Thursday

July 31, 2014

July 31, 2014

Posted by **Jake** on Tuesday, February 13, 2007 at 2:45am.

Use the formula

tan (a + b) = (tan a + tan b)/[1 - tan a tan b)

in two steps. First, let a = b = theta and get a formula for tan (2 theta).

tan (2 theta) = 2 tan theta/[(1 - tan theta)^2]

Then write down the equation for

tan (2 theta + theta)

Are you sure you are supposed to use complex numbers to answer this question? My previous answer used a trigonometric identity. I don't see a way to use complex numbers, but there probably is a way.

Exp(3 i theta) = [Exp(i theta)]^3 --->

cos(3 theta) = c^3 - 3cs^2

sin(3 theta) = 3c^2s - s^3

where c = cos(theta) and s = sin(theta)

This means that:

tan(3 theta) =

[ 3c^2s - s^3]/[c^3 - 3cs^2]

divide numerator and denominator by c^3:

tan(3 theta) =

[ 3t - t^3]/[1 - 3t^2] =

where t = tan(theta)

Thanks everyone for your help. Yes I did have to use complex numbers. Thanks Count I really should have seen your method for myself as I had found cos 3theta and sin 3 theta in an earlier problem.

Thanks again Jake

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