Help with Trig Question?! (1 Viewer)

fartman · Nov 30, 2011

a) Write down, or derive, an expression for cos3x in terms of cosx
b) Explain why, if x = 18, then sin2x = cos3x
c) prove that sin18 = (sqrt(5) - 1)/4

I got part a and b but how to do c

SpiralFlex · Nov 30, 2011

a) $\cos 3x = \cos (2x+x)$

$=\cos 2x \cos x - \sin 2x \sin x$

$=(2\cos^2 x - 1) \cos x - 2 (1-\cos^2 x) \cos x$

$=2 \cos ^3 x - \cos x - 2 \cos x + 2 \cos^3 x$

$=4 \cos ^3 x-3 \cos x$

$\therefore \cos 3x = 4 \cos ^3 x-3 \cos x$

b) $\sin 36=\cos 54$

This is the case because of complementary angles,

$\sin \theta = \cos (90-\theta)$ (And vise versa.)

c) Let $x=18$

We have shown before,

$\sin 2(18)= \cos 3(18)$

Now,

$2 \sin x \cos x = 4 \cos^3 x - 3 \cos x$

$2 \sin x=4 \cos^2 x - 3$

$2 \sin x =4(1-\sin^2 x)-3$

$4 \sin^2 x +2 \sin x -1=0$

$\sin x = \frac{-2 \pm \sqrt{20}}{8}$

$=\frac{-1 \pm \sqrt{5}}{4}$

It must be positive! Since $\sin 18$ is in the first quadrant!

And before we let $x=18$ . So,

$\therefore \sin 18=\frac{ \sqrt{5}-1}{4}$

fartman · Nov 30, 2011

wow! ur a legend, thanks heaps

Help with Trig Question?! (1 Viewer)

fartman

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SpiralFlex

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fartman

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