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JonaskBest ResponseYou've already chosen the best response.0
\[\sin^266=\sin^2(60+6)\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
\[\cos^248=\cos^24(12)\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
\[\sin^2(60+6)= (\sin 60\cos6+\sin6\cos60)^2\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
\[\cos^248=(\cos6012)^2=(\cos60\cos12+\sin60\sin12)^2\]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
You guys certainly know how to make life difficult don't you? LOL Here's a hint:\[\cos 2\theta =\cos ^{2}\theta \sin ^{2}\] \[\cos 2\theta =2\cos ^{2}\theta 1\] \[\cos 2\theta =12\sin ^{2}\theta\]and finally the mos famous identity of them all\[\sin ^{2}\theta +\cos ^{2}\theta =1\]Use these to help you simplify.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
For example:\[\cos ^{2}(12)=\cos ^{2}2(6)\]Hint!!!!!
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
this seems too long i cant cont. @calculusfunctions
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
@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.
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
let \[\cos^212=(12\sin^26)^2\]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
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?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
Ye Yes sorry\[\cos ^{2}(12)=(12\sin ^{2}6)^{2}\]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?
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
good clean maths needs good variables
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
\[\cos ^{2}48°=[\cos (60°12°)]^{2}\] \[=(\cos 60°\cos 12°+ \sin 60°\sin 12°)^{2}\] \[=(\frac{ 1 }{ 2 }\cos 12°+\frac{ \sqrt{3} }{ 2 }\sin 12°)^{2}\] \[=\frac{ 1 }{ 4 }\cos ^{2}12°+2(\frac{ 1 }{ 2 })(\frac{ \sqrt{3} }{ 2 })\sin 12°\cos 12°+\frac{ 3 }{ 4 }\sin ^{2}12°\] \[=\frac{ 1 }{ 4 }\cos ^{2}12°+\frac{ \sqrt{3} }{ 4 }\sin 2(12°)+\frac{ 3 }{ 4 }\sin ^{2}12°\] Understand so far?
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
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!
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
Make sure you show me all the steps like I did.
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
\[(\sin60\cos6+\sin6\cos60)^2\] \[(\frac{ \sqrt{3} }{ 2 }\cos6+\frac{ 1 }{ 2 }\sin6)^2\] \[\frac{ 3 }{ 4 }\cos^26+\frac{ \sqrt{3} }{ 4 }\sin6\cos6+\frac{ 1 }{ 4 }\sin^26\] \[\frac{ 3 }{ 4 }\cos^26+\frac{ \sqrt{3} }{ 4 }(\frac{ 1 }{ 2 }\sin2(6)+\frac{ 1 }{ 4 }\sin^26\]
 one year ago

JonaskBest ResponseYou've already chosen the best response.0
\[1\frac{ 3 }{ 4 }=\frac{ 1 }{ 4 }\]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
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?
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
\[ \sin^26  \sin^2 12  \sin^2 48 +\sin^2 66 + 2 \] Arrange as \[ \sin^26 +\sin^2 66  \sin^2 12  \sin^2 48 + 2 \]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
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.
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
That last term is\[...\sin ^{2}(60°+6°)\]
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
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.
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
anyway maybe like this : i think we can use the formulas : sin^2 x = (1cos2x)/2 and cos^2 x = (1+cos2x)/2 sin^2 (6)+cos^2 (12)+cos^2 (48)+sin^2 (66) = (1cos12)/2 + (1+cos24)/2 + (1+cos96)/2 + (1cos132)/2 = {4 + cos96 +cos24  (cos132+cos12)}/2 Hint : cosA + cosB = 2cos((A+B)/2)*cos((AB)/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
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
\[\sin \frac{ \theta }{ 2 }=\sqrt{\frac{ 1\cos \theta }{ 2 }}\]and\[\cos \frac{\theta }{ 2 }=\sqrt{\frac{ 1+\cos \theta }{ 2 }}\]This is what @experimentX meant by half angle formula.
 one year ago

RadEnBest ResponseYou've already chosen the best response.2
the last, just put cos(36) = (sqrt(5) + 1)/4 like @calculusfunctions said before
 one year ago

calculusfunctionsBest ResponseYou've already chosen the best response.2
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.3
well .. i think the same :)
 one year ago
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