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Thread: Pythagorean Identities

  1. #1
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    Pythagorean Identities

    I am still having trouble with these, maybe you could explain what I am doing wrong, here are a few I have been working on.
    find tan angle
    tan angle in QII, sec angle=-4/sqrt(5)
    tan 2 angle in QI, cos 2 angle=3/5
    tan 2 angle in QI, csc 2 angle=4/3
    tan angle/2 in QII, Sec angle/2=-4
    sin angle/2=1/3

    here is what I did
    sec angle=-4/sqrt(5), sec=1/cos
    tan^2 angle+1=(-4/sqrt(5))^2
    tan^2 angle +1= 16/5
    since sec is neg is QII, tan=-sqrt(11/5)

    cos2 angle=3/5, cos is pos in QI
    cos^2 angle=1-sin^2 angle
    (3/5)^2=1-sin^2 angle
    sin angle= 4/5
    tan=sin angle/con angle= 4/5/3/5= 4/3
    tan= 2(4/3)= 8/3

    tan2 angle in QI, csc2 angle= 4/3, csc is pos in QI
    csc angle = 1/sin angle
    cos^2 angle=1-1/4/3
    cos^2= 9/4, cos= 3/4
    csc2 angle, cos angle=9/4
    tan^2 angle +1=(1/9/4)^2
    tan^2 angle +1= 16/81
    I can see this is going badly.

    tan angle/2 in QII, sec angle/2=-4, sec in QII is neg
    sec= 1/cos angle
    tan^2 angle+1=(-8)^2
    tan^2 angle=63
    tan=sqrt(63)
    tan angle/2 in QII= -6sqrt(7)

    sin angle/2= 1/3
    +-sqrt(1-cos a/2)= +-sqrt(1-1/3/2)= +-sqrt(1/3)

    Could you please show me what I am doing wrong. I have been working on this for a few hours and I think a thirty minute break to clear my head might do some good before I start doing some more problems. I would appreciate all and any suggestions you can give because I reall want to understand these identities.
    Thank You very much!!!!
    Keith Stevens
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  2. #2
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    Hello, Keither!

    Instead of wrestling with various identities,
    . . why not use the method I've used . . . "build the triangle."


    Find $\displaystyle \tan\theta$

    $\displaystyle (1)\;\theta \text{ in Q2},\;\sec\theta = -\frac{4}{\sqrt{5}}$

    Since $\displaystyle \sec\theta = \frac{hyp}{adj}$, we have: .$\displaystyle \sec\theta \:=\:\frac{4}{-\sqrt{5}} \:=\:\frac{hyp}{adj}$

    From $\displaystyle opp^2 + adj^2\:=\:hyp^2$, we have: .$\displaystyle opp^2 +(-\sqrt{5})^2\:=\:4^2\quad\Rightarrow\quad opp = \pm\sqrt{11}$
    In $\displaystyle Q2$, opposite is positive: $\displaystyle opp = \sqrt{11}$

    We know all three sides of the triangle: .$\displaystyle opp = \sqrt{11},\;adj = -\sqrt{5},\;hyp = 4$

    We can write any function now: .$\displaystyle \tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{\sqrt{11}}{-\sqrt{5}} \:=\:\boxed{-\frac{\sqrt{55}}{5}}$



    $\displaystyle (2)\;2\theta\text{ in Q1},\;\cos2\theta = \frac{3}{5}$

    This time using identities is faster . . .

    We have: .$\displaystyle \sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{1-\cos2\theta}{2}} $

    . . Then: .$\displaystyle \sin\theta\:=\:\pm\sqrt{\frac{1 - \frac{3}{5}}{2}} \:=\:\pm\sqrt{\frac{\frac{2}{5}}{2}} \:=\:\pm\sqrt{\frac{1}{5}}\:=\:+\frac{1}{\sqrt{5}}$ . . . (positive in Q1)


    We have: .$\displaystyle \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}\quad\Rightarrow\quad\cos\theta \:=\:\pm\sqrt{\frac{1 + \cos2\theta}{2}}$

    . . Then: .$\displaystyle \cos\theta\:=\:\pm\sqrt{\frac{1 + \frac{3}{5}}{2}} \:=\:\pm\sqrt{\frac{\frac{8}{5}}{2}}\:=\:\pm\sqrt{ \frac{4}{5}} \:=\:+\frac{2}{\sqrt{5}}$ . . . (positive in Q1)


    Therefore: .$\displaystyle \tan\theta \:=\:\frac{\sin\theta}{\cos\theta} \:=\:\frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} \:=\:\boxed{\frac{1}{2}}$

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  3. #3
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    Quote Originally Posted by kcsteven View Post
    I am still having trouble with these, maybe you could explain what I am doing wrong, here are a few I have been working on.
    find tan angle
    tan angle in QII, sec angle=-4/sqrt(5)
    tan 2 angle in QI, cos 2 angle=3/5
    tan 2 angle in QI, csc 2 angle=4/3
    tan angle/2 in QII, Sec angle/2=-4
    sin angle/2=1/3

    here is what I did
    sec angle=-4/sqrt(5), sec=1/cos
    tan^2 angle+1=(-4/sqrt(5))^2
    tan^2 angle +1= 16/5
    since sec is neg is QII, tan=-sqrt(11/5)
    This one is fine.

    From here on either your notation is inaccurate, or you were trying to answer the wrong questions.

    Quote Originally Posted by kcsteven View Post
    cos2 angle=3/5, cos is pos in QI
    cos^2 angle=1-sin^2 angle
    (3/5)^2=1-sin^2 angle
    sin angle= 4/5
    tan=sin angle/con angle= 4/5/3/5= 4/3
    tan= 2(4/3)= 8/3
    You are given $\displaystyle cos(2x) = \frac{3}{5}$

    So
    $\displaystyle sin^2(2x) = 1 - cos^2(2x) = 1 - \left ( \frac{3}{5} \right )^2 = \frac{16}{25}$

    or
    $\displaystyle sin(2x) = \frac{4}{5}$ (Positive since x is in the first quadrant.)

    Thus
    $\displaystyle tan(2x) = \frac{sin(2x)}{cos(2x)} = \frac{ \frac{4}{5} }{ \frac{3}{5} } = \frac{4}{3}$


    Quote Originally Posted by kcsteven View Post
    tan2 angle in QI, csc2 angle= 4/3, csc is pos in QI
    csc angle = 1/sin angle
    cos^2 angle=1-1/4/3
    cos^2= 9/4, cos= 3/4
    csc2 angle, cos angle=9/4
    tan^2 angle +1=(1/9/4)^2
    tan^2 angle +1= 16/81
    I can see this is going badly.
    $\displaystyle csc(2x) = \frac{4}{3}$

    So
    $\displaystyle sin(2x) = \frac{1}{csc(2x)} = \frac{3}{4}$

    Then, just like the last problem:
    $\displaystyle cos(2x) = \frac{ \sqrt{7} }{4}$

    So
    $\displaystyle tan(2x) = \frac{3}{\sqrt{7}}$


    Quote Originally Posted by kcsteven View Post
    tan angle/2 in QII, sec angle/2=-4, sec in QII is neg
    sec= 1/cos angle
    tan^2 angle+1=(-8)^2
    tan^2 angle=63
    tan=sqrt(63)
    tan angle/2 in QII= -6sqrt(7)
    $\displaystyle sec(x/2) = - 4$

    So
    $\displaystyle cos(x/2) = \frac{1}{sec(x/2)} = -\frac{1}{4}$

    Thus
    $\displaystyle sin(x/2) = \frac{ \sqrt{15} }{4}$ (This is positive since x is in the second quadrant.)

    So
    $\displaystyle tan(x/2) = - \frac{ \sqrt{15} }{1} = -\sqrt{15}$


    Quote Originally Posted by kcsteven View Post
    sin angle/2= 1/3
    +-sqrt(1-cos a/2)= +-sqrt(1-1/3/2)= +-sqrt(1/3)
    Do we want $\displaystyle tan(x/2)$ again? It can't be done until we know what quadrant the angle is in.

    -Dan
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