How can I prove this question concerning trigonometry? Prove that, for some constant $B$,
$$4\cos(x) - 3\sin(x) = 5\cos(x+B).$$
Then, estimate the value of $B$.
 A: Expanding $\cos(x+B)$ and comparing the coefficients of $\cos x,\sin x$,
$4=5\cos B\iff\cos B=?,3=5\sin B\iff\sin B=?$ 
Observe that $\sin B,\cos B>0$
Hope you are aware of All Sin Tan Cos Rule
A: You have
$$ {4\over 5}\cos(x) - {3\over 5}\sin(x) = \cos(x + B).$$
Now put $\theta = \cos^{-1}(4/5)$.  You have $\cos^{-1}(4/5) = \sin^{-1}(3/5)$.
so
$$\cos(\theta)\cos(x) - \sin(x)\sin(\theta) = \cos(x + B).$$
Take $B = \theta.$.  
A: Expanding using the cosine sum identity gives
$$4 \cos x - 3 \sin x = 5 (\cos B \cos x - \sin B \sin x),$$
and comparing coefficients gives
$$5 \cos B = 4, \qquad 5 \sin B = 3.$$
One can use, say, the Pythagorean identity to check that some solution to one equation is also a solution of the other, but we may as well divide the equations to get
$$\tan B = \frac{3}{4}. \qquad{(*)}$$
Now, $\sin$ and $\cos$ have period $2 \pi$ and $\tan$ has period $\pi$, up to addition of multiples of $2 \pi$ (which doesn't affect the value of $\sin$ and $\cos$) the solutions of the original equation are exactly the solutions of ($\ast$) in $[-\pi, \pi)$, which are $B = \arctan \frac{3}{4}$ and $B = -\arctan \frac{3}{4}$, and only the first value gives positive values of $\sin$ and $\cos$, and so only the first value is a solution.
One can estimate it in several ways, and what method is appropriate depends on your context, but one interesting option would be to expand a Maclaurin series for $\arctan$ around $\frac{1}{2}$ to a few terms.
A: Since you received good answers and that you have been asked to estimate $B$, I cannot resist to the pleasure of giving you two approximations I "discovered" recently $$\sin(x) \approx \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}$$ $$\cos(x)\approx \frac{5 \pi ^2}{x^2+\pi ^2}-4$$ Applied to your case, from the sine we obtain $$B_{sine} =\frac{1}{46} \left(23-2 \sqrt{46}\right) \pi=0.644391$$ and from the cosine $$B_{cosine}= \frac{\pi }{2 \sqrt{6}}=0.641275$$ while the exact solution is $$B_{exact}=0.643501$$ while the average of the above estimates is $$B_{average}=\left(\frac{1}{24} \left(6+\sqrt{6}\right)-\frac{1}{\sqrt{46}}\right) \pi=0.642833$$ and solving $4\sin(x)=3\cos(x)$ leads to a solution equal to $0.643265$.
Where this becomes interesting is that these approximations were proposed, almost 1400  years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician.
A much less accurate approximation of my own $$\sin(x)\approx \frac{120 (\pi -x) x}{\pi ^5}$$ would have given $$B=\frac{\pi}{20} \left(10-\sqrt{100-2 \pi ^3}\right)=0.602652$$
