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klbala2006

  • 3 years ago

sin(sin^{-1} \frac{ -sqrt{3} }{ 2 }+cos^{-1} \frac{ 1 }{ 2 })

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  1. ash2326
    • 3 years ago
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    @klbala2006 Do you know the range of \(\sin^{-1} x\) ?

  2. campbell_st
    • 3 years ago
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    looking at it its all 4th quadrant |dw:1352871092096:dw|

  3. klbala2006
    • 3 years ago
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    i need to evaluate this function . @ash2326 no i dont

  4. ash2326
    • 3 years ago
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    Do you know the range and domain of sin (x) ?

  5. klbala2006
    • 3 years ago
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    @ash2326 domain of sin(x) is (R), while its range is [-1,1]

  6. ash2326
    • 3 years ago
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    Good, to have the inverse of sin function, we limit it's range domain= range of sin x= [-1, 1] Range = \(\large [\frac{-\pi}{2}, \frac{\pi}{2}]\) DO you get this?

  7. klbala2006
    • 3 years ago
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    yes i do

  8. ash2326
    • 3 years ago
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    if x is positive, then y lies in \([0, \frac {\pi}{2}]\) if x is negative, then y lies in \([\frac{-\pi}{2}, 0)\) Now what's \(\sin^{-1}(\frac{-\sqrt 3}{2})\) ?

  9. klbala2006
    • 3 years ago
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    \[-45\]

  10. klbala2006
    • 3 years ago
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    \[-\frac{ \pi }{ 4 }\] in radian

  11. klbala2006
    • 3 years ago
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    @ash2326 complete please

  12. ash2326
    • 3 years ago
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    Sorry, I was away. But sin (-45) is not \(\frac{-\sqrt 3}{2}\) it's, \(\frac{-1}{\sqrt 2}\).

  13. klbala2006
    • 3 years ago
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    isnt it the same ?

  14. ash2326
    • 3 years ago
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    What's sin 60 ?

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