# Inverse Trigonometric Functions

• Mar 20th 2009, 12:32 AM
mj.alawami
Inverse Trigonometric Functions
Note:
$\displaystyle arcsin(siny) \neq y$ if y is not in the interval [-1,1]

Question:
$\displaystyle arcsin(sin\frac{3\pi}{4}) \neq \frac{3\pi}{4}$

Why ?

Attempt:
I try doing it with the calculator but i keep getting $\displaystyle \frac{3\pi}{4}$

thank you
• Mar 20th 2009, 12:45 AM
Prove It
Quote:

Originally Posted by mj.alawami
Note:
$\displaystyle arcsin(siny) \neq y$ if y is not in the interval [-1,1]

Question:
$\displaystyle arcsin(sin\frac{3\pi}{4}) \neq \frac{3\pi}{4}$

Why ?

Attempt:
I try doing it with the calculator but i keep getting $\displaystyle \frac{3\pi}{4}$

thank you

$\displaystyle \frac{3\pi}{4} \approx \frac{3 \times 3.14159}{4}$.

What does this equal? Is it in the interval $\displaystyle [-1, 1]$?
• Mar 20th 2009, 02:16 AM
mr fantastic
Quote:

Originally Posted by mj.alawami
Note:
$\displaystyle arcsin(siny) \neq y$ if y is not in the interval [-1,1] Mr F says: This interval is wrong. It should be $\displaystyle {\color{red}\left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]}$.

Question:
$\displaystyle arcsin(sin\frac{3\pi}{4}) \neq \frac{3\pi}{4}$

Why ?

Attempt:
I try doing it with the calculator but i keep getting $\displaystyle \frac{3\pi}{4}$

thank you

$\displaystyle \sin \frac{3 \pi}{4} = \frac{1}{\sqrt{2}}$.

$\displaystyle \arcsin \frac{1}{\sqrt{2}} = \frac{\pi}{4}$ since the range of the arcsin function is, by definition, $\displaystyle \left[ -\frac{\pi}{2}, \frac{\pi}{2}\right]$.