Stuck :( (1 Viewer)

Faera

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Okay guys, it's a pretty stupid question, because of the not being allowed to use the double angle rule, but I just cannot see any other way of doing it!

If you can come up with a solution, it would be absolutely wonderful!

Thanks in advance.


Prove that:
tan[2@] + cot@ = sec[2@] * cot@

Without using double angel results.


thanks!
 

Calculon

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sin2@/cos2@ + cos@/sin@ = cos@/(cos2@sin@)
(sin2@sin@ + cos@cos2@)/sin@cos2@=cos@/cos2@sin@
sin2@sin@ + cos2@cos@ = cos@
sin2@tan@ + cos2@= 1
sin2@tan@ + cos2@ = sin<sup>2</sup>2@ + cos<sup>2</sup>2@
tan@ = sin2@ + cos2@

and now I'm lost. Try the t method maybe?
 
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CM_Tutor

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You could use calculus - show that d/d@(LHS) = d/d@(RHS), and so LHS = RHS + C, for some constant C. Then, by putting @ = pi / 6, show that C = 0.

Note - I haven't written this out, so I'm not sure that you can easily show d/d@(LHS) = d/d@(RHS) without double angle formulae. :)
Originally posted by [FTHRW8]
t method?
This is another possibiliy, but it is arguable as the t-method is really based on double angle formulae.
 

Faera

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Thanks for the help, guys.

CM's method works! (very smart!)

yayyyyyy!

=D

thanks!
 

Affinity

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What about straight trig.

cos(2x - x) = cos(x)

cos(2x)cos(x) + sin(2x)sin(x) = cos(x)

divide both sides by cos(2x)sin(x)

cos(x)/sin(x) + sin(2x)/cos(2x) = cos(x)/sin(x) * 1/cos(2x)

cot(x) + tan(2x) = cot(x) * sec(2x)
 

CM_Tutor

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Originally posted by Affinity
cos(2x - x) = cos(x)

cos(2x)cos(x) + sin(2x)sin(x) = cos(x)
How is using the result cos(2x - x) = cos(2x)cos(x) + sin(2x)sin(x) not using a double angle formula?
 

Affinity

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the double angle formulae are

cos(2x) = [cos(x)]^2 - [sin(x)]^2
sin(2x) = 2sin(x)cos(x)
tan(2x) = 2tan(x)/(1-tan(x)^2)

the result there is a difference of angles formula :D
 

CM_Tutor

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OK, but I think that is hair splitting. :) Anyway ...
 

Xayma

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Hmm great now theres argument about meaning of words in Maths, BOS is going to far.
 

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