sin(pi/12)=(1/2)(sqrt(A-sqrt(B))). I get the part about the half-angle identity sin(x/2)= +-Sqrt((1-cosx)/(2))). But I dont even know where to start with this problem. I'm supposed to figure out what A and B are.

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- Oct 7th 2013, 06:10 PMrhcprule3Can someone walk me through this trig problem?
sin(pi/12)=(1/2)(sqrt(A-sqrt(B))). I get the part about the half-angle identity sin(x/2)= +-Sqrt((1-cosx)/(2))). But I dont even know where to start with this problem. I'm supposed to figure out what A and B are.

- Oct 7th 2013, 06:31 PMSlipEternalRe: Can someone walk me through this trig problem?
Let's write it out. $\displaystyle \sin\left( \dfrac{\pi}{12} \right) = \dfrac{1}{2}\sqrt{A - \sqrt{B}} = \sin\left( \dfrac{x}{2} \right) = \pm \sqrt{\dfrac{1-\cos x}{2}}$. So, let's equate things that look similar. $\displaystyle \sin\left( \dfrac{\pi}{12} \right) = \sin\left( \dfrac{x}{2} \right)$ means that $\displaystyle \dfrac{x}{2} = \dfrac{\pi}{12}$ or $\displaystyle x = \dfrac{\pi}{6}$. So plug that in.

$\displaystyle \dfrac{1}{2}\sqrt{A - \sqrt{B}} = \sqrt{\dfrac{1-\cos \left( \tfrac{\pi}{6} \right)}{2}}$

Multiply both sides by 2:

$\displaystyle \sqrt{A - \sqrt{B}} = 2\sqrt{\dfrac{1-\cos \left( \tfrac{\pi}{6} \right)}{2}}$

Square both sides:

$\displaystyle A - \sqrt{B} = 2\left( 1-\cos \left( \tfrac{\pi}{6} \right) \right) = 2 - 2\cos \left( \tfrac{\pi}{6} \right)$

Give me a possible value for $\displaystyle A$. Once you choose a value for $\displaystyle A$, subtract it from both sides, then solve for $\displaystyle B$. - Oct 7th 2013, 07:15 PMrhcprule3Re: Can someone walk me through this trig problem?
so I can just choose anything for A?

- Oct 7th 2013, 07:21 PMProve ItRe: Can someone walk me through this trig problem?
- Oct 7th 2013, 07:45 PMSlipEternalRe: Can someone walk me through this trig problem?
I disagree with Prove It. You have a single equation with two variables. Given any $\displaystyle A \in \mathbb{R}$, there exists a solution for $\displaystyle B\in \mathbb{R}$. So, based on the problem in your original post, yes, you can choose anything for A.

Edit: I am mistaken. You can choose any $\displaystyle A\in \mathbb{R}$ such that $\displaystyle A \ge 2 - 2\cos\left( \dfrac{\pi}{6}\right) $ since $\displaystyle \sqrt{B}\ge 0$. - Oct 7th 2013, 08:15 PMibduttRe: Can someone walk me through this trig problem?
- Oct 8th 2013, 04:59 AMrhcprule3Re: Can someone walk me through this trig problem?
Thanks guys, I got her!!