Trig equation

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  • #1
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Homework Statement


[itex]\ arcsin \frac{x}{x-1} + 2 arctan \frac{1}{x+1} = \frac{π}{2}[/itex]


Homework Equations




The Attempt at a Solution



if [itex]\ arcsin\frac{x}{(x-1)} = α [/itex]

then [itex] sin α = \frac{x}{(x-1)}[/itex]

if [itex]2 arctan\frac{1}{x+1} = β[/itex]

then [itex]tan(\frac{β}{2}) = \frac{1}{(x+1)}[/itex]

[itex]tan(β) = \frac{2\frac{1}{x+1}}{1-(\frac{1}{x+1})^2}[/itex]

[itex]tan(β) = \frac{2(x+1)}{x(x+2)}[/itex]

[itex]α + β = \frac{π}{2}[/itex]

[itex]sin(α + β) = 1[/itex]

[itex]sin(α + β) = sinαcosβ + cosαsinβ[/itex]

[itex]cosβ = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

[itex]sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

[itex]cosα = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}[/itex] by pythagoras' theorem

[itex](\frac{x}{(x-1)})(\frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) + (\frac{\sqrt{(x-1)^2 - x^2}}{(x-1)})(\frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) = 1[/itex]

[itex]\frac{x^2(x+2) + (\sqrt{(x-1)^2 - x^2)}(2(x+1))}{(x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})} = 1[/itex]

[itex]x^2(x+2) + (\sqrt{(x-1)^2 - x^2})(2(x+1)) = (x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})[/itex]

At this point I figure I half made an error because I can't see a fruitful direction in which to progress and it feels a bit messy compared to the other problems in the exercise.
 

Answers and Replies

  • #2
eumyang
Homework Helper
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I haven't figured the answer myself, but I have a couple of comments:

[itex]\cos β = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

[itex]\sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem
The denominators here will simplify into an irreducible quadratic.

[itex]\cos α = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}[/itex] by pythagoras' theorem
TBH, I don't know if this will be helpful, but the numerator can be simplified a bit (although you won't lose the square root).
 
  • #3
verty
Homework Helper
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198

Homework Statement


[itex]\ arcsin \frac{x}{x-1} + 2 arctan \frac{1}{x+1} = \frac{π}{2}[/itex]


Homework Equations




The Attempt at a Solution



if [itex]\ arcsin\frac{x}{(x-1)} = α [/itex]

then [itex] sin α = \frac{x}{(x-1)}[/itex]

if [itex]2 arctan\frac{1}{x+1} = β[/itex]

then [itex]tan(\frac{β}{2}) = \frac{1}{(x+1)}[/itex]

[itex]tan(β) = \frac{2\frac{1}{x+1}}{1-(\frac{1}{x+1})^2}[/itex]

[itex]tan(β) = \frac{2(x+1)}{x(x+2)}[/itex]

[itex]α + β = \frac{π}{2}[/itex]

[itex]sin(α + β) = 1[/itex]

[itex]sin(α + β) = sinαcosβ + cosαsinβ[/itex]

[itex]cosβ = \frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

[itex]sinβ = \frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}[/itex] by pythagoras' theorem

[itex]cosα = \frac{\sqrt{(x-1)^2 - x^2}}{(x-1)}[/itex] by pythagoras' theorem

[itex](\frac{x}{(x-1)})(\frac{x(x+2)}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) + (\frac{\sqrt{(x-1)^2 - x^2}}{(x-1)})(\frac{(2(x+1))}{\sqrt{(x(x+2))^2 + (2(x+1))^2}}) = 1[/itex]

[itex]\frac{x^2(x+2) + (\sqrt{(x-1)^2 - x^2)}(2(x+1))}{(x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})} = 1[/itex]

[itex]x^2(x+2) + (\sqrt{(x-1)^2 - x^2})(2(x+1)) = (x-1)(\sqrt{(x(x+2))^2 + (2(x+1))^2})[/itex]

At this point I figure I half made an error because I can't see a fruitful direction in which to progress and it feels a bit messy compared to the other problems in the exercise.

You were on the right track at this point:

##α + β = \frac{π}{2}##

but the next step was wrong. Can you think of something better to do here?

Hint: Don't make it any more complicated than it already is.

Use sin and cos
 
  • #4
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Thanks for your comments. After staring at this for a good proportion of the day I noticed that α and β are complementary angles which enabled me to proceed in the following fashion:

[itex]sinα = \frac{x}{x-1} [/itex]

[itex]cosβ = \frac{x}{x-1} [/itex] complementary angle

[itex]cosα = \frac{-2x+1}{x-1} [/itex] pythagoras' theorem

[itex]sinβ = \frac{-2x+1}{x-1} [/itex] complementary angle

[itex]sin(α+β) = 1[/itex]

[itex]sin(α+β) = (\frac{x}{x-1})^2 + (\frac{-2x+1}{x-1})^2 = 1[/itex]

[itex]\frac{5x^2-4x+1}{x-1} = 1[/itex]

[itex]5x^2-4x+1 = x-1[/itex]

[itex]x^2-x = 0[/itex]

[itex]x(x-1) = 0[/itex]

[itex]x = 0[/itex] or [itex]x = 1[/itex]

but [itex] asin \frac{1}{1-1}[/itex] makes no sense

so [itex]x = 0[/itex]

However the process I have deployed here seems no less complicated than the process I was following earlier. Is there a quicker/simpler solution?
 
  • #5
eumyang
Homework Helper
1,347
10
Really? I think the above is so much simpler than your first attempt, radicals and all. I want to kick myself for not seeing it earlier.
 

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