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farah2318

  • 3 years ago

why is 9sin(pi/3)= 9sqrt3/2 and 3tan(pi/4)=3 and 2 sec(pi/3)=4 and 5 cos(pi/4)= 5sqrt2 how would i figure these out?

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  1. timo86m
    • 3 years ago
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    try using the unit circle

  2. farah2318
    • 3 years ago
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    how would i memorize all of that

  3. NotTim
    • 3 years ago
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    Continuously draw the unit circle?

  4. farah2318
    • 3 years ago
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    lol

  5. NotTim
    • 3 years ago
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    That's how it works for martial arts, biology and combat situations :|

  6. farah2318
    • 3 years ago
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    do they have unit circle for tan and sec?

  7. NotTim
    • 3 years ago
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    eh...leemme check.

  8. NotTim
    • 3 years ago
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    here's what it says in my book

  9. NotTim
    • 3 years ago
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    |dw:1347081765068:dw| That suffice?

  10. farah2318
    • 3 years ago
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    lol wth

  11. hartnn
    • 3 years ago
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    i would advice you to atleast remember these 3 values : \(sin (\pi/6)=0.5\) or sin 30 =0.5 \(sin (\pi/3)=\sqrt{3}/2\) or sin 60 =\(\sqrt{3}/2\) \(sin (\pi/4)=1/\sqrt{2}\) or sin 45 =\(1/\sqrt{2}\) then you can easily derive any ratio with angles 30,45,60 degrees, using standard formulas like sin x= cos (90-x) or cot x =cos x / sin x say cos 60 = sin (90-60)= sin 30 = 0.5 tan 45 = sin 45 / cos 45 = sin 45 / sin (90-45) = sin 45/sin 45 =1

  12. farah2318
    • 3 years ago
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    thank you!

  13. hartnn
    • 3 years ago
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    from your question 2 sec \( \pi/3\)= 2/(cos\( \pi/3\) )= 2/(sin (\(\pi/2-\pi/3))\)=2/sin \( \pi/6\)=2/0.5=2*2=4 5 cos (pi/4)=5 sin (pi/2-pi/4)= 5 sin (pi/4) = 5/sqrt 2

  14. hartnn
    • 3 years ago
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    welcome :)

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