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Thread: 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 $\displaystyle \sin^2 x(1+\cot^2 x)=1$

    Method:
    I figured out that I should convert $\displaystyle \sin^2 x(1+\cot^2 x)=1$ to $\displaystyle \sin^2 x (\csc^2 x)=1$. 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 $\displaystyle \sin x \csc x=1$ 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 $\displaystyle \tan x(\cot x +\tan x)=\sec^2 x$.

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

    3:
    Find the exact value of $\displaystyle \cos 105$ 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: $\displaystyle \cos(\alpha - \beta)=\cos\alpha \cos \beta+\sin \alpha \sin \beta$.
    For \alpha I used value 135 and for \beta I used 30 so $\displaystyle \cos(\alpha - \beta) \rightarrow \cos(135-30)$, therefore the whole formula changes to become: $\displaystyle \cos(135 - 30)=\cos 135 \cos 30 + \sin 135 \sin 30$.
    So, $\displaystyle \cos 135 \cos 30 + \sin 135 \sin 30 \rightarrow -\frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} \rightarrow \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \rightarrow \frac{\sqrt{2} +\sqrt{6}}{4}$, have I gotten the right answer?

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

    Method:
    The book said nothing about the addition formula for $\displaystyle \tan$ so I searched the internet and found this:$\displaystyle \tan(\alpha-\beta)= \frac{\tan \alpha - \tan \beta}{1+ \tan\alpha \tan\beta}$
    I decided to use this formula since I could figure out no other to use, I entered $\displaystyle \tan 45$ for $\displaystyle \alpha$ and $\displaystyle \tan 30$ for $\displaystyle \beta$, 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 $\displaystyle tan(x)$ as $\displaystyle \frac {sin(x)}{cos(x)}$ and $\displaystyle cot(x)$ as $\displaystyle \frac {cos(x)}{sin(x)}$

    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:

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

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

    $\displaystyle \tan(a - b) = \frac{\sin(a - b)}{\cos(a - b)}$
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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:

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

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

    $\displaystyle \tan(a - b) = \frac{\sin(a - b)}{\cos(a - b)}$
    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 $\displaystyle \sin^2 x(1+\cot^2 x)=1$

    Method:
    I figured out that I should convert $\displaystyle \sin^2 x(1+\cot^2 x)=1$ to $\displaystyle \sin^2 x (\csc^2 x)=1$. 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 $\displaystyle \sin x \csc x=1$ 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 $\displaystyle \sin^2 x (\csc^2 x)=1$ because $\displaystyle \sin x (\csc x)=1$?
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  7. #7
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    $\displaystyle

    \sin^2 x (\csc^2 x)=1
    $

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

    or

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

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

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