How do you evaluate $arcsin (\frac{\sqrt{2}}{2})$ $arcsin(sqrt 2/2)$ ?

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How do you evaluate $arcsin (\frac{\sqrt{2}}{2})$ $arcsin(sqrt 2/2)$ ?

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Answer 1 · 2015-05-24T08:29:19

$sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin (pi/4) = sqrt(2)/2$ is the length of one side of the isoceles right angled triangle with sides $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ , $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ and $1$ $1$ , which has internal angles $\frac{π}{4}$ $pi/4$ , $\frac{π}{4}$ $pi/4$ and $\frac{π}{2}$ $pi/2$ .

( $\frac{π}{4}$ $pi/4$ radians = $45^{o}$ $45^o$ and $\frac{π}{2}$ $pi/2$ radians = $90^{o}$ $90^o$ if you prefer)

To show this is right angled, check with Pythagoras:

${(\frac{\sqrt{2}}{2})}^{2} + {(\frac{\sqrt{2}}{2})}^{2}$ $(sqrt(2)/2)^2 + (sqrt(2)/2)^2$

$= \frac{{\sqrt{2}}^{2}}{2^{2}} + \frac{{\sqrt{2}}^{2}}{2^{2}}$ $= sqrt(2)^2/2^2 + sqrt(2)^2/2^2$

$= \frac{2}{4} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1 = 1^{2}$ $= 2/4 + 2/4 = 1/2+1/2 = 1 = 1^2$

So since $sin (\frac{π}{4}) = \frac{\sqrt{2}}{2}$ $sin (pi/4) = sqrt(2)/2$ and $\frac{π}{4}$ $pi/4$ is in the

required range for $arcsin$ $arcsin$ viz $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ $-pi/2 <= theta <= pi/2$ , we find

$arcsin (\frac{\sqrt{2}}{2}) = \frac{π}{4}$ $arcsin (sqrt(2)/2) = pi/4$

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How do you evaluate $arcsin (\frac{\sqrt{2}}{2})$ $arcsin(sqrt 2/2)$ ?

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