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How do i solve this trig question? identities etc?

  1. Mar 20, 2009 #1
    How do i solve this trig question? identities etc?
    4*[cos(B) + 3*sin(B)]=1+ 2*[3*sin(39) - cos(39)]


    i can get it to
    cosB + 3sinB=0.08054076 using basic algebra, but how do i find a value for B
     
  2. jcsd
  3. Mar 20, 2009 #2

    danago

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    Gold Member

    Can you think of a way to perhaps combine the sine and cosine terms of the left into a single trig function?
     
  4. Mar 20, 2009 #3
    no, thats what i need to somehow do
     
  5. Mar 20, 2009 #4

    danago

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    Gold Member

    One common method is to use the 'auxiliary angle method', which is to combine the sine and cosine terms into a single sine (or cosine) term of the form:

    R sin (B+a).

    The reason why we can write it in such a form is because this is equivalent to: (using an appropriate trig identity)

    R sin(B)cos(a) + R cos(B)sin(a)

    So by comparing coefficients, you can see that:

    R cos a = 12
    R sin a = 4

    From which you can solve to find a and R. See if you can continue from there.
     
  6. Mar 20, 2009 #5
    so we have
    R cos a = 4
    R sin a = 12

    tan a = 3
    a=71.565
    R=12.65

    am i meant to be saying that

    R sin (B+a) = R sin(B)cos(a) + R cos(B)sin(a) ??????????

    therefore

    R sin (B+a) = 1+ 2*[3*sin(39) - cos(39)]
    sin(B+a)=0.25467

    sin(c)=0.25467
    c=14.75

    c=B+71.565=14.75
    B= (-56.81)

    but this doesnt come right

    cos(-56.81) + 3sin(-56.81)= -1.963
    where i need
    cosB + 3sinB=0.8054076
     
  7. Mar 20, 2009 #6

    danago

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    Gold Member

    Yea the reason is because i made a mistake in my post but have since edited it out, and i think you have carried through that same mistake.

    Comparing the two forms:

    R sin(B)cos(a) + R cos(B)sin(a) = 4cos(B) + 12sin(B)

    On the left hand side, the coefficient of cos(B) is R sin(a), and on the right hand side it is 4, and so we must have:

    R sin(a) = 4

    And by similar reasoning,:

    R cos(a) = 12

    So i had originally typed the 4 and 12 in the wrong places. I guess you read my post before i got a chance to edit it out (or perhaps coincidentally made the exact same mistake haha? :P)
     
  8. Mar 20, 2009 #7
    so...
    R sin(a) = 4
    R cos(a) = 12

    tan(a)=1/3
    (a)=18.34
    R=12.65

    R sin (B+a) = 1+ 2*[3*sin(39) - cos(39)]
    sin(B+a)=0.25467

    sin(c)=0.25467
    c=14.75

    c=B+18.34=14.75
    B= (-3.59)

    cos(-3.59) + 3sin(-3.59)= -0.81
    thanks!!
     
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