Inverse Trigonometric Functions

**mj.alawami** · March 20th 2009, 12:32 AM

Note:
$arcsin(siny) \neq y$ if y is not in the interval [-1,1]

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

Why ?

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

thank you

**Prove It** · March 20th 2009, 12:45 AM

Originally Posted by mj.alawami

Note:
$arcsin(siny) \neq y$ if y is not in the interval [-1,1]

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

Why ?

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

thank you

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

What does this equal? Is it in the interval $[-1, 1]$ ?

**mr fantastic** · March 20th 2009, 02:16 AM

Originally Posted by mj.alawami

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

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

Why ?

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

thank you

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

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

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