Solve Trig Substitution Homework with P

AI Thread Summary
The discussion focuses on solving trigonometric expressions using the substitution of sin54 with P. For cos18, it is expressed as 2p√(1-p²), derived from sin108. The second expression simplifies to tan54, which is also expressed as p/√(1-p²). Participants clarify notation and correct minor typographical errors in the calculations. The thread emphasizes the importance of clear mathematical expressions and accurate substitutions in trigonometric identities.
DERRAN
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Homework Statement


If sin54=P, express the following in terms of p, without using a calculator.
1] cos18
2] \frac{tan27+cot63}{1+tan207.cot117}


Homework Equations





The Attempt at a Solution



1] cos18=sin(90-18)
=sin(72)

sin72=sin(180-108)
=sin(180-108)
=sin180.cos108-cos180.sin108
=0-(-sin108)
=sin108

sin108=2.sin54.cos54
=2pcos54.\sqrt{1-p^{2}}



2] =\frac{2tan27}{1-tan27.tan27}

=\frac{2tan27}{1-tan^{2}27}

=tan2(27)

=tan54

=\frac{p}{\sqrt{1-p^{2}}}
 
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DERRAN said:

Homework Statement


If sin54=P, express the following in terms of p, without using a calculator.

"P" or "p"? You shouldn't use both to mean the same thing.

1] cos18
2] \frac{tan27+cot63}{1+tan207.cot117}


Homework Equations





The Attempt at a Solution



1] cos18=sin(90-18)
=sin(72)

sin72=sin(180-108)
=sin(180-108)
=sin180.cos108-cos180.sin108
=0-(-sin108)
=sin108

sin108=2.sin54.cos54
=2pcos54.\sqrt{1-p^{2}}
Well, you mean, of course,
sin(108)= 2p\sqrt{1- p^2}
Also you haven't finished the problem. Yes, cos(18) is equal to that but you should write that explicitely:
cos(18)= 2p\sqrt{1- p^2}
2] =\frac{2tan27}{1-tan27.tan27}

=\frac{2tan27}{1-tan^{2}27}

=tan2(27)

=tan54

=\frac{p}{\sqrt{1-p^{2}}}
I see nothing wrong with (2).
 
Thanks, I made a few typing errors.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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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