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Math Help - Help with solving these Trig Identities...

  1. #1
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    Help with solving these Trig Identities...

    How do I solve:

    Using Sum & Difference Identities

    1. tan ( - pi/12)
    2. tan (alpha + beta); alpha (-3/5) ends QIII; Beta (-1/4) ends QII
    3. cos (-1/5 + -1/4)
    4. (sin A - sin B)^2 + (cos A - cos B)^2


    Answers:
    1. -2 + sq.root(3)
    2.[ 192- 25 sq.root(15) ] / -119
    3.[ 1 + 6 sq.root(10) ] / 20
    4. -2 cos (A - B)



    Thank you very much for your time! =)
    Last edited by mr fantastic; April 5th 2009 at 04:11 AM. Reason: Questions moved from original thread
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  2. #2
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    Quote Originally Posted by coolacu View Post
    How do I solve:

    Using Sum & Difference Identities

    1. tan ( - pi/12)
    2. tan (alpha + beta); alpha (-3/5) ends QIII; Beta (-1/4) ends QII
    3. cos (-1/5 + -1/4)
    4. (sin A - sin B)^2 + (cos A - cos B)^2


    Answers:
    1. -2 + sq.root(3)
    2.[ 192- 25 sq.root(15) ] / -119
    3.[ 1 + 6 sq.root(10) ] / 20
    4. -2 cos (A - B)



    Thank you very much for your time! =)
    Check out the trigonometric sum formulas (google them for a list). These questions seem like simply picking the correct one and then working through the gears. For example here's the first one

    \tan(\frac{- \pi}{12}) = \tan(\frac{\pi}{4} - \frac{\pi}{3})

    Looking at the sum and difference formulas for tan we see that this is equal to

    \frac{\tan(\frac{\pi}{4})-\tan(\frac{\pi}{3})}{1+\tan{\frac{\pi}{4}}\tan{\fr  ac{\pi}{3}}}

    =\frac{1-\sqrt{3}}{1+\sqrt{3}}

    =\frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})} = \frac{1-2\sqrt{3}+3}{-2}

    = -2 + \sqrt{3}
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  3. #3
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    Hello, coolacu!

    There must be a typo in #3.
    Also in #4 . . .


    (4)\;\;(\sin A - \sin B)^2 + (\cos A - \cos B)^2

    Answer: . -2\cos (A - B) . . . . no

    \text{Multiply out: }\;\sin^2\!A - 2\sin A\sin B + \sin^2\!B + \cos^2\!A-2\cos A\cos B + \cos^2\!B


    \text{Rearrange: }\;\underbrace{\sin^2\!A + \cos^2\!A}_{\text{This is 1}} \;+\; \underbrace{\sin^2\!B + \cos^2\!B}_{\text{This is 1}} \;-\; 2\underbrace{(\cos A\cos B + \sin A\sin B)}_{\text{This is }\cos(A-B)}

    . . \text{and we have: }\;{\color{blue}2 - 2\cos(A - B)}

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  4. #4
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    Hello coolacu
    Quote Originally Posted by coolacu View Post
    How do I solve:

    Using Sum & Difference Identities

    1. tan ( - pi/12)
    2. tan (alpha + beta); alpha (-3/5) ends QIII; Beta (-1/4) ends QII
    3. cos (-1/5 + -1/4)
    4. (sin A - sin B)^2 + (cos A - cos B)^2


    Answers:
    1. -2 + sq.root(3)
    2.[ 192- 25 sq.root(15) ] / -119
    3.[ 1 + 6 sq.root(10) ] / 20
    4. -2 cos (A - B)



    Thank you very much for your time! =)
    I don't understand what questions 2 and 3 mean, and I think the answer to 4 is wrong. Here's what I get for that one:

    (\sin A - \sin B)^2 + (\cos A - \cos B)^2

    = \sin^2A - 2\sin A\sin B + \sin^2B + \cos^2A - 2\cos A\cos B + \cos^2 B

    = \sin^2 A + \cos^2A + \sin^2B +\cos^2 B -2(\cos A\cos B + \sin A \sin B)

    = 2 - 2\cos(A-B)

    Can you explain what you mean by questions 2 and 3?

    Grandad
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