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IRINA_888 [86]
3 years ago
11

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

Mathematics
1 answer:
PSYCHO15rus [73]3 years ago
8 0

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}}}

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