# $a \cos(v) + b\sin(v) = A\sin(u+v)$ proof

I'm trying to find an $A$ and $u$ that satisfy: $a\cos(v) + b\sin(v) = A\sin(u+v)$. However, my result gets me $\sqrt{(a^2 + b^2)}\sin\big(v+\tan^{-1}(a/b)\big)$ which is incorrect according to the internet. The angle $u$ should apparently be $\tan^{-1}(b/a)$. What am I doing wrong?

Here's the relevant part of my proof to find the angle u.

$$a\cos(v) + b\sin(v) = A\cos(u)\sin(v) + A\cos(v)\sin(u)$$

$$a = A\sin(u)$$ $$b = A\cos(u)$$

$$\frac a{\sin(u)} = \frac{b}{\cos(u)}$$

$$\frac a b = \frac{\sin(u)}{\cos(u)} = \tan(u)$$

$$\frac{a}{b} = \tan(u)$$ $$\tan^{-1}\left(\frac a b\right) = u$$

Where am I wrong?

• You have a typo: too many "v" and too few "u". Check it out. – Moritz Jul 26 '15 at 14:49
• in $a\cos(u)+b\sin(v)$ are the arguments $u,v$? – Dr. Sonnhard Graubner Jul 26 '15 at 14:51

Multiply and divide, $a\cos(v) + b\sin(v)$, by $\sqrt {a^2+b^2}$, you will get that $$\sqrt {a^2+b^2}\left(\frac {a}{\sqrt {a^2+b^2}}\cos(v)+\frac{b}{\sqrt {a^2+b^2}}\sin(v)\right)$$, Now take, $\sin(u)=\frac {a}{\sqrt {a^2+b^2}},\cos(u)=\frac{b}{\sqrt {a^2+b^2}}$

The above expression will be reduces to

$$\sqrt {a^2+b^2}\{\sin(u)\cos(v)+\cos(u)\sin(v)\}$$

$$=\sqrt {a^2+b^2}\sin(u+v)=A\sin(u+v)$$

where, $\tan(u)=\frac{a}{b},\sqrt {a^2+b^2}=A$

$$a\sin x +b\cos x=\\ a(\sin x +\frac{b}{a}\cos x)=\overset{\frac{b}{a}=\tan(\theta )}{\rightarrow}\\a (\sin x + \tan (\theta)\cos x) \\=a(\sin x +\frac{\sin (\theta)}{\cos (\theta)}\cos x)\\=a \frac{\sin x \cos (\theta)+\sin (\theta)\cos x}{\cos (\theta)}\\= a \frac{\sin(x+\theta)}{\cos (\theta)}\\=$$ now turn $\cos(\theta)$ $$\cos ^2(\theta)=\frac{1}{1+\tan^2(\theta)}=\frac{1}{1+(\frac{b}{a})^2} \rightarrow \cos (\theta)=\pm \frac{a}{\sqrt{a^2+b^2}}$$ so we have $$a \frac{\sin(x+\theta)}{\cos (\theta)}\\= \pm \sqrt{a^2+b^2} \sin(x+\theta)\\=\frac{|a|}{a}\sqrt{a^2+b^2}\sin(x+\theta)\\ \space \\ \sqrt{a^2+b^2}\sin(x+\theta) , \space or ,-\sqrt{a^2+b^2}\sin(x+\theta)=\sqrt{a^2+b^2}\sin(-x-\theta)\\$$

• Use \sin x, \cos x, \tan x, \csc x, \sec x, \cot x in math mode to produce, respectively $\sin x$, $\cos x$, $\tan x$, $\csc x$, $\sec x$, $\cot x$. – N. F. Taussig Jul 27 '15 at 16:43

It would be nice if $a=\sin u$ and $b=\cos u$, but that would only make sense if $$a^2+b^2= \sin^2 u +\cos^2 u =1$$, but in general, that's not true. So perhaps we could multiply by $\frac{A}{A}$ so that $A((\frac{a}{A})^2+(\frac{b}{A})^2)=A(\cos^2 u + \sin^2 u)$

Now we just need to determine what $A$ is.