta+\alpha)=R\cos\theta\cos\alpha-R\sin\theta\sin\alpha$$

Therefore $$R\cos\alpha=A$$ and $$R\sin\alpha=1$$

Hence $$R=\sqrt{A^2+1}$$ and $$\tan\alpha=\frac 1A$$

Can you solve it now?

• This answer was a bit harder for me to follow, but in the end, proved to be the most useful. Thank you! – abelenky May 3 '16 at 16:44

hint: Square both sides: $A^2(1-\sin^2\theta) =A^2\cos^2 \theta = B^2 + 2B\sin \theta + \sin^2 \theta$, and you simplify this to get a quadratic equation in $\sin \theta$. Can you finish it ?

• You should avoid squaring both sides as this will create false solutions – David Quinn Apr 29 '16 at 19:35

$$\frac{(A \cdot \cos\theta-1 \cdot \sin\theta)}{\sqrt{A^2+1}}=\frac{B}{\sqrt{A^2+1}}$$

with proper substitution as indicated

$$\sin \beta \cos \theta - \cos \beta \sin \theta = c <1$$

$$sin( \beta - \theta ) = c$$

You can make $\theta$ as subject of equation at LHS.