Proving ... equals to pi/2

**circuscircus** · September 27th 2007, 10:04 AM

How do I prove that

$sin^{-1}(\frac{x}{3}) - 2tan^{-1}sqrt{\frac{3+x}{3-x}} = -\frac{\pi}{2}$

How do I simplify the tan or sin so they would somehow cancel?

**ticbol** · September 27th 2007, 11:31 AM

You try to make the RHS equal to zero.
cos(-pi/2) = 0
So take the cosines of both sides of your original equation.

Let A = arcsin(x/3)
And B = 2arctan[sqrt{(x+3)/(x-3)}]

So your original equation is in the form
A -B = -pi/2
Take the cosines of both sides,
cos(A-B) = 0
cosAcosB +sinAsinB = 0 ------------***

A = arcsin(x/3)
So,
sinA = x/3 ----------------(i)
cosA = sqrt(9 -x^2) /3 ------(ii)

B = 2arctan[sqrt{(x+3)/(x-3)}]
So,
sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
sinB = sqrt(9-x^2) /18 -------------------------(iii)

cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
cosB = -x/18 ----------------------------(iv)

Hence,
cosAcosB +sinAsinB = 0 ------------***
[(sqrt(9 -x^2) /3)*(-x/18)] +[(x/3)*(sqrt(9 -x^2) /18)] =? 0
[-x*sqrt(9 -x^2) /54] +[x*sqrt(9 -x^2) /54 =? 0
0 =? 0
Yes.

Therefore, your original equation is proven.

**circuscircus** · September 27th 2007, 11:40 AM

cosA = sqrt(9 -x^2) /3 ------(ii)

B = 2arctan[sqrt{(x+3)/(x-3)}]
So,
sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
sinB = sqrt(9-x^2) /18 -------------------------(iii)

cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
cosB = -x/18 ----------------------------(iv)

I got really lost over here... where do u get cosA from? also I was a bit confused witht he arctan part

**ticbol** · September 27th 2007, 11:16 PM

Originally Posted by circuscircus

cosA = sqrt(9 -x^2) /3 ------(ii)

B = 2arctan[sqrt{(x+3)/(x-3)}]
So,
sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
sinB = sqrt(9-x^2) /18 -------------------------(iii)

cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
cosB = -x/18 ----------------------------(iv)

I got really lost over here... where do u get cosA from? also I was a bit confused witht he arctan part

Originally Posted by circuscircus

cosA = sqrt(9 -x^2) /3 ------(ii)

B = 2arctan[sqrt{(x+3)/(x-3)}]
So,
sinB = 2sin(arctan[sqrt{(x+3)/(x-3)}]) * cos(arctan[sqrt{(x+3)/(x-3)}])
sinB = sqrt(9-x^2) /18 -------------------------(iii)

cosB = cos^2(arctan[sqrt{(x+3)/(x-3)}]) -sin^2(arctan[sqrt{(x+3)/(x-3)}])
cosB = -x/18 ----------------------------(iv)

I got really lost over here... where do u get cosA from? also I was a bit confused witht he arctan part

I see. Sorry. I thought you could follow those because your questions are all a little bit advanced or complicated.

-------------------------------

A = arcsin(x/3)
So,
sinA = x/3 ----------------(i)
cosA = sqrt(9 -x^2) /3 ------(ii)

Draw the right triangle reference of angle A.
A = arcsin(x/3)
So, sinA = x/3
That means the leg opposite A is x
The hypotenuse is 3
So, tThe leg adjacent to A is sqrt(3^2 -x^2) = sqrt(9 -x^2)
Hence, cosA = (adjacent side)/hypotenuse = sqrt(9 -x^2) /3

----------------------------------------------------------------
The arctangent?

B = 2arctan[sqrt{(x+3)/(x-3)}]

, My! I wrote it incorrectly. It should have been
B = 2arctan[sqrt{(3+x)/(3-x)}]
as shown in your original equation.

Zeez, I think I'm getting older. Transferring from scratch paper to computer needs some thinking too. And if you're in a hurry and you're getting older, then your mind cannot catch up easily.

Okay, let angle C = arctan[sqrt(3+x) /sqrt(3-x)]

Draw the right triangle reference of angle C.
C = arctan[sqrt(3+x) /sqrt(3-x)]
So, tanC = sqrt(3+x) /sqrt(3-x)
That means the leg opposite C is sqrt(3+x)
The leg adjacent to C is sqrt(3-x)
So, the hypotenuse is sqrt[(sqrt(3+x))2 +(sqrt(3-x))^2]
= sqrt[(3+x) +(3-x)]
= sqrt(6)

Hence,
sinC = sqrt(3+x) /sqrt(6)
cosC = sqrt(3-x) /sqrt(6)

So,
sinB = 2sin(arctan[sqrt{(3+x)/(3-x)}]) * cos(arctan[sqrt{(3+x)/(3-x)}])
sinB = 2[sqrt(3+x) /sqrt(6)]*[sqrt(3-x) /sqrt(6)]
sinB = 2[sqrt((3+x)(3-x)) /6]
sinB = sqrt(9 -x^2) /3 -------------------------(iii)

cosB = cos^2(arctan[sqrt{(3+x)/(3-x)}]) -sin^2(arctan[sqrt{(3+x)/(3-x)}])
cosB = [sqrt(3-x) /sqrt(6)]^2 -[sqrt(3+x) /sqrt(6)]^2
cosB = (3-x)/6 -(3+x)/6
cosB = -2x/6
cosB = -x/3----------------------------(iv)

--------------------------------------------------------------

Hence,
cosAcosB +sinAsinB = 0 ------------***
[(sqrt(9 -x^2) /3)*(-x/3)] +[(x/3)*(sqrt(9 -x^2) /3)] =? 0
[-x*sqrt(9 -x^2) /9] +[x*sqrt(9 -x^2) /9 =? 0
0 =? 0
Yes.

Therefore, your original equation is proven.

Whew!

**Soroban** · September 28th 2007, 04:14 AM

Hello, circuscircus!

Let me give it a try . . .

Prove: . $\sin^{-1}\!\left(\frac{x}{3}\right) - 2\!\cdot\!\tan^{-1}\!\left(\sqrt{\frac{3+x}{3-x}}\right) \;= \;-\frac{\pi}{2}$

Let: . $\alpha \:=\:\sin^{-1}\!\left(\frac{x}{3}\right)\;\text{ and }\;\beta \:=\:\tan^{-1}\!\left(\sqrt{\frac{3+x}{3-x}}\right) \quad\Rightarrow\quad \alpha - 2\beta \:=\:-\frac{\pi}{2}$

We have: . $\sin\alpha \:=\:\frac{x}{3} \:=\:\frac{opp}{hyp}$ . . . and Pythagorus gives us: . $adj \:=\:\sqrt{9-x^2}$
Hence: . $\cos\alpha \:=\:\frac{\sqrt{9-x^2}}{3}$

We have: . $\tan\beta \:=\:\frac{\sqrt{3+x}}{\sqrt{3-x}}\:=\:\frac{opp}{adj}$ . . . and Pythagorus gives us: . $hyp \:=\:\sqrt{6}$
Hence: . $\sin\beta \:=\:\frac{\sqrt{3+x}}{\sqrt{6}},\;\;\cos\beta\:=\ :\frac{\sqrt{3-x}}{\sqrt{6}}$

The equation is: . $\alpha - 2\beta \:=\:-\frac{\pi}{2}$

Take the sine of the left side: . $\sin(\alpha - 2\beta) \;=\;\sin(\alpha)\cos(2\beta) - \sin(2\beta)\cos(\alpha)$

. . $= \;\sin(\alpha)\left[1-2\sin^2\!(\beta)\right] - \left[2\sin(\beta)\cos(\beta)\right]\cos(\alpha)$

. . $= \;\frac{x}{3}\left(1 - 2\!\cdot\!\frac{3+x}{6}\right) - 2\left(\frac{\sqrt{3+x}}{\sqrt{6}}\right)\left(\fr ac{\sqrt{3-x}}{\sqrt{6}}\right)\left(\frac{\sqrt{9-x^2}}{3}\right)$

. . $= \;\frac{x}{3}\left(-\frac{x}{3}\right) - 2\left(\frac{\sqrt{9-x^2}}{6}\right)\left(\frac{\sqrt{9-x^2}}{3}\right)$

. . $= \;-\frac{x^2}{9} - \frac{9-x^2}{9} \;=\; \frac{-9}{9} \;=\;-1$

Since $\sin(\alpha - 2\beta) \:=\:-1$ , then: . $\alpha - 2\beta \;=\;-\frac{\pi}{2}$

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