An elusive trig proof I can't seem to get

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To prove the equation (sin(x) + tan(x)) / (cos(x) + 1) = tan(x), one can start by multiplying both sides by cos(x), leading to the expression (sin(x)cos(x) + sin(x)) = sin(x). The discussion clarifies that tan(x) can be expressed as sin(x) / cos(x), simplifying the proof. By factoring tan(x) from the numerator, the expression reduces to cos(x) + 1, which cancels with the denominator. The proof ultimately requires only a couple of algebraic steps.
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Homework Statement



Prove that:

(sin(x)+ tan(x))/(cos(x)+ 1)= tan(x)

Homework Equations



There are just trig identities that we can use.

The Attempt at a Solution


I've attempted every possible way I can think of and it would just look like jibberish here.
 
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What you wrote, sin(x)+ tan(x)/cos(x)+ 1= tan(x). isn't true! For example, if x= 0, then sin(x)= sin(0)= 0, tan(x)/cos(x)= tan(0)/cos(0)= 0/1= 0 so the left side is 1. But the right side, tan(x)= tan(0), is 0.

What I think you meant was (sin(x)+ tan(x))/(cos(x)+ 1)= tan(x).

If you multiply both sides of the equation by cos(x) you get
(sin(x)cos(x)+ sin(x))/(cos(x)+ 1)= sin(x).
 
Sorry yeah, that's what I meant, my mistake. I'll change that right now.
 
Use the fact that tan(x)=sin(x)/cos(x). The proof should just involve two lines of algebra.
 
factor out tan(x) from numerator and u get cos(x)+1 which cancels with denominator
 
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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