Trigonometric identities

Please help with any of the following..

Prove these identities

cos2x = (1 - tan^2 x)/(1 +tan^2 x)


cos2x = cos^4 x - sin^4 x


cosec2x + cot2x = cot x


sin2x = 2tanx / (1 + tan^2 x)


tan x = sqrt ((1 - cos2x) / (1 + cos2x))


Thankyou for your time and help

Quote:

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Originally Posted by BeckyDWood

Please help with any of the following..

Prove these identities

cos2x = (1 - tan^2 x)/(1 +tan^2 x)

note that 1 + tan^2 x = sec^2 x


cos2x = cos^4 x - sin^4 x

factor the right side


cosec2x + cot2x = cot x

left side ...

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

[1 + cos(2x)]/sin(2x)

now use cos(2x) = 2cos^2 x - 1 and sin(2x) = 2sinxcosx



sin2x = 2tanx / (1 + tan^2 x)

same hint as the first


tan x = sqrt ((1 - cos2x) / (1 + cos2x))

on the right side, try these substitutions ...

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

denominator ... cos(2x) = 2cos^2 x - 1

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use this identity: (cos^2 x + sin^2 x = 1)

i) cos2x = cos^4 x - sin^4 x, use factoring by difference of 2 squares;

RHS = (cos^2 x - sin^2 x)(cos^2 x + sin^2 x) = cos^2 x - sin^2 x = cos 2x = LHS


ii) csc 2x + cot 2x = cot x,

use: csc y = 1/sin y & cot y = cos y/sin y;

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

use these identities: (2cos^2 x = 1 + cos 2x) & (sin 2x = 2sin x cos x)

LHS = (2cos^2 x)/(sin 2x)

LHS = 2(cos x)(cos x)/(2 sin x cos x)

LHS = 2(cos x)/(2 sin x)

LHS = (cos x)/(sin x)

LHS = cot x

LHS = RHS

RHS = sin2x = 2tanx / (1 + tan^2 x) = (2tan x)/(sec^2 x) = (2sin x/cos x)/[(1/cos x)(cos x)] = (2 sin x / cos x)(cos x)(cos x)

= (2 sin x cos x) = sin 2x = LHS