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Prove this identity: tan x = (sin (x + 2pi))/(cos(x - 2pi))
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The sine and cosine functions are periodic with a periodicity equal to 2*pi. For any angle x, sin x = sin (x + n*2*pi) and cos x = (cos x + n*2*pi) with n being any integer.
sin (x + 2*pi) = sin x
cos (x - 2*pi) = cos x
(sin (x + 2*pi)) /cos (x - 2*pi)
=> sin x / cos x
=> tan x
This proves the given identity.
Posted by justaguide on April 17, 2011 at 11:43 AM (Answer #1)
High School Teacher
Best answer as selected by question asker.
tan(x) = sin(x+2pi) / cos(x-2pi)
We will use trigonometric identities to prove the equality.
We know that:
sin(a+ b) = sina*cosb+ cosa*sinb
==> sin(x+2pi) = sinx*cos2pi + cosx*sin2pi
==> sin(x+2pi) = sinx*1 + cosx* 0 = sinx,..........(1)
Now we know that:
cos(a-b) = cosa*cosb- sina*sinb
==> cos(x-2pi) = cosx*cos2pi + sinx*sin2pi
= cosx*1 + sinx*0 = cosx.............(2)
Now, from (1) and (2) we conclude that:
sin(x+2pi)/cos(x-2pi) = sinx/cos x = tanx .........q.e.d
==> Then we prove that tanx = sin(x+2pi)/cos(x-2pi)
Posted by hala718 on April 17, 2011 at 11:46 AM (Answer #2)
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