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Math Help - 6 Trigonometry Questions needing help...

  1. #1
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    6 Trigonometry Questions needing help...

    So I have like 6 problems right now left or so on my Calculus homework sheet (somewhere around like 150 problems total) and I'm having trouble showing these solutions.

    Problem #1: Express sin(arccos x) in a form that contains no trigonometric functions or their inverses.

    Problem #2: Use the trigonometric identity, sin(A+B) = (sin A cos B) + (cos A sin B), to verify that, sin2x = 2sin x cos x

    Problem #3: Show that sin⁵ x cos⁴ x = sin⁵ x - 2sin⁷ x + sin⁹ x.

    Problem #4: Show that sec³ x tan⁵ x = sec x tan x(sec⁶ x - 2sec⁴ x + sec² x).

    Problem #5: Show that sin⁴ x = ³⁄₈ - ½cos 2x + ¹⁄₈cos 4x

    Problem #6: Show that (1/1+ sin x) = sec² x - sec x tan x


    And just to restate, that I'm sorry if it seems that I'm posting alot of questions on here but there is alot of problems and I do not have a book or any example problems to work off)

    Thanks for any and all help.

    Okay so I know that I'm a new member and all, and I'm sorry if it seems like
    Last edited by forkball42; August 12th 2007 at 05:15 PM. Reason: Spelling error
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  2. #2
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    1) You know that \sin x=\sqrt{1-\cos^2x}, now compute \sin(\arccos x)

    2) \sin2x=\sin(x+x), now apply the given formula.

    It doesn't understand very well the exponents of some identities, try to use LaTeX
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  3. #3
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    Hello, forkball42!

    Welcome aboard . . . Here's some help . . .


    1) Express \sin(\arccos x) in a form
    that contains no trigonometric functions or their inverses.
    Let \theta \:=\:\arccos x\quad\Rightarrow\quad \cos\theta \:=\:x \:=\:\frac{x}{1} \:=\:\frac{adj}{hyp}

    So \theta is an acute angle in a right triangle
    . . with: . adj \,= \,x,\:hyp \,= \,1

    Using Pythagorus, we find that: . opp \,=\,\sqrt{1-x^2}
    . . Hence: . \sin\theta \:=\:\frac{opp}{hyp} \:=\:\frac{\sqrt{1-x^2}}{1} \:=\:\sqrt{1-x^2}

    Therefore: . \sin(\arccos x) \:=\:\sqrt{1-x^2}



    2) Use the identity: . \sin(A+B) \:= \:\sin A\cos B + \cos A\sin B
    to verify that: . \sin2x \:= \:2\sin x \cos x
    We have: . \sin2x \;=\;\sin(x + x) \;= \;\sin x\cos x + \sin x\cos x \;=\;2\sin x\cos x


    3) Show that sin⁵ x cos⁴ x = sin⁵ x - 2sin⁷ x + sin⁹ x.
    As you can see, we can't read the exponents.
    . . But I'll take a guess . . .


    Is it: . \sin^2\!x\cos^4\!x \;=\;\sin^6\!x - 2\sin^4\!x + \sin^2\!x ?


    We need the identity: . \sin^2\!\theta + \cos^2\!\theta \:=\:1\quad\Rightarrow\quad\cos^2\!\theta \:=\:1 -\sin^2\!\theta

    We have: . \sin^2\!x\cos^4\!x \;=\;\sin^2\!x\left(\cos^2\!x\right)^2 \;=\;\sin^2\!x\left(1 - \sin^2\!x\right)^2

    . . = \;\sin^2\!x\left(1 - 2\sin^2\!x + \sin^4\!x\right) \;=\;\sin^2\!x - 2\sin^4\!x + \sin^6\!x

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  4. #4
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    Thank you for the help on problems 1 and 2. However I don't understand why my powers will not show up for your computers but I have managed to correctly solve 3 of the 4 other problems.

    The one I still need is problem 5.

    Here it is again and I'll make it to where I know you can read it.

    Problem #5: Show that sin^4 x = (3/8) - (1/2)cos 2x + (1/8)cos 4x.

    Sorry if this seems harder to read but it is the only way I can post the problem.

    Thanks again for everything!
    Last edited by forkball42; August 13th 2007 at 01:28 PM. Reason: Spelling
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  5. #5
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    Hello, forkball42!

    For #5, we will need these two double-angle identities:

    . . \sin^2\!\theta \:=\:\frac{1-\cos2\theta}{2}\qquad\qquad\cos^2\!\theta \:=\:\frac{1+\cos2\theta}{2}


    5) Show that: . \sin^4\!x \:=\:\frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x

    We have: . \sin^4\!x \:=\:\left(\sin^2\!x\right)^2 \;=\;\left(\frac{1-\cos2x}{2}\right)^2\;=\;\frac{1}{4}\left(1 - 2\cos2x + \cos^2\!2x\right)

    . . = \;\frac{1}{4}\left(1 - 2\cos2x + \frac{1+\cos4x}{2}\right) \;=\;\frac{1}{4}\left(1 - 2\cos2x + \frac{1}{2} + \frac{1}{2}\cos4x\right)

    . . = \;\frac{1}{4}\left(\frac{3}{2} - 2\cos2x + \frac{1}{2}\cos4x\right) \;=\;\frac{3}{8} - \frac{1}{2}\cos2x + \frac{1}{8}\cos4x

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  6. #6
    Eater of Worlds
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    If you don't mind, here's another approach. More long connected than Soroban's, but just to show there's more than one way to skin a trig identity.

    we will use cos(2x)=cos^{2}(x)-sin^{2}(x) and

    cos(4x)=cos^{2}(2x)-sin^{2}(2x)

    Start with the right side:

    \frac{3}{8}-\frac{4cos(2x)}{8}+\frac{cos(4x)}{8}

    \frac{3}{8}-\left(\frac{4cos^{2}(x)-4sin^{2}(x)}{8}\right)+\left(\frac{cos^{2}(2x)-sin^{2}(2x)}{8}\right)

    \frac{3-4cos^{2}(x)+4sin^{2}(x)+cos^{2}(2x)-sin^{2}(2x)}{8}

    \frac{3-4(1-sin^{2}(x))+4sin^{2}(x)+(1-sin^{2}(2x))-sin^{2}(2x)}{8}

    \frac{3-4+4sin^{2}(x)+4sin^{2}(x)+1-sin^{2}(2x)-sin^{2}(2x)}{8}

    \frac{8sin^{2}(x)-2sin^{2}(2x)}{8}

    But 2sin^{2}(2x)=8sin^{2}(x)cos^{2}(x)

    \frac{8sin^{2}(x)-8sin^{2}(x)cos^{2}(x)}{8}

    \frac{8sin^{2}(x)(1-cos^{2}(x))}{8}

    sin^{4}(x)
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  7. #7
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    Hello, Galactus!

    Good job . . . lovely work!

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