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mathmath333

  • one year ago

Trignometry question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & x\cos \theta -\sin \theta =1 \hspace{.33em}\\~\\ & x^{2}+(1+x^{2})\sin \theta \ \text{equals} \hspace{.33em}\\~\\ & a.)\ 0 \hspace{.33em}\\~\\ & b.)\ 2 \hspace{.33em}\\~\\ & c.)\ 1 \hspace{.33em}\\~\\ & d.)\ -1 \hspace{.33em}\\~\\ \end{align}}\)

  2. phi
    • one year ago
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    one thought is to set theta to a particular value. for example let theta=0 (the idea is if w have an identity, it works for all values of theta) using that idea, from the first equation x cos theta - sin theta = 1 x = 1 and x=1 in the 2nd equation gives 1

  3. mathmath333
    • one year ago
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    is this question has to solved in 1 min there must be a trick here

  4. mathmath333
    • one year ago
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    beside \(\theta =0\)

  5. phi
    • one year ago
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    your answer choices are all numbers. that suggests the answer is independent of theta (otherwise we would have choices that are expressions with theta) so if theta does not matter, pick a value that is easy to evaluate.

  6. mathmath333
    • one year ago
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    ok

  7. phi
    • one year ago
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    unfortunately, if we solve for x using the first equation x= (1+ sin A)/ cos A and plug that into the 2nd equation, we get a messy trig function wolfram can plot it. http://www.wolframalpha.com/input/?i=x^ {2}%2B%281%2Bx^{2}%29\sin+\theta+where+x+%3D+%281%2Bsin+\theta%29%2Fcos+\theta and I don't see how you get a single number out of that. in other words, I don't really understand how they get a constant number independent of theta... that appears bogus...

  8. phi
    • one year ago
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    the link did not post properly

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