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How do you solve cot parenthesis inverse of sin 4 over 7 closed parenthesis?
Kiwi215
The inverse sin function I write as arcsin x.
Make use of the trignometric relationships:
cos2θ + sin2θ = 1
⇒ cosθ = √(1 - sin2θ)
cotθ = cosθ/sinθ
= √(1 - (sinθ)2)/sinθ
sin(arcsin x) = x
Then:
cot(arcsin(x)) = √(1 - (sin(arcsin(x))2)/sin(arcsin(x))
= √(1 - x2)/x
⇒ cot(arcsin(4/7)) = √(1 - (4/7)2)/(4/7)
= √(49/72 - 16/72) ÷ 4/7
= √(49 - 16) x 1/7 x 7/4
= 1/4 x √33
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∙ 12y ago
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