How to find Exact Value of Sin (18 Degrees)?

Question

Anyone can help me out for this?
The answer is (-1 + √5 / 4) but I'm not sure what is the method to get it.

Answer 1

18* = x
So sin 3x = cos 2x
So 3sin x - 4sin^3 x = 1-2sin^2 x
So 4sin^3 x -2sin^2 x - 3sin x + 1 =0
Clearly sin x = 1 is a soln, but not the one we need.
So factorizing,
4sin^3 x - 4sin^2 x +2sin^2 x - 2sin x -sin x +1 =0
Let sin x = y
So (y-1)(4y^2 +2y -1)=0
So sin x is a root of 4y^2 +2y -1 =0
Solve the quadratic eqn, and keep in mind that sin 18* is positive.

Answer 2

Sin 18 Degrees

Answer 3

*Sorry for reviving this 9-year old post, but here is a concise answer. All angles here are given in degrees.

The idea is to force the value of sin(18) into a solvable quadratic equation using trigonometric identities:

Angle property: 2 * 18 = 90 - 3 * 18

sin(2 * 18) = sin(90 - 3 * 18) = cos(3 * 18)

Double- and triple-angle identity: 2sin(18)cos(18) = 4cos^3(18) - 3cos(18)

2sin(18)cos(18) - 4cos^3(18) + 3cos(18) = 0

Divide both sides by cos(18) (non-zero): 2sin(18) - 4cos^2(18) + 3 = 0

Pythagorean identity: 2sin(18) - 4(1 - sin^2(18)) + 3 = 0

The equation becomes: 4sin^2(18) + 2sin(18) - 1 = 0

(Allow the substitution B = sin(18).)

4B^2 + 2B - 1 = 0

Quadratic formula: B = (-1 +/- sqrt(5)) / 4

But since B = sin(18) > 0, then B = sin(18) = (-1 + sqrt(5) / 4, and this is the answer.

Answer 4

Dont try calculator to find any value sin(3degree) = sin(75-72) After that remember a formula Sin[x-y]==Cos[y] Sin[x]-Cos[x] Sin[y] == sin(75-72)=cos(75)sin(72)+cos(72)sin(75) sin(75)=Cos[45] Sin[30]+Cos[30] Sin[45] =(1+sqrt3)/(2sqrt2) sin(72) =1/2 sqrt(0.5(5+sprt5)) you can do the rest for cos(75) and cos(72) ==1/16( sqrt2(1+sqrt3)(-1+sqrt5)+2sqrt(5+sqrt5)-... you can simplify the rest sqrt2 means square root of 2

Answer 5

letx=18*
5x=90*
3x=90*-2x
cos3x=cos(90*-2x)=sin2x
4cos^3x-3cosx=2sinxcosx
since cosx is not=0
4cos^2x-3=2sinx
4(1-sin^2x)-3=2sinx
4sin^2x+2sinx-1=0
sinx=[-2+/-rt(2^2-4*4(-1))]/2*4
sin18*=(-1+/-rt5)/4
18* is an acuteangle.so sin18*>0
therefore sinx=(-1+rt5)/4

Answer 6

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sin(36°) = 2sin(18°)cos(18°) = 2sin(18°)sin(90° - 18°) = 2sin(18°)sin(72°) = 2sin(18°)[2sin(36°)cos(36°)] = 4sin(18°)sin(36°)cos(36°) = 4sin(18°)sin(36°)sin(90° - 36°) = 4sin(18°)sin(36°)sin(54°) And we have: sin(36°) = 4sin(18°)sin(36°)sin(54°) Divide both sides by sin(36°). 1 = 4sin(18°)sin(54°) 1/4 = sin(18°)sin(54°)

Answer 7

Use your calculator to find the value of sin 18º :

0.309016994374947424102293417182819

Copy & paste into the Inverse Symbolic Calculator, and browse around. If you select Simple Lookup you get

sqrt( ( 3 - sqrt(5) ) / 8 )

If you select Smart Lookup then K/4 you get

-(1-1*sqrt(5))/16

and after multiplying by 4, get

( sqrt(5) - 1 ) / 4

Please note that you mixed up your brackets:

(-1 + √5 / 4) means -1 + (√5 / 4 ). You should write it as:

(-1 + √5 ) / 4

Answer 8

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.

Cool, eh?

Answer 9

sin of 18 degrees
a triangle.... with hypotenuse of 1 and angle of 18

the endpoint of the hypotenuse is at the point (cos18,sin18)
(.951096 , .3090169)

if so, then Pythagoras says (.951096)^2 + (.3090169)^2 = 1
.904508 + .095491 = (.999999)

sin 18 = .3090169

Answer 10

let x=18deg>>>>5x=90deg

cos3x=sin(90-3x)=sin(5x-3x)

cos3x=sin2x

4cos^3(x)-3cosx =2sinx.cos x

4cos^2(x)-3 =2sinx

-4sin^2(x)-2sinx+1 =0

4sin^2(x)+2sinx-1 =0

sinx= (-2+or- sqrt(4+16))/8

= (-1+sqrt5)/4 (since 18deg in 1st quadrant)

i hope that this helps