[SOLVED] Proof help!

bsothe1 · December 6th 2007, 07:11 PM

cot2x = (cosx+cos3x)/(sin3x+sinx) please list all steps. It's REALLY hard .thx

**ticbol** · December 7th 2007, 12:25 AM

cot(2x) = (cos(x) +cos(3x)) / (sin(3x) +sin(x)) -------(i)
Or, rewriting that,
cot(2x) = [cos(3x) +cos(x)] / [sin(3x) +sin(x)] -------(ii)

RHS of (ii) =
= [cos(3x) +cos(x)] / [sin(3x) +sin(x)]

Using trig identities:
cosA +cosB = 2cos((A+B)/2)cos((A-B)/2)
sinA +sinB = 2sin((A+B)/2)cos((A-B)/2),

= [2cos((3x+x)/2)cos((3x-x)/2)] / [2sin((3x+x)/2)cos((3x-x)/2)]
= cos(2x) / sin(2x)
= cot(2x)
= LHS

So, proven.

**Soroban** · December 7th 2007, 06:52 AM

Hello, bsothe1!

If we are allowed the Sum-to-Product identities, it's easy.

. . $\sin A + \sin B \:=\:2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left( \frac{A-B}{2}\right)$

. . $\cos A + \cos B \:=\:2\cos\!\left(\frac{A+B}{2}\right)\cos\!\left( \frac{A-B}{2}\right)$

$\cot2x \:= \:\frac{\cos3x + \cos x}{\sin3x+\sin x}$

The numerator is: . $\cos3x + \cos x \:=\: 2\cos\!\left(\frac{3x+x}{2}\right)\cos\!\left(\fra c{3x-x}{2}\right) \:=\; 2\cos2x\cos x$

The denominator is: . $\sin3x + \sin x \:=\:2\sin\!\left(\frac{3x+x}{2}\right)\cos\!\left (\frac{3x-x}{2}\right) \:=\:2\sin2x\cos x$

The right side becomes: . $\frac{2\cos2x\cos x}{2\sin2x\cos x} \;=\;\frac{\cos2x}{\sin2x} \:=\:\cot2x$

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