Further Trig question (1 Viewer)

Hermes1 · Aug 10, 2011

This question is from the Fort Street 2010 3U trial paper:

i) show that: $\cos 3x = 4\cos ^{3}x-3\cos x$ - ive done this part

ii) Show that the solution of cos 3x - sin 2x = 0 between 0 and pi/2 is given by

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

Deep Blue · Aug 10, 2011

$4\cos ^3 x - 3cosx - 2sinxcosx = 0$

$\cos x \left ( 4\cos^2 x - 2sin x - 3 \right ) = 0$
$\therefore cosx = 0 \or \4(1 - sin^2 x ) -2sinx - 3 = 0$

Then solve for sinx = 0

Hermes1 · Aug 10, 2011

Deep Blue said:
$4\cos ^3 x - 3cosx - 2sinxcosx = 0$

$\cos x \left ( 4\cos^2 x - 2sin x - 3 \right ) = 0$
$\therefore cosx = 0 \or \4(1 - sin^2 x ) -2sinx - 3 = 0$

Then solve for sinx = 0

lol im such an idiot, tnks for the help.

SpiralFlex · Aug 10, 2011

We can rewrite,

$4\cos^3 x - 3\cos x = \cos x(4\cos^2 x - 3)$

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

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

We know that,

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

Now,

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

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

$\sin x = \frac{-1 + \sqrt{5}}{4}$

Edit: Deep blue is too quick. I am getting old.

Hermes1 · Aug 10, 2011

Okay so the rest of the question was:

verify that x = pi/10 is a solution of cos 3x = sin 2x (I did this using general solutions

and then:
prove $\sin \frac{\pi }{5}\cos \frac{\pi }{10}=\frac{\sqrt{5}}{4}$

this is wat i did:

$\sin \frac{\pi }{5}\cos \frac{\pi }{10} = 2\sin \frac{\pi }{10}\cos^{2} \frac{\pi }{10}$
$2\sin \frac{\pi }{10}\cos^{2} \frac{\pi }{10} = 2\sin \frac{\pi }{10}(1-\sin ^{2}\frac{\pi }{10})$

and then using the result: $sin \frac{\pi }{10} = \frac{\sqrt{5}-1}{4}$
i subbed it in and got the answer
is this the quickest way?

Further Trig question (1 Viewer)

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