Posted by Edwards on Wednesday, August 3, 2011 at 9:27am.
I will use x instead of è
sin 3x - cos 2x = (1-sinx)(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(1-2sin^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
(y-1)(4y^2 + 2y -1) = 0
the other roots are (-1 ± √5)/4
so sinx = (-1 ± √5)/4
This also answers your second-last 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 far-fetched 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.
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