Solving for cos 2v and sin 2v: Quick Guide

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Homework Help Overview

The discussion revolves around solving the equations cos 2v = cos v and sin 2v = sin v, which fall under trigonometric identities and double angle formulas.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of double angle formulas, with one suggesting the transformation of sin 2v into 2(sin v)(cos v). Others discuss the implications of these transformations and the potential for quadratic equations.

Discussion Status

The discussion is active, with various approaches being considered. Some participants provide insights into the use of double angle identities, while others question the necessity of revisiting previously answered queries. There is no explicit consensus on a single method, but multiple interpretations are being explored.

Contextual Notes

One participant notes the importance of considering the range of values for variables involved, specifically mentioning constraints on the variable t in the context of cosine values.

Maria
How do I solve these two:

cos 2v = cos v

sin 2v = sin v
 
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Double angle formulas, e.g. sin 2x = 2(sin x)(cos x)
So your second equation becomes:
2(sin x)(cos x) = sin x
 
cos2v=2(cosv)^2-1
t=cosv and solve equation.
The second I remmember you already asked and were properly answered by matt and mods, so why "repeat" it ?:confused:

[dont forget to choose only t's in the range [-1,1] only]
 
Maria said:
How do I solve these two:

(1) cos 2v = cos v

(2) sin 2v = sin v

Does it go like this?

(1) cosv = cos2v = cos( v + v) = cosv cos v - sinv sinv

= cosv [cos v] - [sinv sinv] = cosv[ 1] - [0] if it is to equal cosv, as given.

Therefore need: [cosv] = 1 and [sinv] = 0; so v = 0.

[PS: Your teacher may prefer Motifs' suggestion where you solve a standard quadratic equation.]

(2) is similar, but more interesting. Good luck.
 
Last edited:

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