Writing A Trig Expression as an Algebraic Expression

AI Thread Summary
The discussion focuses on rewriting the trigonometric expression cos(2arccos 2x) as an algebraic expression. Participants clarify that by letting u = arccos(2x), the expression simplifies to cos(2u). They utilize the double angle formula, cos(2u) = 2cos²(u) - 1, leading to the final result of 8x² - 1. The initial confusion about the presence of the '2' in front of arccos is resolved through this simplification process. Ultimately, the correct algebraic expression is confirmed as 8x² - 1.
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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 theorm 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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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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