Mathematics
posted by Edwards .
1. Show that Sin3è  Cos2è = (1  Sinè)(4Sin^2è + 2Sinè  1).
Without using a calculator, show that è = 18 degrees is an exact solution of the equation Sin3è = Cos2è.
Justifying your answer, find the exact values of:
i) Sin18 degrees.
ii) Sin234 degrees.

I will use x instead of è
sin 3x  cos 2x = (1sinx)(4sin^2 x + 2sinx  1)
RS = 4sin^2 x + 2sinx  1  4sin^3 x + 2sin^2 x + sinx
= 4sin^3 x + 2sin^2 x + 3sinx  1
LS = sin(2x+x)  cos 2x
= (sin 2x)(cosx) + (cos 2x)(sinx)  (1  2sin^2 x)
= 2sinxcosxcosx + sinx(12sin^2 x)  1 + 2sin^2 x
= 2sinxcos^2 x + sinx  2sin^3 x  1 + 2sin^2 x
= 2sinx(1  sin^2 x) + sinx  2sin^3 x  1 + 2sin^2 x
= 2sinx  2sin^3 x + sinx  2sin^3 x  1 + 2sin^2 x
= 4sin^3 x + 2sin^2 x + 3sinx  1
= RS
For your next question, show that 18° is a solution to
sin 3x = cos 2 or
sin 3x  cos 2x = 0
since I expanded this expression in LS above
we have to solve
4sin^3 x + 2sin^2 x + 3sinx  1 = 0 or
4sin^3 x  2sin^2 x  3sinx + 1 = 0
let sinx = y
4y^3  2y^2  3y + 1 = 0
clearly y = 1 is a solution
(y1)(4y^2 + 2y 1) = 0
the other roots are (1 ± √5)/4
so sinx = (1 ± √5)/4
This also answers your secondlast question.
that sin 18° = (√5  1)/4
I now have to show that this equals sin 18° without a calculator, (using my calculator shows me that so far I am correct, since 18 would be a solution using my machine)
Working on this ...... 
sin 18° = cos 72° by the complementary property.
going out on a farfetched limb here .......
draw a pentagon ABCDE
draw diagonals AC and BD to intersect at P
Look at triangle ABC, angle B = 108 and angles A and C are 36° each.
If AB = 1, I happen to know that the diagonal : side = the golden ratio
which is (1 + √5)/2 : 1
so let AB = 2, then AC = √5 + 1
using the cosine law in triangle ABC
2^2 = 2^2 + (√5+1)^2 – 2(2)(√5+1)cos 36°
4 = 4 + 5 + 2√5 + 1 – 4(√5+1)cos 36
cos 36 = (6+2√5)/(4(√5+1))
= (√5 + 1)/4 after rationalizing the denominator.
Now cos 72 = 2cos^2 36°  1 , using cos 2A = 2cos^2 A  1
= 2[(√5+1)/4]^2 – 1
= 2(5 + 2√5 + 1)/16 – 1
= (12 + 4√5)/16 – 16/16
= (4 + 4√5)/16
= (1 + √5)/4
but sin 18 = cos 72, as noted above
so sin 18° = (1 + √5)/4 
for your last one:
234 = 180 + 54
and 54 = half of 108 which is the large angle in the pentagon.
sin234 = sin(180 + 54) = sin180cos54 + cos180sin54
= 0  sin54
= sin 54°
you can find cos 108° using the cosine law in triangle ABC
then
cos 108 = 1  2sin^2 54°
give it a shot.