Check out some similar questions!

orkdork · Nov 29, 2009, 03:20 PM

Arctanx+Arcsinx=pi/2

I did this so far...
Arc sinx=pi/2-Arc tanx

x=sin (pi/2-Arctanx)

I did the sine expansion and it came out to cos(Arctanx)

then u=Arctanx

then x=1/(sqrt 1+x^2) (mutiply both sides by (sqrt 1+x^2) )
x(sqrt1+x^2)=1
x(sqrt1+x^2)-1=0

I don't know what to do now! The book has a problem similar (cosine instead of sine) but it comes out nicer because you can factor out an x... =0. Please help!

ebaines · Dec 1, 2009, 10:49 AM

So far so good. All you need do now is a little light algbra. You have:

$ x \sqrt {1 + x^2} = 1 $

Square both sides, then rearrange:

$ x^2 (1 + x^2) = 1 \\ x^4 + x^2 - 1 = 0 $

Now, let $w = x^2$ :
$ w^2 + w -1 = 0 $

Solve for $w$ using the quadratic equation, then take the square root of that to find the value for $x$ .

Post back and let us know what you get for a final answer.