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Math Help - exact value of cos75, sin15, cot-210

  1. #1
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    exact value of cos75, sin15, cot-210

    ok so i need to find exact value of Cos75, Sin15 and Cot-210

    for the first, i have the identity cos(a-b)=Cos(b)*Cos(a) + Sin(b)*Sin(a)

    and im using 120 = a and 45 = b...cause i figured out cos and sin of those using the 30-60-90 and 45-45-90 triangle rules......but i have like: 1/root2 * -.5 + 1/root2 * root3 /2

    and i dont know how to do this...i keep getting the wrong answer

    so please help and help do the other two

    thanks
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  2. #2
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    Quote Originally Posted by stones44 View Post
    ok so i need to find exact value of Cos75, Sin15 and Cot-210

    for the first, i have the identity cos(a-b)=Cos(b)*Cos(a) + Sin(b)*Sin(a)

    and im using 120 = a and 45 = b...cause i figured out cos and sin of those using the 30-60-90 and 45-45-90 triangle rules......but i have like: 1/root2 * -.5 + 1/root2 * root3 /2

    and i dont know how to do this...i keep getting the wrong answer

    so please help and help do the other two

    thanks
    Your identity \cos{(a-b)} = \cos{b}\cdot{\cos{a}} + \sin{b}\cdot{\sin{a}} works just fine. Let a = 45 degrees and b = 30 degrees.
    Then \cos{15} = \frac{\sqrt{2}}{2}\cdot{\frac{\sqrt{3}}{2}} + \frac{\sqrt{2}}{2}\cdot{\frac{1}{2}}.
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  3. #3
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    Quote Originally Posted by stones44 View Post
    ok so i need to find exact value of Cos75, Sin15 and Cot-210

    for the first, i have the identity cos(a-b)=Cos(b)*Cos(a) + Sin(b)*Sin(a)

    and im using 120 = a and 45 = b...cause i figured out cos and sin of those using the 30-60-90 and 45-45-90 triangle rules......but i have like: 1/root2 * -.5 + 1/root2 * root3 /2

    and i dont know how to do this...i keep getting the wrong answer

    so please help and help do the other two

    thanks
    You will probably want to use the identity \cos{(a+b)} = \cos{a}\cos{b} - \sin{a}\sin{b} for the computation of \cos{75}. And just so you're clear, \cos{75} = \sin{15}.
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  4. #4
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    Hello,

    For the third one, you know that cot is an odd function.
    Hence cot(-210)=-cot(210)

    cot=cos/sin

    Take 210=180+30
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  5. #5
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    hmm ok..im having trouble with csc(-225)

    can people not just give tips but do one out please..visuals help me
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    \left( { - 225^ \circ  } \right) \cong \left( {135^ \circ  } \right)
    If one goes 225degrees clockwise it is the same as going 135degrees counter-clockwise from the positive x-axis.
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  7. #7
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    well can you do one of them for me so i can get a feel for it? i get confused when there are multiple angles and stuff

    mainly the csc one cause there arent identitys that ive learned that work with it
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  8. #8
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    Quote Originally Posted by stones44 View Post
    the csc one cause there arent identitys that ive learned that work with it
    That is an impossible for anyone who is doing a serious study of trigonometric functions.
    \csc ( - 225^ \circ  ) = \csc (135^ \circ  ) = \frac{1}{{\sin (135^ \circ  )}}
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