Writing A Trig Expression as an Algebraic Expression

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Homework Help Overview

The problem involves rewriting a trigonometric expression, specifically cos(2arccos(2x)), as an algebraic expression. The subject area is trigonometry and its relationship with algebraic expressions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of inverse properties and the implications of the coefficient in front of the arccos function. There are attempts to express the problem in terms of a variable u, and some participants explore the relationship between cosine and sine in the context of double angle formulas.

Discussion Status

The discussion includes various interpretations of the problem, with some participants expressing confusion about the necessity of certain steps. There is mention of a specific answer from a textbook, but no consensus on the approach to reach that answer has been established.

Contextual Notes

Some participants question the relevance of the Pythagorean theorem in this context, and there is uncertainty about the implications of the expression cos(2arccos(2x)). The presence of a specific answer in the textbook adds a layer of complexity to the discussion.

themadhatter1
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Homework Statement



Write the Trigonometric Expression as an algebraic expression.

cos(2arccos 2x)

Homework Equations



Probably the inverse properties, I'm not sure.

The Attempt at a Solution



I know I can rewrite this equation as.

u= arccos 2x

cos(2cos u=2x)

I can also say that the adjacent leg is 2x units long and the hypotenuse is 1 unit long. Then using the pythagorean theorem I can figure the opposite leg to be sqrt(1-4x2)

I'm not sure If this is necessary though can someone point me in the right direction? The 2 in front of the arccos is throwing me off because if that wasn't there I would just use the inverse property and cos(arccos 2x) would equal 2x.
 
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If u=cos-1(2x) then you want to find cos(2u).

cos(2u)=cos2u-sin2u=2cos2u-1 = 1-2sin2u

and cos2u = (cosu)2
 
I'm not sure I understand why you'd want to find cos(2u)

The answer is supposed to be 8x2-1 and that's the answer listed in the back of the book.
 
themadhatter1 said:
I'm not sure I understand why you'd want to find cos(2u)

The answer is supposed to be 8x2-1 and that's the answer listed in the back of the book.

cos(2cos-1(2x))

if you put u = cos-1(2x), wouldn't cos(2cos-1(2x)) become cos(2u)?
 
Thanks, now I understand.

Once you have simplified it to 2cos2u-1 all you have to do is simplify it with the u in place

cos2(arcsin 2x)=2x

2(2x)2-1

8x2-1

Thanks!
 

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