# Solving $q = \sin\left(\frac{a}{2}\right)*\cos(B_x)$ for $B_x$

I'm a bit rusty and am having trouble using Trig Identities to solve for $B_x$. Can someone show me how to do this?

$$q = \sin\left(\frac{a}{2}\right)*\cos(B_x)$$ I want to solve for $B_x$

My main hang ups are when to apply $\arccos$ or when to use $\csc()$ or when to use the trig identity for $\sin(x)\cos(y)$.

• this is my first question here. reason for downvotes will help. – scord Feb 16 '15 at 3:42
• There is a vendetta against no work shown posts. They are called PSQ(Problem statement questions) by certain users. It isn't even homework related, it is just about quality. I can see you are a software engineer, so I doubt foul play, but you will need to edit in some attempt of your own unfortunately. – user142198 Feb 16 '15 at 3:43
• ah i see. My main hang ups are when to multiply both sides by arccos or when to use csc() or when to use the trig identity for sin(x)*cos(y) – scord Feb 16 '15 at 3:48
• If this is for a non-mathematical audience (for example, you need to apply the formula within a computer program), you can write $\frac{1}{\sin u}$ instead of $\csc u$ (or $\frac{v}{\sin u}$ instead of $v \csc u$). Even for a mathematical audience it's sometimes preferable to have a sine in the denominator rather than a cosecant in the numerator (viz. spherical law of sines). – David K Feb 16 '15 at 5:45
• You are right to consider arc cos for problems like this, but I hope you don't actually mean to "multiply" by it--that's not how you cancel out a cosine. Perhaps you mean when to "apply" or "take" the arc cos, that is, when to use the fact that $\arccos(\cos u) = u.$ – David K Feb 16 '15 at 5:50