How do I prove that
How do I simplify the tan or sin so they would somehow cancel?
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.
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!
Hello, circuscircus!
Let me give it a try . . .
Let: .Prove: .
We have: . . . . and Pythagorus gives us: .
Hence: .
We have: . . . . and Pythagorus gives us: .
Hence: .
The equation is: .
Take the sine of the left side: .
. .
. .
. .
. .
Since , then: .