# Trigonometry (re-moved to Precalc Math Homework)

#### Crystal037

Homework Statement: (cos2A -1)/sinA +sin2A/(cos2A+1)*cosA=__________________
Homework Equations: sin2A=2sinAcosA
cos2A+1=cos^2A
cos2A-1=sin^2A

(cos2A -1)/sinA +sin2A/(cos2A+1)*cosA
[cos^2(2A)-1 +2sin^2(A)cos^2(A)]/sinA(cos2A+1)
[-sin^2(2A) + 2sin^2(A)cos^2(A)]/sinA(cos2A+1)
I cant proceed any further.
The answer must be 2 sin^2A

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#### FactChecker

Gold Member
2018 Award
The second and third identities that you use are wrong.

#### Crystal037

cos2A+1=2cos^2A
cos2A-1=2sin^2A
Yeah I typed it wrong.
But I am still not getting the answer.
Give me a hint on how to proceed.
(cos2A -1)/sinA +sin2A/(cos2A+1)*cosA
[cos^2(2A)-1 +2sin^2(A)cos^2(A)]/sinA(cos2A+1)
[-sin^2(2A) + 2sin^2(A)cos^2(A)]/sinA(cos2A+1)

#### FactChecker

Gold Member
2018 Award
Your second identity has the wrong sign. $\cos(2A) -1 = -2\sin^2(A)$

#### Crystal037

[cos^2(2A)-1 +2sin^2(A)cos^2(A)]/sinA(cos2A+1)
=[cos^2(2A)-1 + sin^2(2A)/2]/sinA(cos2A+1)
=[cos^2(2A)-1]/2sinA(cos2A+1)
=[cos(2A)-1]/2sinA
=-2sin^2A/2sin(A)
=-sinA
But this isn't correct when I put some values of A for eg. 90 degrees and 45 degrees

#### FactChecker

Gold Member
2018 Award
That is the correct answer. I don't know where the proposed answer of $2\sin^2(A)$ in the original post came from.

#### Crystal037

The answer 2sin^2(A) was given as the answer in the book. But when I put any value of A, like 90 I don't get the LHS and RHS EQUAL. Also when I put A=90 in the answer I got I didn't get LHS = RHS

#### FactChecker

Gold Member
2018 Award
Right. Something is wrong there. Those two are not identical.

#### Crystal037

Can you please tell me where's the mistake

#### FactChecker

Gold Member
2018 Award
Can you please tell me where's the mistake
Make sure that you read the problem correctly. You got the right answer and it is not what you expected from your problem statement. Maybe the book is wrong.

#### Crystal037

What do you mean? Elaborate

#### FactChecker

Gold Member
2018 Award

Functions f and g are identical. Function h is not the same as function f.

#### Crystal037

but f( 90 degrees )=-2 when solved through the LHS part, then why in the graph it is-1

#### FactChecker

Gold Member
2018 Award
You tell me. Are you sure that you are handling the zero of the denominator $(\cos(2A)+1)$ correctly? Actually, the function f is not defined at 90 degrees, but it can be smoothly extended to include that point.

#### Crystal037

How come the L.H.S of an equation is not defined for a point but the RHS value is. Then both the functions are not equal and our equality shouldn't hold

#### mfb

Mentor
Consider e.g. $\displaystyle \frac{x^2}{x} = x$. Is this equation always true? It looks so - but for x=0 the left hand side is undefined while the right hand side is 0. Everywhere else the equation is true, so the two sides are still "sort of" equal. Equal at every point where the left hand side is defined.

#### FactChecker

Gold Member
2018 Award
@Crystal037, Division by zero is not defined. When one starts with a function like $f(x) = \frac {x^2} x$ then, to be accurate, he should specify what to do about $x=0$. Like specifying that $f$ is not defined for $x=0$, or $f(0) = 0$, or $f(0) = 3.14159$. That information should be retained if the function is later reduced to $h(x) = x$.
The same is true of the original function in this thread. To be picky, all the values of A where there would be a division by zero should be ruled out of the definition or function values should be specified. In any case, your calculation of $f( 90 degrees )=-2$ does not fit in smoothly with the rest of the function.

Last edited:

#### Crystal037

Ok thanks

"Trigonometry (re-moved to Precalc Math Homework)"

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