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I have an expression like this that I need to evaluate:

$$16\sin(2\pi/3)$$

According to my book the answer is $8\sqrt{3}$. However, when I'm using my calculator to get this I get an answer like $13.86$. What I want to know, is it possible to make a calculator give the answer without evaluating $\pi$, so that $\pi$ is kept separate in the answer? And the same for in this case, $\sqrt{3}$. If the answer involves a square root, I want my calculator to say that, I don't want it to be evaluated.

I am using the TI-83 Plus if that makes a difference.

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I have used the TI-83 and TI-84 (which is just an upgraded version of the TI-83) extensively, and it is not possible for it to give you an answer in terms of $\pi$ or $\sqrt{3}$. For example, it won't give you $8 \sqrt{3}$ as an answer. It will give you 13.86 (and more digits). What you can always do, in such a situation, is type in $8 \sqrt{3}$ into your calculator as well and see that you get the same decimal as you did when you typed in $16 \sin(2\pi/3)$. By the way, some newer TI calculators may give you answers like you want. The TI-30XS Multiview gives $8 \sqrt{3}$. –  Graphth Feb 27 '12 at 13:05
    
That's interesting. I used to have a calculator that had a button you could press to get the answers in fractions or pi or whatever, and it was very easy, so I figured all calculators would be like that. Guess not (guess I'll buy a new calculator). –  victoriah Feb 27 '12 at 13:07
    
I imagine what you mean is in fractions... not in fractions or pi. Just fractions. The TI-83 can do fractions if you hit Math and then the very first choice is triangle Frac which says turn whatever you have now into a fraction. As far as buying a new calculator, that's up to you. You will need to research which ones do this one feature you are looking for and then you'll need to decide if it's worth \$150 to buy a new calculator for this one feature. If all you care about is the one feature, and if fine with 2 calculators, the TI-30XS Multiview is only around $18, but not a graphing calc –  Graphth Feb 27 '12 at 13:12

3 Answers 3

up vote 5 down vote accepted

Well we don't need a calculator here. There is no need to write anything in terms of $\pi$.

If you look closely at the sin function you will see that sin$(x) = $ sin$(\pi - x)$.

So sin$(\frac{2\pi}{3}) = $ sin$(\frac{\pi}{3})$.

But what is sin$(\frac{\pi}{3})$?

Well the answer is $\frac{\sqrt{3}}{2}$. You can get this by drawing an equilateral triangle of side length $2$ and splitting in half down the middle (to get a right angled triangle with one angle $\frac{\pi}{3}$ and side lengths $1$, $2$ and $\sqrt{3}$).

So the answer to your question is $16(\frac{\sqrt{3}}{2}) = 8\sqrt{3}$.

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How do you get from sin(2pi/3) to sin(pi-pi/3)? I see the (pi-x) thing, but I don't get where the 2 disappears to. Otherwise this answer was incredibly helpful, thanks! :) –  victoriah Feb 27 '12 at 9:53
1  
This is just subtraction of fractions. $\pi - \frac{2\pi}{3} = \frac{3\pi - 2\pi}{3} = \frac{\pi}{3}$. –  fretty Feb 27 '12 at 9:56

You may have learned some value of $\sin \theta$ . For example, $\sin 0= 0$, $\sin \frac{\pi}{6}= \frac{1}{2}$ ,$\sin \frac{\pi}{4}= \frac{1}{\sqrt{2}}$, $\sin \frac{\pi}{3}= \frac{\sqrt{3}}{2}$ and $\sin \frac{\pi}{2}= 1$, Other most useful value can be evaluated using these.

For example, in your case $$\sin\frac{2\pi}{3}= \sin{(\pi-\frac{\pi}{3}})$$ $$=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$$ We know that $\sqrt{3}= 1.732..$, this gives $$16.\sin\frac{2\pi}{3}=\frac{1.732..}{2}\times 16= 8\times 1.732.. = 13.856$$

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I feel obliged to point out that this doesn't actually answer the question being asked. –  Chris Taylor Feb 27 '12 at 9:21
    
@MDCCXXIX, Thanks for pointing out... sorry i misunderstood this.. actually question is about to use the calculator so that output is in terms of $\pi$. Thanks for pointing out..i wrote something different... –  zapkm Feb 27 '12 at 9:25

Here’s something I used to tell students that might help. Among the angles that you’re typically expected to know the trig. values for ($30,$ $45,$ $60$ degrees and their cousins in the other quadrants), the only irrational values for the sine, cosine, tangent have the following magnitudes:

$$\frac{\sqrt{2}}{2}, \;\; \frac{\sqrt{3}}{2}, \;\; \sqrt{3}, \;\; \frac{\sqrt{3}}{3}$$

Note that if you square each of these, you get:

$$\frac{1}{2}, \;\; \frac{3}{4}, \;\; 3, \;\; \frac{1}{3}$$

Now consider the decimal expansions of these fractions:

$$0.5, \;\; 0.75, \;\; 3, \;\; 0.3333…$$

The important thing to notice is that if you saw any of these decimal expansions, you’d immediately know its fractional equivalent. (O-K, most people would know it!)

Now you can see how to use a relatively basic calculator to determine the exact value of $\sin\left(2\pi / 3 \right).$ First, use your calculator to find a decimal for $\sin\left(2\pi / 3 \right).$ Using a basic calculator (mine is a TI-32), I get $0.866025403.$ Now square the result. Doing this, I get $0.75.$ Therefore, I know that the square of $\sin\left(2\pi / 3 \right)$ is equal to $\frac{3}{4},$ and hence $\sin\left(2\pi / 3 \right)$ is equal to $\sqrt{\frac{3}{4}}.$ The positive square root is chosen because I got a positive value for $\sin\left(2\pi / 3 \right)$ when I used my calculator. Finally, we can rewrite this as $\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}.$

What follows are some comments I posted in sci.math (22 June 2006) about this method.

By the way, I used to be very concerned in the early days of calculators that students could obtain all the exact values of the $30,$ $45,$ and $60$ degree angles by simply squaring the calculator output, recognizing the equivalent fraction of the resulting decimal [note that the squares of the sine, cosine, tangent of these angles all come out to fractions that any student would recognize (well, they used to be able to recognize) from its decimal expansion], and then taking the square root of the fraction. As the years rolled by, I got to where I didn't worry about this at all, because even when I taught this method in class (to help students on standardized tests and to help them for other teachers who were even more insistent about using exact values than I was), the majority of my students had more trouble doing this than just memorizing the values!

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