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Show that sin 2x=2tan x/(1+tan^2x)?
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We'll manage the LHS.
We'll write the formula that writes the sine of double angle in terms of sine and cosine of the angle.
sin 2x = 2 sin x*cos x
Now, we'll multiply and divide by cos x the right side, to create the tangent function:
sin 2x = 2 sin x*cos x*cos x/cos x
sin 2x = 2 tan x*`cos^(2)` x
But, from Pythagorean identity, we'll have:
1 + `tan^(2)` x = 1/`cos^(2)` x => `cos^(2)` x = 1/(1+`tan^(2)` x)
sin 2x = 2 tan x/(1 + `tan^(2)` x)
Therefore, the identity sin 2x = 2 tan x/(1 + `tan^(2)` x) is verified.
Posted by giorgiana1976 on August 16, 2011 at 3:01 PM (Answer #1)
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L:H:S = sin2x
= 2sinx.cosx ÷ 1
Devide the numerator and denominator by cos²x
= (2sinx.cosx/cos²x) ÷ (1/cos²x)
= 2sinx/cosx ÷ sec²x
= 2tanx ÷ (1+tan²x)
Posted by lochana2500 on July 5, 2012 at 3:39 PM (Answer #2)
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