Year 11 MIF: Question 15g of Exercise 6.15 (1 Viewer)

Ambility · May 23, 2015

I know, MIF is shit, use cambridge yada yada yada... Get over yourself.

Sums and Differences of Angles
$$Simplify: $2cos^23\alpha-1$

If anyone can guide me in solving this problem, I'd appreciate it.

InteGrand · May 23, 2015

Ambility said:
I know, MIF is shit, use cambridge yada yada yada... Get over yourself.

Sums and Differences of Angles
$$Simplify: $2cos^23\alpha-1$

If anyone can guide me in solving this problem, I'd appreciate it.

$We use the trigonometric identity $\cos ^2 \theta = \frac{1}{2} + \frac{1}{2}\cos 2\theta \Rightarrow 2\cos^2 3\alpha -1 = 2\left(\frac{1}{2}+\frac{1}{2}\cos 6\alpha\right) -1 = \cos 6\alpha$.$

Drongoski · May 23, 2015

InteGrand · May 23, 2015

Actually just realised, this trig. identity is what they really wanted: $\cos 2\theta = 2\cos^2 \theta -1$ (can do it in one step then).
Edit: done above

Ambility · May 23, 2015

Thanks guys. I didn't realise that identity existed. I guess it's just a combination of the pythagorean identity and the double angle cosine identity.

InteGrand · May 23, 2015

Ambility said:
Thanks guys. I didn't realise that identity existed. I guess it's just a combination of the pythagorean identity and the double angle cosine identity.

Correct, and they're important ones to know for HSC ( $\sin^2 x = \frac{1}{2}-\frac{1}{2}\cos 2x$ and $\cos ^2 x = \frac{1}{2}+\frac{1}{2}\cos 2x$ ). (For example, they allow you to integrate squares of sines and cosines.)

Drsoccerball · May 23, 2015

Assuming you're doing extension 1*

Year 11 MIF: Question 15g of Exercise 6.15 (1 Viewer)

Ambility

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InteGrand

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InteGrand

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Ambility

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