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Math Help - Trigonometric proofs and exact values.

  1. #1
    Member MathBlaster47's Avatar
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    Trigonometric proofs and exact values.

    I have a couple questions that I need a little direction on.

    1:
    Prove that \sin^2 x(1+\cot^2 x)=1

    Method:
    I figured out that I should convert \sin^2 x(1+\cot^2 x)=1 to \sin^2 x (\csc^2 x)=1. But beyond that, I'm a little lost....I know there is more to be done though. I just don't quite understand how reciprocal and Pythagorean identities are related, I know that \sin x \csc x=1 but I'm not sure if the same holds true for their Pythagorean counterparts or if that would be a valid proof.

    2:
    Prove that \tan x(\cot x +\tan x)=\sec^2 x.

    Method:
    I made the equation \tan x(\cot x +\tan x)=\sec^2 x into \tan x(\frac{1}{\tan x} + \tan x)= sec^2 x.
    Does the equation become 1+\tan^2=\sec^2, or am I missing something?
    I think I have the right answer, as I see that the Pythagorean identity for \sec^2 x coinsides with my converted equation.

    3:
    Find the exact value of \cos 105 with out using tables or a calculator.
    Be sure to show your work.

    Method:
    Since the question asks for the value of a unusual degree measure I have to use an addition formula, namely: \cos(\alpha - \beta)=\cos\alpha \cos \beta+\sin \alpha \sin \beta.
    For \alpha I used value 135 and for \beta I used 30 so \cos(\alpha - \beta) \rightarrow \cos(135-30), therefore the whole formula changes to become: \cos(135 - 30)=\cos 135 \cos 30 + \sin 135 \sin 30.
    So, \cos 135 \cos 30 + \sin 135 \sin 30 \rightarrow -\frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} \rightarrow \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \rightarrow \frac{\sqrt{2} +\sqrt{6}}{4}, have I gotten the right answer?

    4:
    Find the direct value of \tan 15 without using tables or a calculator.
    Be sure to to show your work.

    Method:
    The book said nothing about the addition formula for \tan so I searched the internet and found this: \tan(\alpha-\beta)= \frac{\tan \alpha - \tan \beta}{1+ \tan\alpha \tan\beta}
    I decided to use this formula since I could figure out no other to use, I entered \tan 45 for \alpha and \tan 30 for \beta, but I dont know where to go from here....

    Thanks.
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  2. #2
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    You first three methods are correct.
    Last one,
    tan15 = (tan45 - tan30)/(1-tan45*tan30)
    Substitute the values to get tan 15.
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  3. #3
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    For the second one, it might help to write tan(x) as \frac {sin(x)}{cos(x)} and cot(x) as \frac {cos(x)}{sin(x)}

    Edit: Ah, never mind, I didn't read your question properly. If you can use those identities in your proof then you're right.
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  4. #4
    Member MathBlaster47's Avatar
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    Quote Originally Posted by sa-ri-ga-ma View Post
    You first three methods are correct.
    Last one,
    tan15 = (tan45 - tan30)/(1-tan45*tan30)
    Substitute the values to get tan 15.
    Ok! Thank you!

    Just curious, but is there a simpler way to do this question?
    I know I can and will do it this way--don't get me wrong, but my text book said nothing about this way of solving the question, so I think that they may want a different method to be used. Like I said, they did not give one though...
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  5. #5
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    Quote Originally Posted by MathBlaster47 View Post
    Ok! Thank you!

    Just curious, but is there a simpler way to do this question?
    I know I can and will do it this way--don't get me wrong, but my text book said nothing about this way of solving the question, so I think that they may want a different method to be used. Like I said, they did not give one though...
    You don't need the addition formula for tan. A slightly longer method is to use the subtraction formulas for sin and cos:

    \sin(a - b) = \sin a \cos b - \cos a \sin b

    \cos(a - b) = \cos a \cos b + \sin a \sin b

    \tan(a - b) = \frac{\sin(a - b)}{\cos(a - b)}
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  6. #6
    Member MathBlaster47's Avatar
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    Quote Originally Posted by icemanfan View Post
    You don't need the addition formula for tan. A slightly longer method is to use the subtraction formulas for sin and cos:

    \sin(a - b) = \sin a \cos b - \cos a \sin b

    \cos(a - b) = \cos a \cos b + \sin a \sin b

    \tan(a - b) = \frac{\sin(a - b)}{\cos(a - b)}
    Thank you!
    I was beginning to wonder if I was going crazy, or I if I just didn't read my book properly....

    Quote Originally Posted by MathBlaster47 View Post

    1:
    Prove that \sin^2 x(1+\cot^2 x)=1

    Method:
    I figured out that I should convert \sin^2 x(1+\cot^2 x)=1 to \sin^2 x (\csc^2 x)=1. But beyond that, I'm a little lost....I know there is more to be done though. I just don't quite understand how reciprocal and Pythagorean identities are related, I know that \sin x \csc x=1 but I'm not sure if the same holds true for their Pythagorean counterparts or if that would be a valid proof.
    OK! just so I'm clear, my answer to the question I have above is correct?
    Am I right in inferring that \sin^2 x (\csc^2 x)=1 because \sin x (\csc x)=1?
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  7. #7
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    <br /> <br />
\sin^2 x (\csc^2 x)=1<br />

    \frac{1}{cscx} \frac{1}{cscx} (\csc^2 x)

    or

    \frac{1}{csc^2x} (\csc^2 x)

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

    \frac{csc^2x}{csc^2x} =1
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