Question

find cos 18, sin 18
help:cos 3x=sin2x then x=18

Answer 1

cos 3x = cos (x+2x)
= cos2xcosx - sin2xsinx
= [2(cosx)^2-1]cosx - 2cosx (sinx)^2
= 2(cosx)^3 - cosx - 2cosx [1-(cosx)^2]
= 2(cosx)^3 - cosx - 2cosx+ 2(cosx)^3
= 4(cosx)^3 - 3cosx

cos 3x=sin2x , where x=18
4(cosx)^3 - 3cosx = 2sinxcosx
since cosx =/= 0
4(cosx)^2 -3 = 2sinx
4[1-(sinx)^2] -3 = 2sinx
-4(sinx)^2 - 2sinx +1 =0
\[sinx = \frac{-2 \pm \sqrt {(-2)^2 - 4(-4)(1)}}{2(-4)}= \frac{-1 \pm \sqrt5}{4}\]
Since sin18 >0
therefore \[sin18 = \frac{-1 + \sqrt5}{4}\]
Can you do cos18 ?

Answer 2

Note: (sinx)^2 + (cosx)^2 =1
\[cos^218 = 1- sin^218 = 1- (\frac{-1+\sqrt5}{4})^2 = 1-\frac{3-\sqrt5}{8}=\frac{5+\sqrt5}{8}\]
\[\cos 18 = \sqrt {\frac{5+\sqrt5}{8}}\]

Answer 3

cool

Answer 4

@mahmit2012 i am seeing this qn posted here for 3rd time :) anyway @Callisto nic work

Answer 5

Hmm... it's the first time I saw this question posted here :|

Answer 6

i had been here around 2 years :)

Answer 7

That only proves that this question is quite difficult :P

Answer 8

ya esp that realisation cos 3x=sin2x ,i saw two other methods also

Answer 9

im a new member so callisto solved the problem and i really appreciate.

Answer 10

the other method is sin3x=cos 2x
and the most interesting one is a geometrical beauty

Answer 11

ok go ahead and give me the new solution

Answer 12

the 2nd method i told u is vry similr to wrk of callisto

Answer 13

lemme try to get that geometrical beauty frm my profile which is most elegant way :)

Answer 14

have you ever seen my last question to find f(1)?

Answer 15

guys here is the second method :

Show that if x= 18 degrees, then cos2x =sin 3x. HENCE find the exact value of sin 18 degrees, and prove that cos 36 - sin 18 =1/2.

The first part is trivial, but how does one use this first part to get to the second part. Note that this DOES NOT involve looking up in tables! I have found the correct answer by using the golden ratio, but how to get the answer from the first part I do not know. I thought that this would be a simple problem, since in the 1882 entrance exam for Sandhurst Military Academy, one was asked to find the exact value of sin 18 degrees. It is also interesting that sin18 (cos 72), sin36, (cos 54), sin54 (cos 36) and sin72 (cos 18) all come out as exact and easily represented ratios, just like sin30, sin45 and sin60, and is this due to 18 degrees =pi/10 and so on?



cos(2x) = sin(3x)
so
cos(2x) = sin(x)cos(2x) + cos(x)sin(2x)
or
cos(2x)- sin(x)cos(2x) = cos(x)(2sin(x)cos(x))
cos(2x)[1 - sin(x)] = 2cos2(x)sin(x) = 2[1 - sin2(x)] sin(x)
Hence
cos(2x)[1 - sin(x)] = 2[1 - sin(x)][1 + sin(x)] sin(x)
and thus
cos(2x) = 2 [1 + sin(x)] sin(x)
Note: You need to check that sin(x) = 1 is not a solution to the original problem.

Finally
1 - 2 sin2(x) = 2 sin(x) + 2 sin2(x)
or
4 sin2(x) +2 sin(x) - 1 = 0

Solve for sin(x).

Answer 16

@Callisto , @mahmit2012

Answer 17

i am trying hard to find the third method

Answer 18

got it !!!!

Answer 19

Another way to do this is to start with a regular pentagon where the length of each side is one unit. I'll outline the steps. (You can fill in the details.)

Call the pentagon ABCDE. Draw diagonals AC and BE, which intersect at point F. Using the fact that each angle of the pentagon is 108 degrees, you can prove the triangles BFC and AFE are isosceles. Therefore the lengths of line segments CF and EF are also 1 unit. Also note the angles for the triangles are 72-72-36.

Now show that ABF and CAB are similar triangles. Since they are similar, this means BE/AB = AB/BF. Since BE = EF + BF, substitute to get (EF + BF)/AB = AB/BF.

Let's define x to be the length of BF. Then, since EF = 1, the equation becomes (1 + x)/1 = 1/x, or x^2 + x - 1 = 0. Use the quadratic equation to get x = (sqrt(5) - 1) / 2.

Finally, let's create an 18-72-90 by drawing the altitude of triangle BFC from vertex C to side BF. Since BC = FC, the altitude hits BF at the midpoint of BF, which we will call point G. So, (a) triangle GBE is an 18-72-90 triangle, (b) BG = BF/2 = x/2.

Using this 18-72-90 triangle, sin 18 degrees = BG / BC = x/2. Therefore sin 18 = (sqrt(5) - 1) / 4.

Answer 20

experience counts :P

Answer 21

Ah.... it's getting complicated :|

Answer 22

anyway ..no medals?

Answer 23

thx i took a lot of time to browse my previuos qns and get these i suggest @mahmit2012 should atleast take a look

Answer 24

I thought you typed it :|
But thanks!!!!

Answer 25

:)

Answer 26

its foolish to retype evrything ,thats y i am supporting the tutorial section

Answer 27

if that came into being these methods would be brought to light to evryone

Answer 28

Sure it would :)