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Trigonometry (re-moved to Precalc Math Homework)

  • Thread starter Crystal037
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83
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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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The second and third identities that you use are wrong.
 
83
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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)
 

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Your second identity has the wrong sign. ##\cos(2A) -1 = -2\sin^2(A)##
 
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[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
 

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

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Right. Something is wrong there. Those two are not identical.
 
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Can you please tell me where's the mistake
 

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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.
 
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What do you mean? Elaborate
 

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temp.png

Functions f and g are identical. Function h is not the same as function f.
 
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but f( 90 degrees )=-2 when solved through the LHS part, then why in the graph it is-1
 

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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.
 
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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
 
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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.
 

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@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.
 
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Ok thanks
 

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