Prove Trig Identity: 2arccos(x) = arccos(2x²-1) | Solutions!

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Discussion Overview

The discussion revolves around proving the trigonometric identity 2arccos(x) = arccos(2x² - 1). Participants explore various approaches to demonstrate this identity, including the use of cancellation laws and known trigonometric identities. The scope includes mathematical reasoning and technical explanations related to trigonometric functions.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests starting by taking the cosine of both sides to derive the equation cos(acos(x) + acos(x)) = 2x² - 1.
  • Another participant proposes letting θ = acos(x) and transforming the expression into 2θ = acos(2(cosθ)² - 1), then using the known identity cos(2θ) = 2cos²θ - 1.
  • A different approach involves using the double angle identity directly, stating that cos(2θ) = 2cos²θ - 1, and manipulating the equation to show the identity holds.
  • One participant applies the cos(A+B) expansion to derive the left-hand side as 2cos²(acos(x)) - 1.
  • Another participant raises a question about the validity of the identity cos(2A) = 2cos²A - 1, expressing uncertainty about whether it holds for all real numbers.
  • A later reply asserts that the identity is indeed valid for all A, referencing the Pythagorean identity and the cosine addition formula.

Areas of Agreement / Disagreement

Participants present multiple approaches to proving the identity, with no consensus on a single method being definitive. There is also a separate discussion regarding the validity of the identity cos(2A) = 2cos²A - 1, with differing levels of certainty expressed.

Contextual Notes

Some participants' reasoning depends on the understanding of trigonometric identities and the cancellation laws, which may not be universally agreed upon. The discussion includes various assumptions about the values of x and A, which are not explicitly defined.

Oxymoron
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How would you prove using the cancellation laws 2arccos(x) = arccos(2x² - 1). I am stumped. Any guidance is appreciated.
 
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Start by taking cos of both sides to get :

cos ( acos(x) + acos(x) ) = 2x^2 -1

It's pretty easy from there, so I'll leave itfor you to have a go. :)
 
What if I let θ = acos(x) which implies that x = cosθ

Then the expression becomes...

2θ = acos(2(cosθ )² - 1)
2θ = acos(2cos²θ - 1)

Then take cos of both sides and use the cancellation law...

cos2θ = 2cos²θ - 1

This is a well known trigonometric identity. So if we start from this and work backwards the identity in the question can easily be shown.
 
Start with the double angle identity

cos(2θ) = 2cos²θ -1

Then take the square outside the brackets

cos(2θ) = 2(cosθ )² -1

Since cos(acos(x)) = x we may write it as

cos(2θ) = cos(acos(2(cosθ )² -1))

Now take inverse cos of both sides

acos(cos(2θ)) = acos(cos(acos(2(cosθ )² -1)))

Cancel out the acos(cos(x)) terms

2θ = acos(2(cosθ )² -1)

Let θ = acos(α) implies α = cosθ

2acos(α ) = acos(2α² - 1)

QED
 
I just used the usual cos(A+B) expansion on the LHS to give :

cos^2( acos(x) ) - sin^2 (acos(x))

and then added zero in the form of cos^2(.) + sin^2(.) - 1 to the LHS to give,

2 cos^2( acos(x) ) - 1
 
Another question.

For what values of A is the trigonometric identity cos2A = 2cos²A - 1 valid? I thought it valid for all real numbers. But there must be a trick??

Any hints would be appreciated. Thanks.
 
Oxymoron said:
For what values of A is the trigonometric identity cos2A = 2cos²A - 1 valid? I thought it valid for all real numbers. But there must be a trick??
It is valid for all A. It follows from the pythagorean identity and the identity cos (A+B) = cos A cos B - sin A sin B. That identity gives cos 2A = cos (A+A) = cos² A - sin² A. Adding on cos² A + sin² A - 1 (which is 0 by the pythagorean identity) to the right side gives the identity cos 2A = 2cos² A - 1
 

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