How do you calculate $sin ({sin}^{- 1} (\frac{3}{8}))$ $sin ( sin^ -1 ( 3/8))$ ?

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Answer 1 · 2018-05-20T08:11:54

$\Rightarrow {0.375529}^{c}$ $color(purple)(=> 0.375529^c$ or ${21.5}^{\circ}$ $color(brown)(21.5^@$

$sin ({sin}^{- 1} (\frac{3}{8})$ $sin (sin ^-1 (3/8)$

${sin}^{- 1} (\frac{3}{8}) = 0.384397$ $sin^-1 (3/8) = 0.384397$

$\Rightarrow sin (0.384397)$ $=> sin (0.384397)$

$\Rightarrow {0.375529}^{c}$ $color(purple)(=> 0.375529^c$ or ${21.5}^{\circ}$ $color(brown)(21.5^@$

Answer 2 · 2018-05-24T17:20:16

$sin arcsin (\frac{3}{8}) = \frac{3}{8}$ $sin arcsin(3/8) = 3/8$

I never use the $- 1$ $-1$ power notation for the inverse trig functions -- it's awful.

I treat $arcsin (x)$ $arcsin(x)$ as a multivalued expression, all the angles whose sine is $x .$ $x.$

Even though $arcsin (x)$ $arcsin(x)$ is multivalued, there's only one value for it's sine,

$sin arcsin x = x$ $sin arcsin x = x$

Here

$sin arcsin (\frac{3}{8}) = \frac{3}{8}$ $sin arcsin(3/8) = 3/8$

How do you calculate sin(sin−1(38))sin ( sin^ -1 ( 3/8))?