Question regarding a trig equation

  • Thread starter rxh140630
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  • #1
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Homework Statement:

cos(nθ) = cos((n-1)θ)cos(θ) - sin((n-1)θ)sin(θ)

Relevant Equations:

no equation given for this trig equation
See the attached image. Apostol gives cos(nθ) = cos((n-1)θ)cos(θ) - sin((n-1)θ)sin(θ), in the middle of the picture, but previous info given does not state how he got this equation.

To me it looks like he used the equation cos(x+y) = cosxcosy-sinxsiny
 

Answers and Replies

  • #2
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If a moderator could please remove the thread I guess it actually was just using a previous equation that was given. I was too tired and understand why this is the result now.
 
  • #3
SammyS
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Homework Statement:: cos(nθ) = cos((n-1)θ)cos(θ) - sin((n-1)θ)sin(θ)
Relevant Equations:: no equation given for this trig equation

See the attached image. Apostol gives cos(nθ) = cos((n-1)θ)cos(θ) - sin((n-1)θ)sin(θ), in the middle of the picture, but previous info given does not state how he got this equation.

To me it looks like he used the equation cos(x+y) = cosxcosy-sinxsiny
A moderator may come along and remove this thread, as you wish. However, this formula comes directly from applying angle addition identities for cos. I guess you suspected as much.

##\cos(n\theta) = \cos((n-1)\theta+\theta)##

## = \cos((n-1)\theta)\cos(\theta)-\sin((n-1)\theta)\sin(\theta)##
 

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