## apple_pi 3 years ago simplify $\sin^2(6)+\cos^2(12)+\cos^2(48)+\sin^2(66)$

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$\sin^266=\sin^2(60+6)$

$\cos^248=\cos^24(12)$

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

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

5. calculusfunctions

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

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

this seems too long i cant cont. @calculusfunctions

8. calculusfunctions

@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.

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

10. calculusfunctions

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

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?

good clean maths needs good variables

just a comment

14. calculusfunctions

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

yes its great

16. calculusfunctions

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

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

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

$1-\frac{ 3 }{ 4 }=\frac{ 1 }{ 4 }$

20. calculusfunctions

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?

yes i see

22. experimentX

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

23. calculusfunctions

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

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

25. calculusfunctions

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

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.

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

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

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

30. calculusfunctions

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

well .. i think the same :)

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