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How do you find the exact value of sin(cot^-1 (5/4))?
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cot^-1(5/4)=x cotx=5/4 cosx/sinx=5/4 cos^2(x)+sin^2(x)=1 sinx=4/sqrt41
Thank you. Could you explain the part after cosx/sinx=5/4 ?I'm not really following it.
cotx=cosx/sinx did you know that?
|dw:1366641631895:dw|
I know that cotx=cosx/sinx, I just wasn't really following on what you did after that.
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cos^2 x+sin^2 x=1 cosx=5m and sinx=4m 25m^2+16m^2=1 41m^2=1 m=1/sqrt41 so sinx=4m=4/sqrt41 do you understand now?
Yes, that makes more sense. Thank you.
welcome. if i could help, i'm so happy :)
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