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Math Help - Trig Identities and finding real numbers

  1. #1
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    Trig Identities and finding real numbers

    I am currently finishing a review of precal for my cal I class. It has been a little while since i have used this kind of math so i am at a stand still. I have five questions.

    1. find and equivalent expression involving only sines and cosines, and then simplify it.

    sin^2(x) +sin(x)-1+cos^2(x)

    2. same as above except:

    sec(x) +sin(x)

    3.find sec(x) if tan(x)=3/4 and x is in quadrant one.

    4. find all real numbers in the interval [0,2pi] that satisfy the equation:

    2sin(x)cos(x)+cos(x)=0

    5. same as #4 except:

    sin(x)=1-2sin^2(x)

    I struggled with these kind of problems when i was taking pre-cal. I would greatly appreciate it if someone could help me solve these and teach me a little more about the subject.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by davisjl1 View Post
    I am currently finishing a review of precal for my cal I class. It has been a little while since i have used this kind of math so i am at a stand still. I have five questions.

    1. find and equivalent expression involving only sines and cosines, and then simplify it.

    sin^2(x) +sin(x)-1+cos^2(x)
    Hint: note that -1 + \cos^2 x = -(1 - \cos^2 x) = - \sin^2 x

    2. same as above except:

    sec(x) +sin(x)
    Hint: \sec x = \frac 1{\cos x}

    3.find sec(x) if tan(x)=3/4 and x is in quadrant one.
    Hint: since we are in the first quadrant, the secant is positive. now note that \sec^2 x = 1 + \tan^2 x


    4. find all real numbers in the interval [0,2pi] that satisfy the equation:

    2sin(x)cos(x)+cos(x)=0
    Hint: factor out the common cosine, you get

    \cos x (2 \sin x + 1) = 0

    \Rightarrow \cos x = 0 or 2 \sin x + 1 = 0 \implies \sin x = - \frac 12

    now find x ....you should know the values by heart that do this

    5. same as #4 except:

    sin(x)=1-2sin^2(x)
    Hint: note that you have

    2 \sin^2 x + \sin x - 1 = 0

    this is quadratic in \sin x (if you let y = \sin x you would have 2y^2 + y - 1 = 0)

    thus we can solve this for \sin x like a quadratic equation. then we can find what x is.
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  3. #3
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    Thank You Jhevon,

    Ok now i got number one.

    Number two i found a common denominator and ended up with sin+cos/sin*cos can this be simplified even more?

    Number three i am still in the dark on this one.

    Number four are the answers 2pi/3 and 4pi/3
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