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mathmath333

  • one year ago

Trignometry question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Find the maximum value of }\hspace{.33em}\\~\\ & \sin^{4}\theta +\cos^{4}\theta \hspace{.33em}\\~\\~\\ & a.)\ 3\hspace{.33em}\\~\\ & b.)\ \dfrac13\hspace{.33em}\\~\\ & c.)\ 1 \hspace{.33em}\\~\\ & d.)\ 2 \hspace{.33em}\\~\\ \end{align}}\)

  2. ChillOut
    • one year ago
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    Rewrite either the sin or cos function. I.E \(sin^{2}(x)=1-cos^{2}(x)\)

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \sin^{4}\theta +\cos^{4}\theta \hspace{.33em}\\~\\ & (1-\cos^{2}\theta)^{2} +\cos^{4}\theta \hspace{.33em}\\~\\ \end{align}}\)

  4. ChillOut
    • one year ago
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    Yup. Keep going.

  5. mathmath333
    • one year ago
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    next wht

  6. ChillOut
    • one year ago
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    You now have \(cos^{2}(x)-2cos^{2}(x)+cos^{4}(x)+1\), right?

  7. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \sin^{4}\theta +\cos^{4}\theta \hspace{.33em}\\~\\ &= (1-\cos^{2}\theta)^{2} +\cos^{4}\theta \hspace{.33em}\\~\\ &= 1-2\cos^{2}\theta+2\cos^{4}\theta \hspace{.33em}\\~\\ \end{align}}\)

  8. ChillOut
    • one year ago
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    Oh, yeah, I missed a 2 there.

  9. ChillOut
    • one year ago
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    Let's call \(x=1-2cos^{2}(\theta)+2cos^{4}(\theta)\), so \(\frac{x}{2}=\frac{1}{2}-cos^{2}(\theta)+cos^{4}(\theta)\). Well, I gotta think this for a bit (Assuming you can't use the obvious approach which is sin(x)+cos(x)=1)

  10. ChillOut
    • one year ago
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    All right... We have to rewrite that. Nevermind what I did last post. 1-2cos²(θ)=-cos(2θ). We can work from here.

  11. anonymous
    • one year ago
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    How about using the fact that\[1^2=(\sin^2 x+\cos^2 x)^2=\sin^4 x+\cos^4 x +2 \sin^2 x \cos^2 x\]

  12. mathmath333
    • one year ago
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    \(\sin^{4}x+\cos^{4}x=1-2\sin^{2}x\cos^{2}x\)

  13. mathmath333
    • one year ago
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    is 1 the answrer

  14. ChillOut
    • one year ago
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    Yes.

  15. anonymous
    • one year ago
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    and\[\text{Your expression}=1-\frac{1}{2} (\sin(2x))^2\]yes the answer is 1

  16. ganeshie8
    • one year ago
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    Notice \(\sin^4x\le \sin^2x\) and \(\cos^4x\le \cos^2x\). that implies \(\sin^4x+\cos^4x\le \sin^2x+\cos^2x=1\)

  17. anonymous
    • one year ago
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    Nice observation gane +1

  18. ganeshie8
    • one year ago
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    :) i answered this problem yesterday haha!

  19. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \text{calculate} \hspace{.33em}\\~\\ & \tan 4^{\circ} \tan 43^{\circ} \tan 47^{\circ} \tan 86^{\circ} \hspace{.33em}\\~\\ \end{align}}\) i have to solve this within a minute

  20. ganeshie8
    • one year ago
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    First notice that 4+86 = 43+47 = 90

  21. ganeshie8
    • one year ago
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    next recall the indentity involving tan and cot : \[\tan(x) = \cot(90-x) = \dfrac{1}{\tan(90-x)}\]

  22. mathmath333
    • one year ago
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    1 is the answer

  23. mathmath333
    • one year ago
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    this was easy but looked hard at first sight

  24. ganeshie8
    • one year ago
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    reason like this - you're supposed to solve the problem in minute, so they are "forced" to put only simple problems in the test

  25. ganeshie8
    • one year ago
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    that helps in approaching the problem because when you knw upfront that they wont be asking questions that cannot be solved within a minute, you wont get distracted trying all the fancy methods..

  26. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \text{If}\ \sin \theta +\sin^{2} \theta =1,\ \hspace{.33em}\\~\\ & \text{and}\ \cos^{2} \theta +\cos^{4} \theta =x,\ \hspace{.33em}\\~\\ & \text{then the value of }x= \hspace{.33em}\\~\\ &a.)\ \dfrac{\cos^{2} \theta}{\sin \theta} \hspace{.33em}\\~\\ &b.)\ \text{None} \hspace{.33em}\\~\\ & c.)\ 1 \hspace{.33em}\\~\\ &d.)\ \dfrac{\sin \theta}{\cos^{2} \theta} \hspace{.33em}\\~\\ \end{align}}\)

  27. ganeshie8
    • one year ago
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    compare the first equation,\(\color{red}{\sin\theta}+\sin^2\theta=1\), with the identity : \(\color{red}{\cos^2\theta}+\sin^2\theta=1\)

  28. mathmath333
    • one year ago
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    \(\sin \theta =\cos^2 \theta \)

  29. ganeshie8
    • one year ago
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    Yep, plug that in second equation

  30. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & x=\sin \theta +\sin ^{2} \theta \hspace{.33em}\\~\\ \end{align}}\)

  31. ganeshie8
    • one year ago
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    \(\cos^2\theta + \cos^4\theta=x\) \(\sin\theta + \sin^2\theta=x\) \(1=x\)

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

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