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Thread: 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!

    $\displaystyle sec \theta cos\theta$

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

    $\displaystyle \frac{1}{cos\theta}*\frac{cos\theta}{1}=1$
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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 $\displaystyle \frac{1 - sin^2(u)}{cos(u)}$ and not as $\displaystyle 1 - \frac{sin^2(u)}{cos(u)}$.

    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 $\displaystyle sin^2(u) + cos^2(u) = 1$ and
    $\displaystyle sin(2u) = 2sin(u)cos(u)$ and $\displaystyle cos(2u) = 1 - 2sin^2(u) = 2cos^2(u) - 1 = cos^2(u) - sin^2(u)$.

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

    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

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

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

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

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

    Final answer: .$\displaystyle 1$



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

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

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


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

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