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Math Help - Trig equation help please!

  1. #1
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    Trig equation help please!

    I think I'm having a massive blonde moment...



    How would I begin?

    edit: also this question:



    Here's my working so far:

    \frac{1+\frac{sin^2x}{cos^2x}}{cos^2x}

    \frac{1+\frac{sin^2x}{1-sin^2x}}{1-sin^2x}

    I'm stuck here. I don't actually know what to do with multi-layered fractions, never been taught it.

    I know the answer is

    So clearly, I have to multiply by 1-sin^2x

    How do I get there?


    Also, this question:

    part b,

    Here's my working, I'm starting to think the question is wrong..?

    \frac{1}{cos\theta} + \frac{sin\theta}{cos\theta}=\frac{cos\theta}{1-sin\theta}

    \frac{cos\theta+sin\theta cos\theta}{cos^2\theta}

    \frac{cos\theta(1+sin\theta)}{1-sin^2\theta}

    What do I do from here?
    Last edited by Bude8; April 8th 2014 at 02:08 AM.
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  2. #2
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    Re: Trig equation help please!

    9)

    let $u=x^2$

    $\sin(u)=\dfrac 1 2$ for $-\pi \leq u \leq \pi$

    can you solve that? There are two solutions in between $-\pi$ and $\pi$. You should be able to solve that.
    Then solve $u=x^2$ This will give you 4 solutions total.

    10)

    hint: $1+\tan^2(x)=\sec^2(x)$

    2nd hint: (you shouldn't need this one) $\cos^4(x)=\left(\cos^2(x)\right)^2=\left(1-sin^2(x)\right)^2$

    11)

    $\dfrac 1 {\cos(\theta)}+\tan(\theta)=\dfrac{1+\sin(\theta)} {cos(\theta)}$

    Now multiply that by $1$ in the form of $\dfrac{1-\sin(\theta)}{1-\sin(\theta)}$ and simplify it.
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  3. #3
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    Re: Trig equation help please!

    Quote Originally Posted by romsek View Post
    x
    Thanks for your help

    Can I just confirm something for question 10 though?

    \frac{\frac{1}{cos^2x}}{cos^2x}

    That simplifies to \frac{1}{(cos^2x)^2} right? That's how I get to the answer? Sorry - I don't fully understand multiple layered fractions.

    Edit: For 11, \frac{cos\theta}{1-sin\theta} = \frac{1+sin\theta}{cos\theta}? Confused :s
    Last edited by Bude8; April 8th 2014 at 03:06 AM.
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  4. #4
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    Re: Trig equation help please!

    Quote Originally Posted by Bude8 View Post
    Thanks for your help

    Can I just confirm something for question 10 though?

    \frac{\frac{1}{cos^2x}}{cos^2x}

    That simplifies to \frac{1}{(cos^2x)^2} right? That's how I get to the answer? Sorry - I don't fully understand multiple layered fractions.
    you need to get the basics down cold. Yes that's correct.

    Edit: For 11, \frac{cos\theta}{1-sin\theta} = \frac{1+sin\theta}{cos\theta}? Confused :s
    sigh.. it's just algebra

    $\dfrac 1 {\cos(\theta)} + \tan(\theta) = \dfrac 1 {\cos(\theta)}+\dfrac{\sin(\theta)}{\cos(\theta)}= $

    $\dfrac{1+\sin(\theta)}{\cos(\theta)}=\dfrac{1+ \sin(\theta)}{\cos(\theta)}\cdot \dfrac{1- \sin(\theta)}{1- \sin(\theta)}=$

    $\dfrac{1-\sin^2(\theta)}{\cos(\theta)(1-\sin(\theta))}=\dfrac{\cos^2(\theta)}{\cos(\theta) (1-\sin(\theta))}=$

    $\dfrac{\cos(\theta)}{1-\sin(\theta)}$
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