Inverse Functions (2 Viewers)

[3unitz]

Sigh, another question:

sin^-1(2sqrt2)/3 - tan^-12

I get sin^-1((2sqrt2)-2)/3sqrt5

The answer is tan^-1(18-10sqrt2)/31

So can there be two answers to this question?
i.e. you can use sin(alpha - beta) or you could use tan(alpha - beta) right?

they are the same

 

[lyounamu]

Sigh, another question:

sin^-1(2sqrt2)/3 - tan^-12

I get sin^-1((2sqrt2)-2)/3sqrt5

The answer is tan^-1(18-10sqrt2)/31

So can there be two answers to this question?
i.e. you can use sin(alpha - beta) or you could use tan(alpha - beta) right?

The answer would be same for all even if you used different method (i.e. sine formula and tan formula).

If you check with your calculator, you answer is same as the answer (your answer is 7.0933054... and the answer is also 7.0933054...)

By the way, I got your answer as well.

 

[lacklustre]

The answer would be same for all even if you used different method (i.e. sine formula and tan formula).

If you check with your calculator, you answer is same as the answer (your answer is 7.0933054... and the answer is also 7.0933054...)

By the way, I got your answer as well.

Ty for verifying that.

I'm really sorry to bother you guys but i have another question:

Evaluate: tan^-1x + tan^-11/x

I don't understand this q, when you simplify it, you get a zero in the denominator.

 

[lyounamu]

Ty for verifying that.

I'm really sorry to bother you guys but i have another question:

Evaluate: tan^-1x + tan^-11/x

I don't understand this q, when you simplify it, you get a zero in the denominator.

You get 0 as your denominator because if you add those together (whatever the value of x is), you get 90 degrees. tan 90 degrees is undefined as you know.

The answer being undefined basically means that x = all reals except 0.

 

[lacklustre]

You get 0 as your denominator because if you add those together (whatever the value of x is), you get 90 degrees. tan 90 degrees is undefined as you know.

The answer being undefined basically means that x = all reals except 0.

Ok thanks that was a weird q for me. But i kinda get it.

 

[lacklustre]

Can anyone assist me with this simple differentiation?:

d/dx sin(tan^-1x)



I get (cos(tan^-1x))/(1+x²)

The answer is different.
Cheers

 

[3unitz]

Can anyone assist me with this simple differentiation?:

d/dx sin(tan^-1x)



I get (cos(tan^-1x))/(1+x²)

The answer is different.
Cheers

its right, try to simplify cos(tan^-1x) to get the answer in the book.

 

[lacklustre]

y = cos^-1(1-x) <-- how do you sketch this? I thought i did it right, but when i checked on the computer the graph seems to be flipped upside down from a normal cos^-1 graph.

Any help?

 

[lyounamu]

y = cos^-1(1-x) <-- how do you sketch this? I thought i did it right, but when i checked on the computer the graph seems to be flipped upside down from a normal cos^-1 graph.

Any help?

y = cos^-1(1-x)
cos y = 1-x
0 <_ y <_ pi

-1 <_ 1-x <_ 1
-2 <_ -x <_ 0
2 >_ x >_0

They are the domains & range. Draw the diagram according to this.

The shape of it will be like cot graph.

 

[lacklustre]

y = cos^-1(1-x)
cos y = 1-x
0 <_ y <_ pi

-1 <_ 1-x <_ 1
-2 <_ -x <_ 0
2 >_ x >_0

They are the domains & range. Draw the diagram according to this.

I did get the domain and range (same as yours) but I didn't know the graph would be upside down to a normal inverse cos curve.

This is how i pictured it:

 

[lyounamu]

I did get the domain and range (same as yours) but I didn't know the graph would be upside down to a normal inverse cos curve.

This is how i pictured it:

Yeah the shape is same for both inverse and non-inverse. Only difference is where the graph is situated. You just draw the graph according to the domain & range.