Let sin (SiNx) = t Then: SiNx = arcsint So: x = arcsin (arcsint) =sint Why?

Let sin (SiNx) = t Then: SiNx = arcsint So: x = arcsin (arcsint) =sint Why?

The last step is wrong
x=arcsin(arcsint)
You can't simply remove arc
First of all, the problem of defining domains is not considered
And arcsin (arcsint) is not arcsint
The inverse function of

What is the specific formula for the number of inverse trigonometric functions?

There are three main anti trigonometric functions: y = arcsin (x), definition domain [- 1,1], range [- π / 2, π / 2] y = arccos (x), definition domain [- 1,1], value range [0, π] y = arctan (x), definition domain (- ∞, + ∞), value domain (- π / 2, π / 2) y = arccot (x), definition domain (- ∞, + ∞), value domain (0, π)

Find the value arcsin1 of inverse trigonometric function

It should be π / 2, because the value range of arcsinx function is [- π / 2, π / 2], where only the value of sin is 1
π / 2, so arcsin1 = π / 2

What is the value of an inverse trigonometric function

0° 30° 45° 60° 90°arcsin 0 1/2 √2/2 √3/2 1arccos 1 √3/2 √2/2 1/2 0arctan 0 √3/3 1 √3 ∞

Inverse trigonometric function image Arccosx, arctanx, arccotx images One function, one image, clearly marked coordinates

The image of y = arcsinx

How to find the inverse function of x y = 2 - 1 Y is equal to the X of 2 minus 1

X-1 of y = 2
X of 2 = y + 1
x=log2（y+1）
y=log2（x+1）

Find the inverse function of function y = (0.2 ^ - x) + 1

y=(0.2^-x)+1
y-1=0.2^-x
ln(y-1)=ln(0.2)^-x=-xln(0.2)
x=-ln(y-1)/ln(0.2)
So the inverse function is
y=-ln(x-1)/ln(0.2) (x>1)

Mathematical problems about inverse function Let f (x) satisfy f (x-1) = 2x + 1 / X-2, and the image of function g (x) and function f inverse (x + 1) are symmetric with respect to the straight line y = x, then what is the value of G (11)? This is a multiple-choice question, but what I did was 27 / 11, there was no answer,

The image of G (x) and the inverse of function f (x + 1) is symmetric with respect to the line y = X
y=f-1(x+1)
f(y)=x+1
y=f(x)-1=g(x)
g(11)=f(11)-1
=(2*12+1)/10 -1=5/2-1=3/2

Simple math problem y = √ (x ^ 2-2x + 3) (x

y=√(x^2-2X+3)=√[（x-1)^2+2]．．．（1）　　
It can be obtained from X 〈 = 1
　　　y〉＝√2
For the square of two sides of (1), y ＾ 2 = (x-1) ^ 2 + 2
(X-1)^2=y^2-2
Because of X

A mathematical problem of inverse function It is known that the inverse function of the function y = f (x) is y = F-1 (x). Now, the image of the function y = f (3-2x) is shifted to the left by 1 unit and up by 2 units, and then the inverse function of the function obtained after symmetry about the origin is

After two translation
y-2=f[3-2(x+1)]=f(1-2x)
After symmetry about the origin
-y-2=f(1-2(-x))=f(2x+1)
Its inverse function is
2y+1=f-1(-x-2)