Question

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

Answer 1

\[\sin^266=\sin^2(60+6)\]

Answer 2

\[\cos^248=\cos^24(12)\]

Answer 3

\[\sin^2(60+6)= (\sin 60\cos6+\sin6\cos60)^2\]

Answer 4

\[\cos^248=(\cos60-12)^2=(\cos60\cos12+\sin60\sin12)^2\]

Answer 5

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.

Answer 6

For example:\[\cos ^{2}(12)=\cos ^{2}2(6)\]Hint!!!!!

Answer 7

this seems too long i cant cont. @calculusfunctions

Answer 8

@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.

Answer 9

let \[\cos^212=(1-2\sin^26)^2\]

Answer 10

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?

Answer 11

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?

Answer 12

good clean maths needs good variables

Answer 13

just a comment

Answer 14

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

Answer 15

yes its great

Answer 16

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!

Answer 17

Make sure you show me all the steps like I did.

Answer 18

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

Answer 19

\[1-\frac{ 3 }{ 4 }=\frac{ 1 }{ 4 }\]

Answer 20

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?

Answer 21

yes i see

Answer 22

\[ \sin^26 - \sin^2 12 - \sin^2 48 +\sin^2 66 + 2 \]
Arrange as
\[ \sin^26 +\sin^2 66 - \sin^2 12 - \sin^2 48 + 2 \]

Answer 23

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.

Answer 24

That last term is\[...\sin ^{2}(60°+6°)\]

Answer 25

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.

Answer 26

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.

Answer 27

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

Answer 28

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

Answer 29

the last, just put cos(36) = (sqrt(5) + 1)/4
like @calculusfunctions said before

Answer 30

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.

Answer 31

well .. i think the same :)