Prelim Maths Trig Extension 1 (1 Viewer)

-Dark Light- · Apr 25, 2013

Hey guys, can you help me with this question? Thanks.

$1. $Simplify the following$:\frac{3tan(A/3)-tan^{3}(A/3))}{1-3tan^{2}(A/3)}.$

hit patel · Apr 26, 2013

Try using compound trig method where tan (u) from a/3 is 0<u<120.

-Dark Light- · Apr 26, 2013

hit patel said:
Try using compound trig method where tan (u) from a/3 is 0<u<120.

hm...sorry I dont understand. Can you show me what you mean please?

Drongoski · Apr 26, 2013

Actually this is equivalent to the identity:

$tan3 \theta = \frac {3tan \theta - tan3 \theta}{1-2tan^2 \theta}$

$tan 3 \theta = tan(2 \theta + \theta) = \frac {tan 2\theta + tan \theta}{1-tan 2 \theta tan\theta} = \frac {\frac {2tan \theta}{1-tan^2 \theta} + tan \theta}{1 - tan \theta \frac {2tan \theta}{1-tan^2 \theta}} = \frac {2tan \theta +tan \theta (1-tan^2 \theta)}{1-tan^2 \theta - 2tan^2 \theta}$

= RHS

You can simplify your expression by reversing the steps in the above proof; otherwise it is not obvious how one might simplify it.

-Dark Light- · Apr 26, 2013

Drongoski said:
Actually this is equivalent to the identity:

$tan3 \theta = \frac {3tan \theta - tan3 \theta}{1-2tan^2 \theta}$

$tan 3 \theta = tan(2 \theta + \theta) = \frac {tan 2\theta + tan \theta}{1-tan 2 \theta tan\theta} = \frac {\frac {2tan \theta}{1-tan^2 \theta} + tan \theta}{1 - tan \theta \frac {2tan \theta}{1-tan^2 \theta}} = \frac {2tan \theta +tan \theta (1-tan^2 \theta)}{1-tan^2 \theta - 2tan^2 \theta}$

= RHS

You can simplify your expression by reversing the steps in the above proof; otherwise it is not obvious how one might simplify it.

Thanks. But i havent learnt that formula in school yet. Is there another way of doing it? But thanks for ur help, I also did get the answer too!

hit patel · Apr 27, 2013

Dragon show him what I meant its hard to show it like he shows it you know with the fractions and theta. How do you write what you mean in fractions and using theta????

HeroicPandas · Apr 27, 2013

Drongoski said:
Actually this is equivalent to the identity:

$tan3 \theta = \frac {3tan \theta - tan3 \theta}{1-2tan^2 \theta}$

$tan 3 \theta = tan(2 \theta + \theta) = \frac {tan 2\theta + tan \theta}{1-tan 2 \theta tan\theta} = \frac {\frac {2tan \theta}{1-tan^2 \theta} + tan \theta}{1 - tan \theta \frac {2tan \theta}{1-tan^2 \theta}} = \frac {2tan \theta +tan \theta (1-tan^2 \theta)}{1-tan^2 \theta - 2tan^2 \theta}$

= RHS

You can simplify your expression by reversing the steps in the above proof; otherwise it is not obvious how one might simplify it.

Sir, there is also another method

Step 1: let u = A/3 (for simplicity)

Step 2: change all tans into sin/cos

Step 3: common denominate, ssimplify

Step 4: u will end up with:

$\frac{3\sin\theta - 4\sin^3 \theta}{4\cos^3 \theta - 3\cos\theta}$

Step 5: notice triple angle formulae

(which can be proven by the sine and cosine expansions - compound formulas - just like how Drongoski did for tan3theta

$\sin3\theta = 3\sin\theta - 4\sin^3 \theta$
$\cos3\theta = 4\cos^3 \theta - 3\cos\theta$

$\therefore \frac{3\sin\theta - 4\sin^3 \theta}{4\cos^3 \theta - 3\cos\theta} = \frac{\sin3\theta}{\cos3\theta} = \tan3\theta$

Step 6: finally u change all thetas into A/3, and there u have it

-Dark Light- · Apr 27, 2013

HeroicPandas said:
Sir, there is also another method

Step 1: let u = A/3 (for simplicity)

Step 2: change all tans into sin/cos

Step 3: common denominate, ssimplify

Step 4: u will end up with:

$\frac{3\sin\theta - 4\sin^3 \theta}{4\cos^3 \theta - 3\cos\theta}$

Step 5: notice triple angle formulae (which can be proven by the sine and cosine expansions - compound formulas - just like how Drongoski did for tan3theta

$\sin3\theta = 3\sin\theta - 4\sin^3 \theta$
$\cos3\theta = 4\cos^3 \theta - 3\cos\theta$

$\therefore \frac{3\sin\theta - 4\sin^3 \theta}{4\cos^3 \theta - 3\cos\theta} = \frac{\sin3\theta}{\cos3\theta} = \tan3\theta$

Step 6: finally u change all thetas into A/3, and there u have it

YAY, Thanks Heroic!

Prelim Maths Trig Extension 1 (1 Viewer)

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Drongoski

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Heroic!

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