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- anonymous

Not sure the right approach to solve the identity:
tan3x - tanx = 2sinxsecx

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- anonymous

Not sure the right approach to solve the identity:
tan3x - tanx = 2sinxsecx

- jamiebookeater

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- Owlfred

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- anonymous

sorry i meant
tan3x - tanx = 2sinxsec3x

- anonymous

this is a tricky - you might try converting to functions of sin and cos

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- anonymous

i've spent a lot of time trying diff techniques, splitting 3x in sinx and cosx components but am getting no where fast!

- anonymous

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.

- anonymous

thanks for trying

- anonymous

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

- anonymous

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

- anonymous

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!

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