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sorry i meant tan3x - tanx = 2sinxsec3x
this is a tricky - you might try converting to functions of sin and cos
i've spent a lot of time trying diff techniques, splitting 3x in sinx and cosx components but am getting no where fast!
yeah - i've got a headache trying this one but I've got to go out now. I'll look at it later. Maybe someone else can help meantime. I'd like to see the proof of this one. Meanwhile - the best of luck.
thanks for trying
ok steanson - I've got it first convert to sin and cos: sin3x/cos3x -six/cosx this can be written as (sin3xcosx - sinxcos3x) / cos3xcosx by the 'sums and differences of trig ratios formulae: the numerator = 1/2(sin4x +sin2x) - 1/2(sin4x- sin2x) which reduces to sin2x which we can write as 2sinxcosx sowe have: 2 sinxcosx/cos3xcosx cancelling out cosx = 2sinx/cos3x =2sinxsec3x
the rules i referred to are sinx + siny = 2sin[(x+y/2][cos(x-y)/2] and sinx - siny = 2cos[(x+y/2][sin(x-y)/2] and I used them from' right to left' A good way to remember these is eg the first one 'sinx + siny = 2 sin semi sum cos semi difference' there are similar rules for the sum of differences of cosines. A higher grade maths book should have them
thanks very much!, i think the reason i struggled is i didn't have those sinx + siny identities given in my text, handy to know!