# Math Help - Can you solve this kind of problem on a TI-89 Calc? finding trig angle

1. ## Can you solve this kind of problem on a TI-89 Calc? finding trig angle

I am wanting to know if there is any way that you can solve the following problem by just using your calculator?

Find Exact values using combination of $\frac{\pi}{6}$, $\frac{\pi}{6}$

1. $tan(\frac{\pi}{12})$

2. $sin(\frac{5\pi}{12})$

3. $cos(\frac{11\pi}{12})$

I am looking for a quick and easy way to solve these kinds of problems, just thought that someone could help me!!

Thanks

2. Hello, qbkr21!

Is there any way that you can solve the following problems by using your calculator?

It may have ten decimal places, but it is not exact.

You are expected to know these identities:

. . $\sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad\sin\theta \:=\:\pm\sqrt{\frac{1 - \cos2\theta}{2}}$

. . $\cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}\quad\Rightarrow\quad\cos\theta \:=\:\pm\sqrt{\frac{1 + \cos2\theta}{2}}$

. . $\tan\theta \:=\:\frac{\sin\theta}{\cos\theta} \:=\:\frac{\pm\sqrt{\frac{1-\cos2\theta}{2}}}{\pm\sqrt{\frac{1 + \cos2\theta}{2}}} \:=\: \pm\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}$

$1) \;\tan\left(\frac{\pi}{12}\right)$ . π/12 is in Q1; tangent is positive.

We have:. . $\tan\frac{\pi}{12}\;=\;\sqrt{\frac{1-\cos\frac{\pi}{6}}{1 + \cos\frac{\pi}{6}}}\;=\;\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}}} \;= \;\sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}}$

Rationalize: . $\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}} \;=\;\sqrt{\frac{(2-\sqrt{3})^2}{4-3}} \;=\;2 - \sqrt{3}$

$2)\; \sin\left(\frac{5\pi}{12}\right)$ . 5π/12 is in Q1; sine is positive

We have: . $\sin\frac{5\pi}{12}\;=\;\sqrt{\frac{1 - \cos\frac{5\pi}{6}}{2}}\;=\;\sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{2}\right)}{2}} \;= \;\sqrt{\frac{2 + \sqrt{3}}{4}} \;=\;\frac{\sqrt{2 + \sqrt{3}}}{2}$

$3)\; \cos\left(\frac{11\pi}{12}\right)$ . 11π/12 is in Q2; cosine is negative.

We have: . $\cos\frac{11\pi}{12}\;=\;-\sqrt{\frac{1 + \cos\frac{11\pi}{6}}{2}} \;= \;-\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \;=\;-\sqrt{\frac{2 + \sqrt{3}}{4}} \;=\;-\frac{\sqrt{2+\sqrt{3}}}{2}
$

3. ## What if?

What if the directions were to again:

Find exact values using combinations of $\frac{\pi}{6}$, $\frac{\pi}{4}$? What would be the process if given these directions with the same question for 1,2,3? Is there only 1 answer to each of the problems? Is there no right or wrong way to get the answer? I think she wants us to use the Sum and Differences Formulas, but is it absolutly necessarry?

4. ## Re:

I mean I know how to use the half angle formula, but given values such as $\frac{\pi}{6}$ and $\frac{\pi}{4}$ how can I figure figure out which value to use in the formula. This off all would help me out the most....

Thanks So Much!

5. There is a trick I discovred while in high school so my classmates can find the radical form instead of memorizing all the possible quadrants.

Say you want to find,
$\sin 225^o=?$
Find the value on the calculator,
$\approx -.707...$
Now if this number is rational you are done.
But it is not!
So square this number,
$\approx .499999...$
That means that the value of this sine angle is the square root of $\frac{1}{2}$. But you need to determine whether the sign is negative or positive. The sign is negative because it is in the 3rd quadrant. Thus, the value is,
$-\sqrt{\frac{1}{2}}=-\frac{\sqrt{2}}{2}$
You should rationalize the fraction.

6. ## Re:

But could you please expalin how particularly for the problem $cos(\frac{11\pi}{12})$ and taking the two terms $\frac{\pi}{6}$ and $\frac{\pi}{4}$ I could find a value to put into the half angle formula? What is the trick?

Thanks so much!!!

7. Originally Posted by qbkr21
But could you please expalin how particularly for the problem $cos(\frac{11\pi}{12})$ and taking the two terms $\frac{\pi}{6}$ and $\frac{\pi}{4}$ I could find a value to put into the half angle formula? What is the trick?
Ah! The method no longer works . But it can still be done. You probably are not familar with infinite continued fractions, that is what the method relies on.

8. ## Re:

Can you please show me how I can take the two multiple above and use them through the sum and differences formula to find the answer?

9. Originally Posted by qbkr21
I am wanting to know if there is any way that you can solve the following problem by just using your calculator?

Find Exact values using combination of $\frac{\pi}{6}$, $\frac{\pi}{6}$

1. $tan(\frac{\pi}{12})$
Use the $\tan$ half angle formula (first quadrant version).

$
\tan(A/2)=\sqrt{\frac{1-\cos(A)}{1+\cos(A)}}
$

RonL

10. ## RE: Are these answers correct?

Originally Posted by qbkr21
I mean I know how to use the half angle formula, but given values such as $\frac{\pi}{6}$ and $\frac{\pi}{4}$ how can I figure figure out which value to use in the formula. This off all would help me out the most....

Thanks So Much!
Are these answers right and acceptable?

1. $tan(\frac{\pi}{12})$
= $-(\sqrt{3}-2)$

2. $sin(\frac{5\pi}{12})$
= $\frac{(\sqrt{3}+1)*\sqrt{2}}{4}$

3. $cos(\frac{11\pi}{12})$
= $\frac{-(\sqrt{3}+1)*\sqrt{2}}{4}$

Thanks for the help!!

11. ## Re

I used the TI-Voyage 200 to get these. I typed them in and it spit them back out at me in radians. Correct?

Originally Posted by qbkr21
I mean I know how to use the half angle formula, but given values such as $\frac{\pi}{6}$ and $\frac{\pi}{4}$ how can I figure figure out which value to use in the formula. This off all would help me out the most....

Thanks So Much!

12. Originally Posted by qbkr21
I am wanting to know if there is any way that you can solve the following problem by just using your calculator?
Yes.

According to the headline of your post you are using a TI-89. If you type the given expressions into the calculator you'll get back the exact value.

Originally Posted by qbkr21
Find Exact values using combination of $\frac{\pi}{6}$, $\frac{\pi}{6}$

1. $tan(\frac{\pi}{12})$
2. $sin(\frac{5\pi}{12})$
3. $cos(\frac{11\pi}{12})$
I am looking for a quick and easy way to solve these kinds of problems, just thought that someone could help me!!
Thanks
Hello,

type the problem into your TI-89 and you'll get the exact value:

1. $tan(\frac{\pi}{12})=2-\sqrt{3}$
2. $sin(\frac{5\pi}{12})=\frac{(\sqrt{3} +1)\cdot \sqrt{2}}{4}$
3. $cos(\frac{11\pi}{12})=\frac{-(\sqrt{3} +1)\cdot \sqrt{2}}{4}$

EB