Results 1 to 6 of 6

Math Help - Impossible Trigonometry Questions

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    3

    Post Impossible Trigonometry Questions

    K, so my teacher presented a set of problems and it is basically impossible to get all of them right (it was a competition to see who could do the most). There were a ton of problems and after going to class I still don't understand how to do some of them.

    Disprove the following identity by counterexample:
    #4
    sin^2(x)+cos(x)=0

    Everything that I plug in says that this isn't an identity. What am I doing wrong?

    Prove the following identities:
    #13
    cotx-sinx=(cos^2(x)+cosx-1)csc(x)

    I spent 2 hours trying to prove this, and I am coming with a nothing.

    And the third one

    Solve the following:
    #14
    1-2sin^2(2x)=3sin2x


    I have a bunch of these that I didn't get right that I don't understand, but these 3 are really bugging me. I already know the answers, so I mostly care about how to get that answer because so far I haven't been able to figure it out. Any help will be great!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2009
    Posts
    63
    Quote Originally Posted by jhunt47 View Post

    Prove the following identities:
    #13
    cotx-sinx=(cos^2(x)+cosx-1)csc(x)

    Solve the following:
    #14
    1-2sin^2(2x)=3sin2x
    i'll try "the lucky" #13

    cotx-sinx
    =\frac{cosx}{sinx}-sinx
    =\frac{cosx-sin^2x}{sinx}

    remember:
    sin^2x+cos^2x=1 so,
    sin^2x=1-cos^2x
    =\frac{cosx-(1-cos^2x)}{sinx}
    =\frac{cosx-1+cos^2x}{sinx}

    remember:
    \frac{1}{sinx}=cscx
    =(cosx-1+cos^2x)(cscx)
    ---Q.E.D---
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello, jhunt47!

    Here's #13 . . .


    13. Prove: . \cot x-\sin x\:=\:\left(\cos^2\!x+\cos x-1\right)(\csc x)

    The right side is: . \bigg[\cos x - (1 - \cos^2\!x)\bigg]\cdot\frac{1}{\sin x}

    . . . . . . . . . . . = \;\bigg[\cos x - \sin^2\!x\bigg]\cdot\frac{1}{\sin x}

    . . . . . . . . . . . =\; \frac{\cos x}{\sin x} - \frac{\sin^2\!x}{\sin x}

    . . . . . . . . . . . =\; \cot x - \sin x

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,547
    Thanks
    539
    Hello again, jhunt47!

    Here's #14 . . . solved on the interval [0,\:2\pi)


    Solve: . (14)\;\;1-2\sin^2\!2x\:=\:3\sin2x

    We have: . 2\sin^2\!2x + 3\sin2x - 1 \:=\:0

    Quadratic Formula: . \sin2x \:=\:\frac{-3 \pm\sqrt{3^2 - 4(2)(-1)}}{2(2)} \;=\;\frac{-3\pm\sqrt{17}}{4}


    We have two equations to solve:

    \sin2x \:=\:\frac{-3 - \sqrt{17}}{4} \:=\:-1.7807764067 . . . no real solution

    \sin2x \:=\:\frac{-3 + \sqrt{17}}{4} \:=\:0.280776406 \quad\Rightarrow\quad 2x \:=\:\arcsin(0.280776406)


    . . 2x \:\approx\:\begin{Bmatrix}16.3^o \\ 163.7^o \\ 376.3^o \\ 523.7^o \end{Bmatrix} \quad\Rightarrow\quad x \;\approx\;\begin{Bmatrix}8.15^o \\ 81.85^o \\ 188.15^o \\ 261.85^o \end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    3

    RE

    Thanks!!! You guys are the bomb-diggity!

    I don't know if you guys will be able to help me, but I got halfway through on this problem and I think that I might be doing it wrong.

    (16/81)^sin^2(x) + (16/81)^1-sin^2(x) = (26/27)

    I know that you can substitute sin^2(x) for 1/4 or 3/4 and the problem will work, but the instructions say "Solve" and I don't think I am solving it very well.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by jhunt47 View Post
    K, so my teacher presented a set of problems and it is basically impossible to get all of them right (it was a competition to see who could do the most). There were a ton of problems and after going to class I still don't understand how to do some of them.

    Disprove the following identity by counterexample:
    #4
    sin^2(x)+cos(x)=0

    Everything that I plug in says that this isn't an identity. What am I doing wrong?

    [snip]
    Nothing. You've disproved it, as required.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometry Questions and identities.
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: October 24th 2010, 05:51 PM
  2. Trigonometry Questions.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 5th 2010, 02:11 AM
  3. :(impossible factoring questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 1st 2010, 06:58 PM
  4. 3 Hard Trigonometry Questions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 2nd 2009, 04:21 PM
  5. 6 Trigonometry Questions needing help...
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: August 13th 2007, 04:21 PM

Search Tags


/mathhelpforum @mathhelpforum