Prove that the equation is an identity. State any restrictions.

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The equation cos(x)^2/(1+3sin(x)-4sin(x)^2)=(1+sin(x))/(1+4sin(x)) is being analyzed for identity verification. The denominator has been factored correctly into (1 - sin(x))(1 + 4sin(x)). To simplify the left side, it is suggested to use the identity cos^2(x) = 1 - sin^2(x). Further manipulation of both sides is needed to demonstrate that they are equivalent. The discussion emphasizes the importance of correctly factoring and applying trigonometric identities to prove the equation is an identity.
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Homework Statement



cos(x)^2/(1+3sin(x)-4sin(x)^2)=(1+sin(x))/(1+4sin(x))

Homework Equations


We are taking a topic in math where you rearrange one side of the formula to match the other

The Attempt at a Solution


I have factor 1+3sin(x)-4sin(x)^2 to get (-sin(x)+1)(4sin(x)+1)
 
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if you rewrite 1 + 3 sin x -4 (sin x)^2 using u-sin x you get the quadratic - 4 u^2 + 3u +1 which is fairly easy to factor.
 
goracheski said:

Homework Statement



cos(x)^2/(1+3sin(x)-4sin(x)^2)=(1+sin(x))/(1+4sin(x))


Homework Equations


We are taking a topic in math where you rearrange one side of the formula to match the other


The Attempt at a Solution


I have factor 1+3sin(x)-4sin(x)^2 to get (-sin(x)+1)(4sin(x)+1)

Continue working on the left side of your equation. What can you do with cos2(x). There's an identity you can use to change it.

The factors you found for the denominator are correct, but it would be better to write them as (1 - sin(x))(1 + 4sin(x)).
 
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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