PLEASE HELP!!! I attached a snip of my question

Mathematics

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Inga [223]3 years ago

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The answer is

Step-by-step explanation:

In the attachment!

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Evaluate cos(sin^-1(4/5)

PSYCHO15rus [73]

Step-by-step explanation:

$Let \: \: \theta = \sin^{ - 1} (\frac{4}{5}) \\ \\ \therefore \: \sin\theta = \frac{4}{5} \\ \\ \because \: { \cos}^{2} \theta =1 - { \sin}^{2} \theta \\ \\ \therefore \: { \cos}^{2} \theta = 1 - (\frac{4}{5} )^{2} \\ \\ = 1 - \frac{16}{25} \\ \\ = \frac{25 - 16}{25} \\ \\ = \frac{9}{25} \\ \\ \therefore \:{ \cos}\theta = \pm \: \frac{3}{5} \\ \\ \because \: \theta \: lie \: in \: the \: first \: quadrant \\ \\ \therefore \: { \cos}\theta = \: \frac{3}{5} \\ \\ \implies \theta = { \cos}^{ -1 } \: \frac{3}{5}\\\\\implies \theta = { \cos}^{ -1 } \: \frac{3}{5}= { \sin}^{ -1 } \: \frac{4}{5} \\ \\ \therefore \: \cos( \ {sin}^{ - 1} \frac{4}{5} ) \\ \\ = \cos( \ {cos}^{ - 1} \frac{3}{5} ) \\\\ = \frac{3}{5} \\ \\ \purple{ \boxed{\therefore \: \cos( \ {sin}^{ - 1} \frac{4}{5} ) = \frac{3}{5}}}$