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Math Help - trig identity prob

  1. #1
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    trig identity prob

    -Simplify each expression using the fundamental identities

    2) Sec theta x cos theta

    3) 1/csc^2x + 1/sec^2x

    4) 1-sin^2u/cosu

    5) cos^2x+ (sin x +1)^2/sin x + 1

    Pleeease help, I don't know what to do
    thanks!
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  2. #2
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    Quote Originally Posted by zerostumbleine33 View Post
    -Simplify each expression using the fundamental identities

    2) Sec theta x cos theta

    3) 1/csc^2x + 1/sec^2x

    4) 1-sin^2u/cosu

    5) cos^2x+ (sin x +1)^2/sin x + 1

    Pleeease help, I don't know what to do
    thanks!
    here's number 2!

    sec \theta cos\theta

    sec \theta = \frac{1}{cos\theta}

    \frac{1}{cos\theta}*\frac{cos\theta}{1}=1
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  3. #3
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    Quote Originally Posted by zerostumbleine33 View Post
    -Simplify each expression using the fundamental identities

    2) Sec theta x cos theta

    3) 1/csc^2x + 1/sec^2x

    4) 1-sin^2u/cosu

    5) cos^2x+ (sin x +1)^2/sin x + 1

    Pleeease help, I don't know what to do
    thanks!
    Please use parenthesis! I presume that 4 is to read \frac{1 - sin^2(u)}{cos(u)} and not as 1 - \frac{sin^2(u)}{cos(u)}.

    My first rule of thumb is to get everything in terms of sine and cosine. This will immediately solve 2 and 3.

    Also look for sin^2(u) + cos^2(u) = 1 and
    sin(2u) = 2sin(u)cos(u) and cos(2u) = 1 - 2sin^2(u) = 2cos^2(u) - 1 = cos^2(u) - sin^2(u).

    Expand any expressions under exponents. ie. (sin(x) + 1)^2 = sin^2(x) + 2sin(x) + 1.

    Finally simplify and see if you can factor anything after you have done the above steps.

    -Dan
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  4. #4
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    Hello, zerostumbleine33!

    It looks like you don't know any of the basic identities . . . *sigh*



    Simplify each expression using the fundamental identities

    3)\;\;\frac{1}{\csc^2\!x} + \frac{1}{\sec^2\!x}

    We have: . \left(\frac{1}{\csc x}\right)^2 + \left(\frac{1}{\sec x}\right)^2

    You're expected to know that: . \frac{1}{\csc x} \:=\:\sin x .and . \frac{1}{\sec x} \:=\:\cos x

    \text{So we have: }\;(\sin x)^2 + (\cos x)^2 \;=\;\underbrace{\sin^2\!x + \cos^2\!x)}_{\text{This is 1, right?}}

    Final answer: . 1



    4)\;\;\frac{1 - \sin^2\!u}{\cos u}

    You're supposed to know that: . \sin^2\!u + \cos^2\!u \:=\:1

    . . and you should recognize: . 1 - \sin^2\!u \:=\:\cos^2\!u


    So we have: . \frac{1-\sin^2\!u}{\cos u} \;=\;\frac{\cos^2\!u}{\cos u} \;=\;\cos u

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