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Math Help - Analytic Trig.

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Analytic Trig.

    1. (sin2x + cos2x) ^2=1
    solutions are in the interval [0, 2pi)

    using double angle formulas find sin2u , cos2u, and tan2u in:

    2. cot u = -6, 3pi/2< u < 2pi

    rewrite the expression in terms of the first power of cos:
    3. sin^4x*cos^2x

    i get stuck in between theses problems, its usually the factoring...
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  2. #2
    MHF Contributor arbolis's Avatar
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    using double angle formulas find sin2u , cos2u, and tan2u in:

    2. cot u = -6, 3pi/2< u < 2pi
    You see that sin(2u)=sin(u+u). Then you just have to apply the well known formula sin(a+b)=sin(a)cos(b)+sin(b)cos(a).
    For cos(a+b), we have that it's equal to cos(a)cos(b)-sin(a)sin(b).
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    Hello, >_<SHY_GUY>_<!

    1.\;\;(\sin2x + \cos2x)^2\:=\:1,\quad x \in [0, 2\pi)

    Square it out: . \underbrace{\sin^2\!2x + \cos^2\!2x}_{\text{This is 1}} + 2\sin2x\cos2x \:=\:1

    We have: . 2\sin2x\cos2x\:=\:0 \quad\Rightarrow\quad \sin4x \:=\:0

    Hence: . 4x\:=\:0,\:\pi,\:2\pi,\:3\pi,\:4\pi,\:5\pi,\:6\pi,  \:7\pi

    Therefore: . x \;=\;0,\:\frac{\pi}{4},\:\frac{\pi}{2},\:\frac{3\p  i}{4},\:\pi,\:\frac{5\pi}{4},\:\frac{3\pi}{2},\:\f  rac{7\pi}{4}

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    Hello again, >_<SHY_GUY>_<!

    2.\;\;\cot u \,=\,-6,\quad \frac{3\pi}{2} < u < 2\pi

    Using double angle formulas, find: . \sin2u,\;\;\cos2u,\;\;\tan2u

    We are given: . \cot u \:=\:-\frac{6}{1} \:=\:\frac{adj}{opp}\quad\text{ and }u\text{ is in Quadrant 4.}

    Then: . opp = \text{-}1,\;\;adj = 6\;\;\text{ and }\:hyp = \sqrt{37}

    Hence: . \sin u \:=\:-\frac{1}{\sqrt{37}},\quad \cos u \:=\:\frac{6}{\sqrt{37}}



    \sin2u \:=\:2\sin u\cos u \:=\:2\left(-\frac{1}{\sqrt{37}}\right)\left(\frac{6}{\sqrt{37}  }\right) \:=\:-\frac{12}{37}

    \cos2u \:=\:\cos^2\!u - \sin^2\!u \:=\:\left(\frac{6}{\sqrt{37}}\right)^2 - \left(-\frac{1}{\sqrt{37}}\right)^2 \:=\:\frac{35}{37}

    \tan2u \;=\;\frac{\sin2u}{\cos2u} \;=\;\frac{\text{-}\frac{12}{37}}{\frac{35}{37}} \;=\;-\frac{12}{35}

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