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Math Help - trigomatic

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    17

    trigomatic

    Help on this please!!! Need working or explaination for these questions.
    *** Answer given but some might be wrong. THANKS!


    change to an expression containing only sin and cos.

    1
    a.tan0csc0 Ans:1/cos0
    b. sec0-tan0sin0 Ans:cos0
    c.(tan0/csc0)+(sin/tan0) Ans:1/cos0
    d. cot0+tan0 Ans:1/(sin0cos0)
    e. (sec^sqrt0-1)/(sin^sqrt0) Ans:1/cos^sqrt0
    f. (cot^sqrt0)/(csc^sqrt0) Ans:cos^sqrt0


    2. Simplify
    a. (sin x/csc x)+ (cos x/sec x) Ans:1
    b.1/cot0sec0 Ans:sin0
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, watsonmath!

    You really can't do any of these??


    Okay, I must assume you know the fundamental identities:

    . . \csc\theta = \frac{1}{\sin\theta}\qquad\sec\theta = \frac{1}{\cos\theta} . . . \tan\theta = \frac{\sin\theta}{\cos\theta}\qquad\cot\theta = \frac{\cos\theta}{\sin\theta}

    \sin^2\theta + \cos^2\theta \:=\:1\quad\Rightarrow\quad \sin^2\theta \:=\:1 - \cos^2\theta . . . \Rightarrow\quad\cos^2\theta\:=\:1 - \sin^2\theta


    1. Change to an expression containing only sin and cos.

    a)\;\tan\theta\csc\theta
    We have: \tan\theta\csc\theta\:=\:\frac{\sin\theta}{\cos \theta}\cdot\frac{1}{\sin\theta}\:=\:\frac{1}{\cos  \theta}

    b)\;\sec\theta - \tan\theta\sin\theta
    \sec\theta - \tan\theta\sin\theta \;=\;\frac{1}{\cos\theta} -\frac{\sin\theta}{\cos\theta}\sin\theta \;= . \frac{1 - \sin^2\theta}{\cos\theta}\;=\;\frac{\cos^2\theta}{  \cos\theta} \;=\;\cos\theta

    c)\; \frac{\tan\theta}{\csc\theta} + \frac{\sin\theta}{\tan\theta}
    We have: \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\sin  \theta}} + \frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}}\;  =\;\frac{\sin^2\theta}{\cos\theta} + \cos\theta \;=\;\frac{\sin^2\theta + \cos^2\theta}{\cos\theta} \;= \;\frac{1}{\cos\theta}


    d)\;\cot\theta + \tan\theta
    We have: \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} \;=\;\frac{\cos^2\theta}{\sin\theta\cos\theta} + \frac{\sin^2\theta}{\sin\theta\cos\theta}\;= \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \;=\;\frac{1}{\sin\theta\cos\theta}


    e)\;\frac{\sec^2\theta -1}{\sin^2\theta}
    We have: \frac{\tan^2\theta}{\sin^2\theta}\;=\;\frac{\frac{  \sin^2\theta}{\cos^2\theta}}{\sin^2\theta} \;=\;\frac{1}{\cos^2\theta}


    f)\;\frac{\cot^2\theta}{\csc^2\theta}
    We have: \frac{\frac{\cos^2\theta}{\sin^2\theta}}{\frac{1}{  \sin^2\theta}}\;=\;\cos^2\theta


    2. Simplify

    a)\;\frac{\sin x}{\csc x} + \frac{\cos x}{\sec x}
    We have: \frac{\sin x}{\frac{1}{\sin x}} + \frac{\cos x}{\frac{1}{\cos x}} \;=\;\sin^2x + \cos^2x\;=\;1


    b)\;\frac{1}{\cot\theta\sec\theta}
    We have: \frac{1}{\frac{\cos\theta}{\sin\theta}\cdot\frac{1  }{\cos\theta}} \;=\;\frac{1}{\frac{1}{\sin\theta}} \;=\;\sin\theta<br />
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