Use trig identities to simplify an expression (has sins and cosines)

AI Thread Summary
The discussion focuses on simplifying the expression (sinx)^2 - (cosx)^2 over (sinx)^2 - (sinx cosx) using trigonometric identities. The user initially struggles to simplify the expression and seeks clarification on how the answer "1 + cotx" is derived. Suggestions include factoring the numerator as a difference of squares and multiplying both the numerator and denominator by 1/sin^2(x). Ultimately, the user reports success in understanding the simplification process. The conversation highlights the importance of applying fundamental trigonometric identities effectively.
Nishiura_high
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Homework Statement



Use fundamental identities to simplify the expression:

(sinx)^2 - (cosx)^2
____________________
(sinx)^2 - (sinx cosx)*note: it's a numerator and denominator. The underscore line is the fraction line.

*note: The answer in the back of the book is "1 + cotx" but I would like to know how it got there.

Homework Equations



(sinx)^2 + (cosx)^2 = 1

other trig identities

The Attempt at a Solution



(sinx)^2 - (cosx)^2
_________________
sinx(sinx - cosx)

I factored out sinx out of the bottom, but I don't really see any identies that would simplify sinx-cosx. (I have a chart of identities.) I tried to simplify the top using the relevant identity I already listed.

Thanks for any help!
 
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Nishiura_high said:

Homework Statement



Use fundamental identities to simplify the expression:

(sinx)^2 - (cosx)^2
____________________
(sinx)^2 - (sinx cosx)

*note: it's a numerator and denominator. The underscore line is the fraction line.

*note: The answer in the back of the book is "1 + cotx" but I would like to know how it got there.

Homework Equations



(sinx)^2 + (cosx)^2 = 1

other trig identities


The Attempt at a Solution



(sinx)^2 - (cosx)^2
_________________
sinx(sinx - cosx)

I factored out sinx out of the bottom, but I don't really see any identies that would simplify sinx-cosx. (I have a chart of identities.) I tried to simplify the top using the relevant identity I already listed.

Thanks for any help!
Factor the numerator as a difference of squares.

or ...

Starting with the original expression, multiply the numerator and denominator by 1/sin2(x)
 
Thanks. I got it now. :)
 
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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