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our assignment was give 10 best proved trigonometric identities. and our teacher said that it must be an above normal trig identity that is already proven...

what websites can you recommend? or if anyone can help me prove this:
2tanx/1-tan(pow2)x + 1/2cos(pow2)x-1 = cosx+sinx/cosx-sinx

thanks!!

2007-11-01 18:04:51 · 1 answers · asked by AeXanzzZ_gurL 2 in Science & Mathematics Mathematics

1 answers

Required to Prove :
2tan(x) / [1-tan^2(x)] + 1 / [2cos^2(x) -1]
= [cos(x) + sin(x)] / [cos(x) - sin(x)]
--------------------
First fiddle with the left-hand side :

LHS = tan(2x) + 1 / cos(2x), as these are known identities.

= sin(2x) / cos(2x) + 1 / cos(2x), known identity for tan.

= [1 + sin(2x)] / cos(2x)
--------------------
Now for the right-hand side :

Multiply RHS by [cos(x) - sin(x)] / [cos(x) - sin(x)],
that is, multiply numerator and denominator by the same.

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

= cos(2x) / [1 - sin(2x)]
--------------------
Now see if LHS = RHS :

Let [1 + sin(2x)] / cos(2x) = cos(2x) / [1 -sin(2x)]

Cross-multiply :

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

cos^(2x) = 1 - sin^2(2x)

Add sin^2(2x) to each side :

sin^2(2x) + cos^2(2x) = 1

This we know is an identity for any angle,
so the assumption was correct that LHS = RHS.

This was the best site I could find for identities :
http://web.mit.edu/wwmath/trig/identities02.html

2007-11-03 00:06:02 · answer #1 · answered by falzoon 7 · 0 0

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