# Help simplifying the result to a trig question to become one of the accepted answers needed.

#### Joshua David

##### New member
The question asks "what is the exact value of $\sin(-105^\circ)$ using the half-angle identities." I tried solving it and came to the answer $-\frac{\sqrt{2+\sqrt{3}}}{2}$. This answer is correct, but in the quiz it is represented as $\frac{- \sqrt{2} - \sqrt{6}}{4}$. I'm not sure how to go from $-\frac{\sqrt{2+\sqrt{3}}}{2}$ to $\frac{- \sqrt{2} - \sqrt{6}}{4}$ and would appreciate help! Thank you !

I've tried multiplying $\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$ with the fraction, only to receive $-\frac{1}{2\sqrt{2-\sqrt{3}}}$. If I multiply this fraction by $\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$, I arrive where I started xD!

#### The Highlander

##### Full Member
The question asks "what is the exact value of $\sin(-105^\circ)$ using the half-angle identities." I tried solving it and came to the answer $-\frac{\sqrt{2+\sqrt{3}}}{2}$. This answer is correct, but in the quiz it is represented as $\frac{- \sqrt{2} - \sqrt{6}}{4}$. I'm not sure how to go from $-\frac{\sqrt{2+\sqrt{3}}}{2}$ to $\frac{- \sqrt{2} - \sqrt{6}}{4}$ and would appreciate help! Thank you !

I've tried multiplying $\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2-\sqrt{3}}}$ with the fraction, only to receive $-\frac{1}{2\sqrt{2-\sqrt{3}}}$. If I multiply this fraction by $\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$, I arrive where I started xD!
Have you tried multiplying the top and the bottom by 2 as a starting point?

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#### skeeter

##### Elite Member
[imath]-\dfrac{\sqrt{4} \cdot \sqrt{2+\sqrt{3}}}{2\cdot 2} = -\dfrac{\sqrt{8 + 4\sqrt{3}}}{4} = -\dfrac{\sqrt{2 + 4\sqrt{3} + 6}}{4} = -\dfrac{\sqrt{2(1 + 2\sqrt{3} + 3)}}{4} = -\dfrac{\sqrt{2(1+\sqrt{3})^2}}{4} = -\dfrac{\sqrt{2}(1 + \sqrt{3})}{4} = -\dfrac{\sqrt{2}+\sqrt{6}}{4}[/imath]

check ...

[imath]\sin(-105) = -\sin(105) = -\sin(45+60) = -\left[ \sin(45)\cos(60)+\cos(45)\sin(60) \right] = -\left(\dfrac{\sqrt{2} + \sqrt{6}}{4}\right)[/imath]

#### Steven G

##### Elite Member
I want to point out how important what @The Highlander said about multiply by 1 as 2/2. 1st of all, multiplying an expression by something that equals 1 will not change the value of the original expression but will change the way it looks--and we certainly want that!
Multiplying by 2/2 will cause the denominator to be equal to the denominator of the end expression.
Getting an expression to look closer to another equal expression will almost always help in showing that the two expressions are in fact equal.

#### Dr.Peterson

##### Elite Member
The question asks "what is the exact value of [imath]\sin(-105^\circ)[/imath] using the half-angle identities." I tried solving it and came to the answer [imath]-\frac{\sqrt{2+\sqrt{3}}}{2}[/imath]. This answer is correct, but in the quiz it is represented as [imath]\frac{- \sqrt{2} - \sqrt{6}}{4}[/imath]. I'm not sure how to go from [imath]-\frac{\sqrt{2+\sqrt{3}}}{2}[/imath] to [imath]\frac{- \sqrt{2} - \sqrt{6}}{4}[/imath] and would appreciate help! Thank you !
What you want to do is to denest a radical, This concept is demonstrated here:

I am curious, though, why they would tell you to use half-angle identities, which naturally result in the form you got, but then give an answer in a form naturally obtained by angle-sum identities. I suspect the solution writer didn't read the problem carefully, and didn't do what it said to do.

(I'd also like to see your actual work, just to see how it compares to mine, and whether there is some point along the way where it would be natural to change the form of the answer.)