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Math Help - Please help with trigonometric functions!

  1. #1
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    Please help with trigonometric functions!

    1)Prove this identity
    \frac{1-cos\theta}{sin\theta}= \frac{sin\theta}{1+cos\theta}

    2)Solve this for -\Pi\leq\theta\leq\Pi

    sin2\theta=cos2\theta

    thanks if ya can
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  2. #2
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    Quote Originally Posted by Confuzzled?
    1)Prove this identity
    \frac{1-cos\theta}{sin\theta}= \frac{sin\theta}{1+cos\theta}

    2)Solve this for -\Pi\leq\theta\leq\Pi

    sin2\theta=cos2\theta

    thanks if ya can
    Hello,

    to 1): Multiply both sides of the equation by sin\theta and 1+cos\theta. You'll get:

    1-cos^2\theta= sin^2\theta

    By the law (rule?) of Pythagoras both sides are equal.

    to 2) sin(x) and cos(x) are equal for x=\frac{\pi}{4}+k \cdot \pi. So in your case you get:

    2\theta=\frac{\pi}{4}+k \cdot \pi
    \theta=\frac{\pi}{8}+k \cdot \frac{\pi}{2}

    For the given domain: \theta = \frac{5\pi}{8} \vee \theta = - \frac{3\pi}{8} \vee \theta = \frac{\pi}{8}

    Greetings

    EB
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  3. #3
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    Quote Originally Posted by earboth
    Hello,

    to 1): Multiply both sides of the equation by sin\theta and 1+cos\theta. You'll get:

    1-cos^2\theta= sin^2\theta

    By the law (rule?) of Pythagoras both sides are equal.

    to 2) sin(x) and cos(x) are equal for x=\frac{\pi}{4}+k \cdot \pi. So in your case you get:

    2\theta=\frac{\pi}{4}+k \cdot \pi
    \theta=\frac{\pi}{8}+k \cdot \frac{\pi}{2}

    For the given domain: \theta = \frac{5\pi}{8} \vee \theta = - \frac{3\pi}{8} \vee \theta = \frac{\pi}{8}

    Greetings

    EB
    Umm... You cannot multiply an identity and make it true and conclude that the hypothesis was true- That is called taking the converse of the statement.
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  4. #4
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    Quote Originally Posted by Confuzzled?
    1)Prove this identity
    \frac{1-cos\theta}{sin\theta}= \frac{sin\theta}{1+cos\theta}

    2)Solve this for -\Pi\leq\theta\leq\Pi

    sin2\theta=cos2\theta

    thanks if ya can
    1) Let A = theta, so,

    (1 -cosA)/sinA = sinA/(1 +cosA)

    Easy is if we cross multiply,
    (1 -cosA)(1 +cosA) = sinA*sinA
    1 -cos^2(A) = sin^2(A)
    sin^2(A) = sin^2(A)
    and that is true.

    The usual way, by developing one side of the equation to become exactly the other side,
    LHS =
    = (1 -cosA)/sinA
    Multiply both numerator and denominator by (1 +cosA),
    = [(1 -cosA)(1 +cosA)]/[sinA(1 +cosA)]
    = [1 -cos^2(A)]/[sinA(1 +cosA)]
    = [sin^2(A)]/[sinA(1 +cosA)]
    = sinA/(1 +cosA)
    = RHS

    Therefore, proven.

    ----------------------------
    2) Solve for angle A, if -pi < A < pi.

    sin(2A) = cos(2A)
    Divide both sides by cos(2A),
    tan(2A) = 1
    Positive tangent, so angle 2A is in the 1st or 3rd quadrants.
    2A = arctan(1) ------***

    In the interval/domain "-pi < A < pi", all measured counterclockwise,
    ----the 1st quadrant is from 0 to pi/2.
    ----the 3rd quadrant is from -pi to -pi/2.


    2A = pi/4, in the 1st quadrant, so,
    A = pi/8 -----------------------------***

    2A = -pi +pi/4 = -3pi/4, in the 3rd quadrant, so,
    A = -3pi/8 -----------------------------------------***

    Check when A = pi/8,
    sin(2*pi/8) =? cos(2*pi/8)
    sin(pi/4) =? cos(pi/4)
    1/sqrt(2) =? 1/sqrt(2)
    Yes, so, OK.

    Check when A = -3pi/8,
    sin(2* -3pi/8) =? cos(2* -3pi/8)
    sin(-3pi/4) =? cos(-3pi/4)
    using a calculator,
    -0.7071 =? -0.7071
    Yes, so, OK.

    Therefore, A or theta = -3pi/4 or pi/4, in radians. ---------answer.
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  5. #5
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    thank you guys for your help, it makes much more sense now. But i've stumbled across another question that I'm struggling with. Can anyone help me with this one:
    Solve for 0<\theta<2\pi
     tan3\theta(tan3\theta+1)=2
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  6. #6
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    Quote Originally Posted by Confuzzled?
    thank you guys for your help, it makes much more sense now. But i've stumbled across another question that I'm struggling with. Can anyone help me with this one:
    Solve for 0<\theta<2\pi
     tan3\theta(tan3\theta+1)=2
    You should have just posted this one separately. Adding separate or new problems to an existing question is not a good practice.

    But let me answer this new one here, for I am about to leave for work and so I cannot wait for a new/separate posting.

    tan(3A)*[tan(3A) +1] = 2
    tan^2(3A) +tan(3A) -2 = 0
    That is like x^2 +x -2 = 0 if x=tan(3A), so, factoring,
    [tan(3A) +2]*[tan(3A) -1] = 0

    tan(3A) +2 = 0
    tan(3A) = -2
    Negative tangent, so angle 3A is in 2nd or 4th quadrants
    3A = arctan(-2) -----***
    Using a calculator,
    3A = -1.10715 radians ---in the 4th quadrant
    In normal measuring, counterclockwise, that is (2pi -1.10715) = 5.176 rad
    Hence,
    A = 5.176/3 = 1.72535 radians ----------**

    3A = -1.10715 -pi = -4.2491 rad ----in the 2nd quadrant
    Counterclocwise, that is 2pi -4.2491 = 2.0341 rad
    hence, A = 2.0341/3 = 0.678 rad --------**

    Check when A = 1.72535 rad,
    tan(3*1.72535)[tan(3*1.72535) +1] =? 2
    (-2)[-2 +1] =? 2
    -2[-1] =? 2
    2 =? 2
    Yes, so, OK.

    Check when A = 0.678 rad,
    tan(3*0.678)[tan(3*0.678) +1] =? 2
    (-2)[-2 +1] =? 2
    -2[-1] =? 2
    2 =? 2
    Yes, so, OK.

    Therefore, A or theta = 0.678 or 1.72535, in radians. -----answer.

    --------------
    Ooppss, I forgot about the other factor.

    tan(3A) -1 = 0
    tan(3A) = 1
    Positive tangent, so angle 3A is in 1st or 3rd quadrants
    3A = arctan(1) ---***

    3A = pi/4, in the 1st quadrant, so,
    A = pi/4/3 = pi/12 rad ------**

    3A = pi +pi/4 = 5pi/4, in the 3rd quadrant, so,
    A = 5pi/4/3 = 5pi/12 rad -----**

    Check when A = pi/12 rad,
    tan(3*pi/12)[tan(3*pi/12) +1] =? 2
    tan(pi/4)[tan(pi/4) +1] =? 2
    (1)[1 +1] =? 2
    1[2] =? 2
    2 =? 2
    Yes, so, OK.

    Check when A = 5pi/12 rad,
    tan(3*5pi/12)[tan(3*5pi/12) +1] =? 2
    tan(5pi/4)[tan(5pi/4) +1] =? 2
    (1)[1 +1] =? 2
    1[2] =? 2
    2 =? 2
    Yes, so, OK.

    Therfore, altogether,
    A or theta = pi/12, 0.678, 5pi/12, or 1.72535, all in radians. ---answer.
    Last edited by ticbol; March 7th 2006 at 11:28 AM. Reason: not complete.
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