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Math Help - In need of trig help..love ya!

  1. #1
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    In need of trig help..love ya!

    I couldnt figure out some of my trig problems...so Im having a bit of trouble wasting time on these problems. Any help would be great. Thanks

    Evaluate the expressions:
    cos(arcsin 1/4) couldnt figure out what arcsin 1/4 was...maybe 22.5 degrees?

    csc(arccsc -2)

    arccos(cos pi/2)

    cos(arcsin 3/5 +arccos sqrt3/2)

    sec^2x-2=tan^2x


    sin^2+sin x=0

    Use a calculator to find the value. Give answers as real numbers.

    sin(arccos .8324)

    tan(arcsin 2)

    Write the following as an algebraic expression in u, u>0

    cos(arctan u)

    sin 2theta =-1/2

    Solve the equation for the interval [0,2pi)

    cos^2x + 2cos x+1-0

    Solve the equation for solutions in the interval [0, 360). Round to the nearest degree.

    sin 2theta =cos theta

    sin 2theta= -1/2

    sqrt3 sec2theta =2

    Solve the equation for x

    y=8 cos 3x

    Solve the equation for exact solutions.

    arcsin(y-pi/6) = pi/3

    arctanx = arcsin 7/25

    Solve the equation.

    arcsin 2x +2 arccos x=pi
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  2. #2
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    \sec^2x - 2 = \tan^2x.

    Use the identity 1+\tan^2x = \sec^2x

    1+\tan^2x - 2 = \tan^2x

    \tan^2x - 1 = \tan^2x

    -1 = 0.

    This equation is nonsense, so there does not exist an x that satisfies this equation.



    \arccos(\cos{\frac{\pi}{2}}) = \frac{\pi}{2}.



    \sin^2x + \sin x = 0

    \sin x(\sin x + 1) = 0

    \sin x = 0 or \sin x + 1 = 0.

    If \sin x = 0, x = \{0, \pi\} + 2n\pi, n \in \mathbf{Z}.

    If \sin x + 1 = 0, \sin x = -1, x = \frac{3\pi}{2} + 2n\pi, n \in \mathbf{Z}.

    So the solution to this equation is \{0, \pi, \frac{3\pi}{2}\} + 2n\pi.
    Last edited by Prove It; December 2nd 2008 at 01:23 AM.
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