# Remembering /Deriving the values of sine and cosine of 18 degrees,36 degrees,54 degrees,72 degrees

I need to remember the values of sine and cosine of 18 degrees,36 degrees,54 degrees,72 degrees. That is multiples if 18 degrees.Is it possible to derive them in about a minute or so ? Do you use any particular method to remember them ?

Yes I do know that we can solve the equation $\sin{5\theta}=\pi/2$ and similar ones but that's way too lengthy (during exams).

• well at the very least you can just memorize the $sin$ values and then call upon your good friend Pythagoras? – still_learning Mar 21 '16 at 18:52
• You can remember $\sin 18^\circ$, and then use Pythagorean identity and addition formulas to derive all the rest. – Wojowu Mar 21 '16 at 18:58
• What horrible exam requires you to remember such horrible things!? – fosho Mar 21 '16 at 18:59
• @Daniel JEE.It's an Indian undergraduate entrance exam :-P.We have to solve 30 math problems in an hour.So it becomes necessary to remember all possible formulae :-P – user220382 Mar 21 '16 at 19:02
• @Wojowu good idea btw – user220382 Mar 21 '16 at 19:03

By route of rote memorization, you need only remember two expressions as ordered below. Picture the negative signs making a single "line" of text and the positive signs making the next line. $$\dfrac{\sqrt{5}\mp 1}{4}, \sqrt{\dfrac{5\mp \sqrt{5}}{8}}.$$ Reading as if it were plain English (i.e. left-to-right and top-to-bottom), we get the sines in order: $\sin 18°,\sin 36°,\sin 54°,\sin 72°$.

Going the opposite direction (bottom-to-top and right-to-left), we similarly get the cosines in order: $\cos 18°,\cos 36°,\cos 54°,\cos 72°$.

Regardless of how one obtains these values for efficiency's sake, the underlying concepts and intuition should always be retained in mind or—at the very least—appreciated.

This is the method I used during my high school days.

Note that $$\sin 18^\circ = \frac{\sqrt5-1}{4}$$ and $$\sin 36^\circ = \sqrt{\frac{5-\sqrt5}{8}}$$

Now you just have to remember this much.

For $\cos 18^\circ$, you will have the same expression as $\sin 36^\circ$ but only with the minus sign replaced by a plus sign.

Similarly,for $\cos 36^\circ$, you will have the same expression as $\sin 18^\circ$ but only with the minus sign replaced by a plus sign.

And we know that $54^\circ$ and $36^\circ$ are complementary just as $72^\circ$ and $18^\circ$ are. So you can calculate them using the rule: $$\sin (90^\circ - x) = \cos x$$ and $$\cos (90^\circ - x) = \sin x$$

I think this derivation is fairly memorable:

Let $x=\pi/10$. Then

$$e^{5ix}=i.$$

Hence

$$e^{3ix}=ie^{-2ix}.$$

In particular $Re(e^{3ix})=-Im(e^{-2ix})$.

DeMoivre lets us compute these in terms of $\cos(x)$ and $\sin(x)$:

$$\cos(x)^3-3\cos(x)\sin(x)^2=2\cos(x)\sin(x)$$

Divide by $\cos(x)$:

$$\cos(x)^2-3\sin(x)^2-2\sin(x)=0.$$

Get $\cos(x)^2=3 \sin(x)^2+2\sin(x)$. Now Pythagoras gives

$$1-\sin(x)^2=3\sin(x)^2+2\sin(x)$$

or

$$4\sin(x)^2+2\sin(x)-1=0.$$

Hence

$$\sin(x)=\frac{-2+\sqrt{20}}{8}=\frac{-2+2\sqrt{5}}{8}=\frac{\sqrt{5}-1}{4}.$$

Then use Pythagoras to get $\cos(x)$. Then use deMoivre to get the rest.

Is this too long to do over?

The easiest way I can think of to derive these results is to start with a regular pentagon ABCDE. Add in diagonals AC and AD. Then triangle ABC has angles of 36, 36 and 108 degrees with $AB:BC:AC = 1:1:(1+\sqrt{5})/2$, and ACD has angles of 72, 72 and 36 degrees with $AC:AD:CD=(1+\sqrt{5})/2:(1+\sqrt{5})/2:1$. Bisect each isosceles triangle at the apex angle to make a pair of right triangles and apply the usual right-triangle relations.

Draw a A 36 degree, B 72degree, C 72degree triangle.

Halve the angle at B until it meets AC at a new point D.

B 36 degrees D 72 degrees C 72 degrees is simmilar to ABC.

The remaining triangle is A 36 degree, B 36 degrees D 144.

Let AB = 1 Let BC = x. Then BD = x and therefore AD = x.

Therefore DC = 1-x Then by similar triangles, $$\frac{AB}{BC} = \frac{BC}{DC}$$ $$\frac{1}{x} = \frac{x}{1-x}$$ $$x^2 = 1-x$$ $$x^2 + x - 1 = 0$$

Giving $$x = \frac{-1 + \sqrt{5}}{2}$$

Then bissect BC at E to make A 18 degrees, B 72 degrees E 90 degrees.

Then, $$sin(18^\circ) = \frac{BE}{AB} = \frac{x}{2} = \frac{-1 + \sqrt{5}}{4}$$ $$cos(72^\circ) = \frac{AE}{AB} = \frac{\sqrt{1-x^2}}{2} = \sqrt{\frac{5-\sqrt5}{8}}$$