Cos 10 cos 30 cos 50 cos 70=

     

Homework Statement

What is the value of ## frac1cos^2 10^circ+frac1cos^2 20^circ+frac1cos^2 40^circ-frac1cos^2 45^circ ## ?A. 1B. 5C. 10D. 15E. 20

Homework Equations

cos^2 x + sin^2 x = 1sin 2x = 2 sin x cos xcos 2x = cos^2 x-sin^2 x

The Attempt at a Solution

I"m only able lớn determine cos 45 degrees which is √2 / 2I have no idea about cos 10, cos 20, và cos 40 degreesPlease help me
Are you just expected lớn look up the cos values in the tables or is an analytic solution being asked for ?Have you done any basic trigonometry in your classes ?
http://www.industrialpress.com/ext/StaticPages/Handbook/TrigPages/mh_trig.aspAre you just expected to look up the cos values in the tables or is an analytic solution being asked for ?Have you done any basic trigonometry in your classes ?
No calculators or tables are allowed in the test.I have to vị it using trigonometric identities, and formulas
Since it"s a multiple choice test you only have lớn estimate the answer. So how far vị you think are cos 40° and cos 45° apart? and where vị cos 10° and cos 20° lie? do you know the graph of cos? Are you sure about the possible answers?

Homework Statement

What is the value of ## frac1cos^2 10^circ+frac1cos^2 20^circ+frac1cos^2 40^circ-frac1cos^2 45^circ ## ?A. 1B. 5C. 10D. 15E. 20
Are you supposed to give the range of that value? As in "between C và D" (just as an example).Because none of these five values are anywhere close.

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Since it"s a multiple choice test you only have khổng lồ estimate the answer. So how far vị you think are cos 40° và cos 45° apart? & where vì cos 10° và cos 20° lie?
Are you supposed lớn give the range of that value? As in "between C và D" (just as an example).Because none of these five values are anywhere close.That works really good khổng lồ get an estimate of the value.
Hmm.. You"re right.It has no right option (in calculator I get 1.86)..Sorry for posting this nonsense problem :(
Pity it is garbled, there is an interesting puzzle lurking behind it.cos(40), cos(80) and -cos(20) are the three roots of 8x3-6x+1. If you change the cos(10) lớn cos(80) the sum gives 34.
Likesterryds
Pity it is garbled, there is an interesting puzzle lurking behind it.cos(40), cos(80) và -cos(20) are the three roots of 8x3-6x+1. If you change the cos(10) to cos(80) the sum gives 34.
Ok, I"ve found a correction that makes the problem work. Leave the 10 as is but change the 20 & 40 to 50 và 70. That yields one of the given answers.
Ok, I"ve found a correction that makes the problem work. Leave the 10 as is but change the 20 và 40 to lớn 50 và 70. That yields one of the given answers.

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Very good haru !Of course, that"s the same as ##displaystyle frac1cos^2 10^circ+frac1sin^2 20^circ+frac1sin^2 40^circ-frac1cos^2 45^circ ##.
Very good haru !Of course, that"s the same as ##displaystyle frac1cos^2 10^circ+frac1sin^2 20^circ+frac1sin^2 40^circ-frac1cos^2 45^circ ##.
How true.Terry, as a clue on how khổng lồ approach such a problem...Note that the given angles are awkward in the sense that they are not multiples of 3 degrees. So we need to relate them lớn angles which are. A very useful formula is ##cos(3 heta)=4cos^3( heta)-3cos( heta)##. There"s a similar one for sine.Since the given angles are multiples of 10 degrees, that gives you cubic equations relating their cosines khổng lồ well known cosines.Using the corrected angles, you discover that all three satisfy the same cubic.The next trick is khổng lồ consider how the sum of the inverse squares of the roots of a cubic relates lớn the coefficients in the cubic.
LikesSamy_A
How true.Terry, as a clue on how to approach such a problem...Note that the given angles are awkward in the sense that they are not multiples of 3 degrees. So we need lớn relate them to angles which are. A very useful formula is ##cos(3 heta)=4cos^3( heta)-3cos( heta)##. There"s a similar one for sine.Since the given angles are multiples of 10 degrees, that gives you cubic equations relating their cosines to well known cosines.Using the corrected angles, you discover that all three satisfy the same cubic.The next trick is to lớn consider how the sum of the inverse squares of the roots of a cubic relates to the coefficients in the cubic.

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Okay, so now the problem becomes##frac1cos^210^ circ+frac1cos^250^ circ+frac1cos^270^ circ-frac1cos^245^ circ##I still get stuck..How vày I use the formula ## cos (3 heta) = 4 cos ^3 ( heta) - 3 cos ( heta) ## ??Neither 50 nor 70 are the triple of 10Please tell me the steps khổng lồ solve it, since I don"t have any idea at all
Okay, so now the problem becomes##frac1cos^210^ circ+frac1cos^250^ circ+frac1cos^270^ circ-frac1cos^245^ circ##I still get stuck..How vày I use the formula ## cos (3 heta) = 4 cos ^3 ( heta) - 3 cos ( heta) ## ??Neither 50 nor 70 are the triple of 10Please tell me the steps lớn solve it, since I don"t have any idea at all
cos (3*50) = 4 cos^3(50) - 3 cos(50)- √3 / 2 = 4 cos^3(50) - 3 cos(50)cos(3*70°)= 4 cos^3(70) - 3 cos(70)- √3 / 2 = 4 cos^3(70) - 3 cos(70)cos (3*10) = 4 cos^3(10) - 3 cos(10)√3 / 2 = 4 cos^3(10) - 3 cos(10)Then, I still don"t get the idea..How khổng lồ get cos(50), cos(70) & cos(10) ? It"s a cubic function..And, at first I think cos(50) equals to lớn cos(70) due khổng lồ the same cos(3x), but they"re different..Please help
cos (3*50) = 4 cos^3(50) - 3 cos(50)- √3 / 2 = 4 cos^3(50) - 3 cos(50)cos(3*70°)= 4 cos^3(70) - 3 cos(70)- √3 / 2 = 4 cos^3(70) - 3 cos(70)cos (3*10) = 4 cos^3(10) - 3 cos(10)√3 / 2 = 4 cos^3(10) - 3 cos(10)Then, I still don"t get the idea..How to get cos(50), cos(70) và cos(10) ? It"s a cubic function..And, at first I think cos(50) equals to cos(70) due khổng lồ the same cos(3x), but they"re different..Please help
The idea is not lớn solve the cubic equation, but to use properties of sums/products (and combinations thereof) of its roots.I made use of Vieta"s formulas, and some more computations. There may well be an easier way that escaped me.
cos (3*50) = 4 cos^3(50) - 3 cos(50)- √3 / 2 = 4 cos^3(50) - 3 cos(50)cos(3*70°)= 4 cos^3(70) - 3 cos(70)- √3 / 2 = 4 cos^3(70) - 3 cos(70)cos (3*10) = 4 cos^3(10) - 3 cos(10)√3 / 2 = 4 cos^3(10) - 3 cos(10)Then, I still don"t get the idea..How khổng lồ get cos(50), cos(70) và cos(10) ? It"s a cubic function..And, at first I think cos(50) equals khổng lồ cos(70) due to lớn the same cos(3x), but they"re different..Please help
Yes, in my post #30 I missed a step out.You need the cosines to be the three roots of the same cubic, but the three cubics you found are not quite the same. Since the cosines are all squared in the target formula, cos(50) will be equivalent lớn -cos(50)=cos(130) etc. That works, because 3*130=390=360+30, giving the same cubic as for cos(10). Vị the same to lớn find a substitute for 70.As you say, the cosines of 10, 130 và the third angle are all different, but they satisfy the same cubic, so they must be the three roots of that cubic.Having done that, our target expression, the sum of the inverses of the squares of those cosines, is a-2+b-2+c-2 where a, b và c are all roots of the cubic 4x3-3x-cos(30)=0.In general, if a cubic is px3+qx2+rx+s=0, there are several simple formulae relating the roots to lớn the coefficients. S/p=abc, for example. Using these, can you find a way khổng lồ express a-2+b-2+c-2 in terms of p, q, r và s?