# Problem getting a Trig question check in My Calc

• Dec 2nd 2006, 11:38 PM
qbkr21
Problem getting a Trig question check in My Calc
For example if I am given:

$\displaystyle \text{G:}Cos(x)=\frac{-1}{6}$ and $\displaystyle x$ is in $\displaystyle \text{Quadrant III}$

$\displaystyle \text{U:}Sin\frac{x}{2}=?$ in quadrant III

I worked it out in by using the half angle formula and got $\displaystyle \frac{-\sqrt 21}{6}$

I usually use the old inverse trick on the calculator and check to see if the decimal I get matches the answer I got. But on several half angle problems I am getting an undefined message on my calc. What is the deal and how can I go about checking these...

Like what PerfectHacker did on this post http://www.mathhelpforum.com/math-he...i-89-calc.html

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Also one other question.... When I am using the double angle and half angle formulas, when I get what the single trig value is equal to such as $\displaystyle Cos(x)=\frac{-1}{6}$ and I am going to stick the value in a problem do I want to insert the negative or positive sign along with it depending on which quadrant I am talking about? For example in the problem above we are in the 3rd quadrant so both $\displaystyle Sin$ & $\displaystyle Cos$ both are negative. Would I want to insert them into the formulas as negative? But suppose we are in quadrant 1 where both $\displaystyle Sin$ & $\displaystyle Cos$ are positive. Would I want to put both values into the double or half-angle formulas as a postive?

I could use some real help with both questions if you could...

Thanks as always for the help!
• Dec 3rd 2006, 05:49 AM
Soroban
Hello, qbkr21!

Be careful! . . .

Quote:

Given: .$\displaystyle \cos x =-\frac{1}{6}$ and $\displaystyle x$ is in $\displaystyle Q3$

Find: .$\displaystyle \sin\frac{x}{2}$

Identity: .$\displaystyle \sin\frac{\theta}{2}\:=\:\pm\sqrt{\frac{1-\cos\theta}{2}}$

So we have: .$\displaystyle \sin\frac{x}{2} \:=\:\pm\sqrt{\frac{1 - (-\frac{1}{6})}{2}} \:=\:\pm\frac{\sqrt{21}}{6}$
. . . which you already knew

Since $\displaystyle x$ is in $\displaystyle Q3\!:\;\;\pi \:\leq \:x \:\leq \:\frac{3\pi}{2}\quad\Rightarrow\quad \text{then: }\frac{\pi}{2} \:\leq \:\frac{x}{2} \:\leq \:\frac{3\pi}{4}$

. . Hence, $\displaystyle \frac{x}{2}$ is in Quadrant 2, where sine is positive.

Therefore: .$\displaystyle \sin\frac{x}{2}\:=$ +$\displaystyle \frac{\sqrt{21}}{6}$

• Dec 3rd 2006, 05:58 PM
qbkr21
Re:
Thanks Soroban,

So can I check all questions like these in my calculator after doing the work to ensure that my answers are correct. If the decimals matches the irrational fraction I got then I guess that my answers are correct?