algebra-precalculustrigonometry
I'm learning Trigonometry right now with myself and at current about general solution. I have a question which is confusing me from some time. The question is –
$If \cos(A-B) = 1/2$ and $\sin(A+B) =1/2$, find the smallest positive value of A and B and also their most general values.
I am getting $A=105^\circ$ and $B=45^\circ$,the answer of general value given in my book is 
Please help. Thankyou in advance.
$$x=r\cos\theta$$ the radius here is $r=1$ $$x=\cos\theta$$ $$y=r\sin\theta$$ $$y=\sin\theta$$
One of the ways of defining the trigonometric functions is geometrically.
See Wikipedia.
Taking the Unit Circle (the circle of radius $1$ with center the origin, describable by the equation $x^2+y^2=1$), given a particular angle $\theta$, the value of cosine of $\theta$ is the $x$ coordinate for where the ray with angle $\theta$ above the positive $x$-axis intersects the circle.
Notice that for $90^\circ<\theta<180^\circ$ you will be in the top left quadrant, and in particular will have negative $x$ coordinates. Similarly for $-180^\circ<\theta<-90^\circ$ it will lie in the bottom left quadrant. In particular for something like $-100^\circ$.
(there are more complicated definitions which fix some inconsistencies with this definition as well as extend cosine to more abstract settings, but for now this definition is good enough)
Best Answer
When you find $\cos \theta =\cos x$ just remember that $\theta= 2n\pi \pm x$ and if $\sin\theta =\sin x$ then $\theta = n\pi+(-1)^n x$. In both the cases $n$ is an integer.
In the given problem $$A-B=2n\pi \pm \frac{\pi}{3}$$ and $$A+B=m\pi + (-1)^m\frac{\pi}{6}$$
Hope you can solve it now.