[PreCalculus] Proving Identities

AI Thread Summary
To prove the identity 1 + cos² x / sin² x = 2csc² - 1, the left-hand side can be rewritten as csc² x + cot² x by separating the terms. The identity cos² x + sin² x = 1 is crucial for manipulation. Users have reported multiple attempts at solving the equation, often getting stuck midway. Utilizing known trigonometric identities can help in reaching the solution. Engaging with additional resources can enhance understanding of these identities.
Snowglober
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Homework Statement


Prove that this is an identity.

1 + cos² x / sin² x = 2csc² - 1

Homework Equations


cos²x + sin²x = 1 (manipulative equation)
tan²x = sec² - 1
cot²x = csc²x - 1
etc..

The Attempt at a Solution


I attempted this equation more than 10+ times. Each time, I find a way but I get half way there and it doesn't work out.
 
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Hi Snowglober, welcome to PF
The problem should be
(1 + cos^2x)/sin^2x = 2csc^2x - 1
Now LHS can be written as
1/sin^2x + cos^2x/sin^2
= csc^2x + cot^2x
Now use the identities to get the result.
 
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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