without using calculator , prove that sin 54 sin18 = 0.25?

Answer 1

that's a really hard proof. Here goes:

sin54°sin18°

use cofunction identity sin54° = cos (90°-54°) = cos36°
= cos36°sin18°

rewrite sin18° as (sin18°cos18°)/cos18°
=cos36°(sin18°cos18°)/cos18°

Use product to sum formula to convert sin18°cos18°: sinUcosV = .5(sin(U+V) - sin(U-V) --> sin18°cos18°= .5(sin36° - sin0°) = .5 sin36°


=.5*cos36°sin36°/cos18°

Use same product to sum formula to convert cos36°sin36 = .5(sin72°-sin0°) = .5sin72°

.5*.5sin72° / cos18°

Use cofunction identity to convert cos18° to sin (90°-18°) = sin 72°.

.5*.5*sin72°/sin72° = .5*.5 = .25

Answer 2

We need to prove the following claim:
sin 18 = ¼(-1 + √5).
(Note: all angles are in degrees here.)
First, let's use this to prove your result:
We have cos 36 = cos(54-18) = cos 54 cos 18 + sin 54 sin 18.
and
cos (72) = cos 54 cos 18 - sin 54 sin 18.
So
cos 36 - cos 72 = 2 sin 54 sin 18.
Next,
cos 72 = sin 18
and cos 36 = 1 - 2 sin² 18.
So
2 sin 54 sin 18 = 1- 2 sin² 18 -sin 18.
Now let's plug in sin 18 = ¼(-1 + √5).
This gives us
2 sin 54 sin 18 = 4/4 -2(6-2√5)/16 -
¼(-1 +√5) = 0.5
Which gives us
sin 54 sin 18 = 0.25.
Now we will prove our claim:
First sin 72 = 2 sin 36 cos 36 = 4 cos 18 sin 18 cos 36
So
cos 18 = 4 cos 18 sin 18 cos 36.
Now cos 18 > 0, so we can divide it out.
Also cos 36 = 1 - 2 sin² 18.
To simplify the writing, let's let x = sin 18.
Then
1 = 4x-8x³,
8x³ - 4x + 1 = 0.
With a bit of trial and error, we find that x = ½
is a root of this equation.
But sin 30 = ½, so ½ is not the answer we seek.
However, x-½ is a factor of 8x³-4x+1
and, by long division or synthetic division the
quotient is 8x²+4x-2.
So we have to solve
8x²+4x-2 = 0
or
4x²+2x-1 = 0.
Since sin 18 is positive, we must take the
positive root and we finally get
x = sin 18 = ¼(-1+√5).
It would be interesting to know if x = 18 is the only
x between 0 and 90 such that sin nx* sin x
is rational for some integer n > 1.

Answer 3

cos18=sin72=2*sin36*cos36

cos18=2(2*sin18*cos18)sin(90-36)

Cancel cos18 from either sides

Answer 4

We had to find these values (sin54 and sin18) in my alg2/trig class last year, but I don't remember how it was done. It ended up being fairly simple though.