Prove Inverse Trigonometric Identities

Section 4.7

Can you do (a) as a guidefor me to do the rest?
Thanks.

 

I did d), leaving a) to you

Let sin^-1(x)=θ

=>x=sin(θ)=cos(π/2-θ)

=>cos^-1(x)=π/2-θ=π/2-sin^-1(x)

=>sin^-1(x)+cos^-1(x)=π/2

 

Can you tell me what the secret is to doing the proofs here? For example, why did you equate sin^-1(x) to theta as step one? What data in the problem told you to do that? Sometimes, getting started is the hardest part of the question.

For part (a), must I equate the left or right side to theta? Must I equate any side to theta? If so, why theta? Can you provide me with a video link showing how this is done in step by step fashion?

 

sin^-1(x) is equal to the angle (for example theta) whose sine is x:

For part (a) manipulate one side and prove it is equal to the other side
hint: let sin^-1(-x) be equal to y
y=sin^-1(x)

 

I will play with part (a) later on my break or tomorrow morning.

 

Part (a)

You said let y = arcsin(-x).

This takes me to y = -arcsin (x).

Give me a hint for the next step.

 

Prove arcsin (-x) =-arcsin x

Let u = arcsin (- x) and v = - arcsin( x)
Then, sin (u) = - x and -v = arcsin(x) =>x = sin(-v) = - sin( v)
or x = - sin( u )

Comparing the two, x = - sin (u) = - sin (v) => u = v

Therefore, arcsin (-x) = -arcsin(x)

 

You are really something special. Can you do part (c)? I will endeavor to do parts (b) and (d).

 

c.

prove: arctan(x)+arctan(1/x)=pi/2, x>0


Consider the point P=(1,x), in the first quadrant, with corresponding angle 0< α <π/2.

Let β=π/2-θ.

Then, also, 0<β <π/2 and tan(β)=tan(π/2-α )=cot(α )=1/tan(α )=1/x

It follows that arctan(x)+arctan(1/x)=α + β=π/2

 

Sorry but I still don't get it. I haven't found one decent video clip that explains this clearly enough for me to fully grasp. Sorry.