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STEP I 2003 q

I was looking through a solution someone on here posted to a q and I'm not sure I follow...

Can someone pls explain the highlighted step and how you'd go on from there?

Thank you! :smile:

Original post by Horizontal 8

STEP I question 3

2cos\left(\frac{x}{2}\right) = cosx \iff 2cos\left(\frac{x}{2}\right) = 2cos^2\left(\frac{x}{2}\right)-1

\iff 2cos^2\left(\frac{x}{2}\right)-2cos\left(\frac{x}{2}\right)-1 =0

\displaystyle \iff cos\left(\frac{x}{2}\right) = \frac{2 \pm \sqrt{12}}{4} = \frac{1 \pm \sqrt{3}}{2}

But

\displaystyle \left|\frac{1 + \sqrt{3}}{2} \right|>1

Hence no solutions arise from this.

Also,

\displaystyle \frac{1 - \sqrt{3}}{2} < 0

So we use

\displaystyle \frac{\sqrt{3}-1}{2}

to find the other angles.

Cosx is -ve for

\displaystyle \frac{\pi}{2}<x<\frac{3 \pi}{2}

when considering 0<x<2pi

Therefore the angles that we want are:

\displaystyle \frac{x}{2} = (2n+1)\pi \pm cos^{-1}\left(\frac{\sqrt{3} -1}{2} \right)

\implies \displaystyle x = (4n+2)\pi \pm 2cos^{-1}\left(\frac{\sqrt{3} -1}{2} \right)

And this is equal to

Unparseable latex formula:

\boxed{\displaystyle (4n+2)\pi \pm 2 \phi \right)}

As required.

Reply 1

Ree69

Notice that

\cos\theta \equiv \cos(-\theta)

Reply 2

shanban

hmm I still don't understand why you wouldn't just use the cosx instead of cos(-x) to work through the remainder of the question???

Reply 3

z0tx

Original post by shanban

hmm I still don't understand why you wouldn't just use the cosx instead of cos(-x) to work through the remainder of the question???

I think it is because the interval you are dealing with in the equation is

[0;\pi/2]

. Since

\displaystyle \frac{1 - \sqrt{3}}{2} < 0

, you rule it out and use the equivalent positive value, which is

\displaystyle \frac{\sqrt{3}-1}{2}

(edited 11 years ago)

Reply 4

shanban

Original post by z0tx

I think it is because the interval you are dealing with in the equation is

[0;\pi/2]

. The absolute value gives two possibilities. Since

\displaystyle \frac{1 - \sqrt{3}}{2} < 0

, you rule it out and use the other (positive) value, which is

\displaystyle \frac{\sqrt{3}-1}{2}

oh of course! got it, thank you :smile:

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STEP I 2003 q

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