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  • 2 years ago

simplify \[\sin^2(6)+\cos^2(12)+\cos^2(48)+\sin^2(66)\]

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  1. Jonask
    • 2 years ago
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    \[\sin^266=\sin^2(60+6)\]

  2. Jonask
    • 2 years ago
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    \[\cos^248=\cos^24(12)\]

  3. Jonask
    • 2 years ago
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    \[\sin^2(60+6)= (\sin 60\cos6+\sin6\cos60)^2\]

  4. Jonask
    • 2 years ago
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    \[\cos^248=(\cos60-12)^2=(\cos60\cos12+\sin60\sin12)^2\]

  5. calculusfunctions
    • 2 years ago
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    You guys certainly know how to make life difficult don't you? LOL Here's a hint:\[\cos 2\theta =\cos ^{2}\theta -\sin ^{2}\] \[\cos 2\theta =2\cos ^{2}\theta -1\] \[\cos 2\theta =1-2\sin ^{2}\theta\]and finally the mos famous identity of them all\[\sin ^{2}\theta +\cos ^{2}\theta =1\]Use these to help you simplify.

  6. calculusfunctions
    • 2 years ago
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    For example:\[\cos ^{2}(12)=\cos ^{2}2(6)\]Hint!!!!!

  7. Jonask
    • 2 years ago
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    this seems too long i cant cont. @calculusfunctions

  8. calculusfunctions
    • 2 years ago
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    @Jonask it'ts not that you don't have the right idea. It's just that you're struggling with finding the most efficient method.

  9. Jonask
    • 2 years ago
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    let \[\cos^212=(1-2\sin^26)^2\]

  10. calculusfunctions
    • 2 years ago
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    NO, @Jonask give me a few minutes to type up and explain what you should have done with what you started because as I said, your mind was sort of in the right place. Alright?

  11. calculusfunctions
    • 2 years ago
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    Ye Yes sorry\[\cos ^{2}(12)=(1-2\sin ^{2}6)^{2}\]OK? Now as I said give me a few few minutes to type up an explanation as to where you went wrong with your previous idea. Alright?

  12. Jonask
    • 2 years ago
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    good clean maths needs good variables

  13. Jonask
    • 2 years ago
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    just a comment

  14. calculusfunctions
    • 2 years ago
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    \[\cos ^{2}48°=[\cos (60°-12°)]^{2}\] \[=(\cos 60°\cos 12°+ \sin 60°\sin 12°)^{2}\] \[=(\frac{ 1 }{ 2 }\cos 12°+\frac{ \sqrt{3} }{ 2 }\sin 12°)^{2}\] \[=\frac{ 1 }{ 4 }\cos ^{2}12°+2(\frac{ 1 }{ 2 })(\frac{ \sqrt{3} }{ 2 })\sin 12°\cos 12°+\frac{ 3 }{ 4 }\sin ^{2}12°\] \[=\frac{ 1 }{ 4 }\cos ^{2}12°+\frac{ \sqrt{3} }{ 4 }\sin 2(12°)+\frac{ 3 }{ 4 }\sin ^{2}12°\] Understand so far?

  15. Jonask
    • 2 years ago
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    yes its great

  16. calculusfunctions
    • 2 years ago
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    Similarly\[\sin ^{2}66°=[\sin (60°+6°)]^{2}\]Noting that\[\sin (A +B)=\sin A \cos B +\cos A \sin B\]Simplify\[\sin ^{2}66°\]Go ahead!

  17. calculusfunctions
    • 2 years ago
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    Make sure you show me all the steps like I did.

  18. Jonask
    • 2 years ago
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    \[(\sin60\cos6+\sin6\cos60)^2\] \[(\frac{ \sqrt{3} }{ 2 }\cos6+\frac{ 1 }{ 2 }\sin6)^2\] \[\frac{ 3 }{ 4 }\cos^26+\frac{ \sqrt{3} }{ 4 }\sin6\cos6+\frac{ 1 }{ 4 }\sin^26\] \[\frac{ 3 }{ 4 }\cos^26+\frac{ \sqrt{3} }{ 4 }(\frac{ 1 }{ 2 }\sin2(6)+\frac{ 1 }{ 4 }\sin^26\]

  19. Jonask
    • 2 years ago
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    \[1-\frac{ 3 }{ 4 }=\frac{ 1 }{ 4 }\]

  20. calculusfunctions
    • 2 years ago
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    We could have also used the fact that\[\sin 18°=\frac{ \sqrt{5}-1 }{ 4 }\]and\[\cos 36°=\frac{ \sqrt{5}+1 }{ 4 }\]These are less known. Are you familiar with them?

  21. Jonask
    • 2 years ago
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    yes i see

  22. experimentX
    • 2 years ago
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    \[ \sin^26 - \sin^2 12 - \sin^2 48 +\sin^2 66 + 2 \] Arrange as \[ \sin^26 +\sin^2 66 - \sin^2 12 - \sin^2 48 + 2 \]

  23. calculusfunctions
    • 2 years ago
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    So then I think this would work better because\[\sin ^{2}6°+\cos ^{2}12°+\cos ^{2}48°+\sin ^{2}66°=\sin ^{2}6°+\cos ^{2}12°+\cos ^{2}(36°+12°)+\sin ^{2}(60°+6°)\]This seems more efficient then previously.

  24. calculusfunctions
    • 2 years ago
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    That last term is\[...\sin ^{2}(60°+6°)\]

  25. calculusfunctions
    • 2 years ago
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    I don't think it's still the most efficient method. I think @experimentX 's method now seems easier. Let's see how that pans out.

  26. experimentX
    • 2 years ago
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    change it into half angle formula \[ 2- 4 ( \cos (12) + \cos (132) ) - 2 + 4 (\cos (24) + \cos (96)) + 2 \] seems that all values are in 120, 72,144 == 18, 36 ... that should solve it.

  27. RadEn
    • 2 years ago
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    anyway maybe like this : i think we can use the formulas : sin^2 x = (1-cos2x)/2 and cos^2 x = (1+cos2x)/2 sin^2 (6)+cos^2 (12)+cos^2 (48)+sin^2 (66) = (1-cos12)/2 + (1+cos24)/2 + (1+cos96)/2 + (1-cos132)/2 = {4 + cos96 +cos24 - (cos132+cos12)}/2 Hint : cosA + cosB = 2cos((A+B)/2)*cos((A-B)/2) = {4 + 2cos60cos36 - (2cos72cos60)}/2 = {4 + 2*1/2*cos36 - 2*cos72*1/2}/2 = {4 + cos36 - cos72}/2 = {4 + cos36 - cos2(36)}/2 = {4 + cos36 - (2cos^2(36) - 1)}/2 = {4 + cos36 - 2cos^2(36) + 1}/2 = {5 + cos36 - 2cos^2(36) }/2

  28. calculusfunctions
    • 2 years ago
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    \[\sin \frac{ \theta }{ 2 }=\sqrt{\frac{ 1-\cos \theta }{ 2 }}\]and\[\cos \frac{\theta }{ 2 }=\sqrt{\frac{ 1+\cos \theta }{ 2 }}\]This is what @experimentX meant by half angle formula.

  29. RadEn
    • 2 years ago
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    the last, just put cos(36) = (sqrt(5) + 1)/4 like @calculusfunctions said before

  30. calculusfunctions
    • 2 years ago
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    Any way you look at it the problem is tedious! @RadEn 's solution is also valid. The bottom line is we're all correct in our methods, and they all appear to be tedious! @RadEn I gave the medal to @experimentX earlier but you also clearly deserve it. I wish you could award a medal to more than one person.

  31. experimentX
    • 2 years ago
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    well .. i think the same :)

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