How do you find the exact value of #cos58# using the sum and difference, double angle or half angle formulas?

Question

Quinn Anderson · Accepted Answer

It's exactly one of the roots of #T_{44}(x) = -T_{46} (x)# where #T_n(x)# is the #n#th Chebyshev Polynomial of the first kind. That's one of the forty six roots of:

#8796093022208 x^44 - 96757023244288 x^42 + 495879744126976 x^40 - 1572301627719680 x^38 + 3454150138396672 x^36 - 5579780992794624 x^34 + 6864598984556544 x^32 - 6573052309536768 x^30 + 4964023879598080 x^28 - 2978414327758848 x^26 + 1423506847825920 x^24 - 541167892561920 x^22 + 162773155184640 x^20 - 38370843033600 x^18 + 6988974981120 x^16 - 963996549120 x^14 + 97905899520 x^12 - 7038986240 x^10 + 338412800 x^8 - 9974272 x^6 + 155848 x^4 - 968 x^2 + 1 = -( 35184372088832 x^46 - 404620279021568 x^44 + 2174833999740928 x^42 - 7257876254949376 x^40 + 16848641306132480 x^38 - 28889255702953984 x^36 + 37917148110127104 x^34 - 38958828003262464 x^32 + 31782201792135168 x^30 - 20758645314682880 x^28 + 10898288790208512 x^26 - 4599927086776320 x^24 + 1555857691115520 x^22 - 418884762992640 x^20 + 88826010009600 x^18 - 14613311324160 x^16 + 1826663915520 x^14 - 168586629120 x^12 + 11038410240 x^10 - 484140800 x^8 + 13034560 x^6 - 186208 x^4 + 1058 x^2 - 1 ) # #58^circ# is not a multiple of #3^circ#. Multiples of #1^circ# that are not multiples of #3^circ# are not constructible with a straightedge and compass, and their trig functions are not the result of some composition of integers using addition, subtraction, multiplication, division and square rooting. That doesn't mean we can't write down some expression for #cos 58^circ#. Let's take the degree sign to mean a factor of #{2pi}/360#. #e^{i 58^circ} = cos 58^circ + i \ sin 58^circ # #e^{-i 58^circ} = cos 58^circ - i \ sin 58^circ # # e^{i 58^circ} + e^{-i 58^circ} = 2 \ cos 58^circ # #cos 58^circ = 1/2 ( e^{i 58^circ} + e^{-i 58^circ} ) # Not very beneficial. We can try to write down a polynomial equation one of whose roots is #cos 58^circ# but it's probably going to be too big to fit. #theta=2^circ# is #180#th of a circle. Since #cos 88^circ = -cos 92^circ# that means #cos 2^circ# satisfies #cos (44 theta) = -cos (46 theta) # #cos(180^circ -44 theta) = cos(46 theta) # Let's solve this for #theta# first. #cos x=cos a# has roots #x=pm a + 360^circ k,# integer #k#. # 180^circ -46 theta = \pm 44 theta - 360^circ k # #46 theta \pm 44 theta = 180 ^circ + 360^circ k # #theta = 2^ circ + 4^circ k or theta = 90^circ + 180^circ k# That's a lot of roots, and we see #theta=58^circ# among them. The polynomials #T_n(x)#, called the Chebyshev Polynomials of the first kind, satisfy #cos(n theta) = T_n(cos theta)#. They have integer coefficients. We know the first few from the double and triple angle formulas: #cos(0 theta) = 1 quad quad # so# quad quad T_0(x)=1# #cos(1 theta) = cos theta quad quad # so# quad quad T_1(x)=x# #cos (2 theta) = 2cos^2 theta - 1 quad quad # so # quad quad T_2(x)=2x^2-1# #cos (3 theta) = 4cos^3 theta - 3 cos theta quad quad # so # quad quad T_3(x)=4x^4-3x# We can confirm the following nice recursion relation: # T_{n+1}(x) = 2x T_{n} (x) - T_{n-1}(x)# So in theory we can generate these for as big #n# as we care to. If we let #x= cos theta,# our equation #cos (44 theta) = -cos (46 theta) # becomes #T_{44}(x) = -T_{46} (x)# I'll write the equation just to test the math rendering, and Wolfram Alpha will be happy to tell us what those are: #8796093022208 x^44 - 96757023244288 x^42 + 495879744126976 x^40 - 1572301627719680 x^38 + 3454150138396672 x^36 - 5579780992794624 x^34 + 6864598984556544 x^32 - 6573052309536768 x^30 + 4964023879598080 x^28 - 2978414327758848 x^26 + 1423506847825920 x^24 - 541167892561920 x^22 + 162773155184640 x^20 - 38370843033600 x^18 + 6988974981120 x^16 - 963996549120 x^14 + 97905899520 x^12 - 7038986240 x^10 + 338412800 x^8 - 9974272 x^6 + 155848 x^4 - 968 x^2 + 1 = -( 35184372088832 x^46 - 404620279021568 x^44 + 2174833999740928 x^42 - 7257876254949376 x^40 + 16848641306132480 x^38 - 28889255702953984 x^36 + 37917148110127104 x^34 - 38958828003262464 x^32 + 31782201792135168 x^30 - 20758645314682880 x^28 + 10898288790208512 x^26 - 4599927086776320 x^24 + 1555857691115520 x^22 - 418884762992640 x^20 + 88826010009600 x^18 - 14613311324160 x^16 + 1826663915520 x^14 - 168586629120 x^12 + 11038410240 x^10 - 484140800 x^8 + 13034560 x^6 - 186208 x^4 + 1058 x^2 - 1 ) # Yes, this answer is getting long, thanks Socratic. Anway, one of the roots of that 46th degree polynomial with integer coefficients is #\cos 58^circ#.

Samantha Abate · Answer

undefined Using the half-angle formula for cosine, you can find the exact value of cos 58 degrees. The formula states that:

$$ \cos(\frac{\theta}{2}) = \sqrt{\frac{1 + \cos(\theta)}{2}} $$

Substituting $ \theta = 116^\circ $ (twice the angle):

$$ \cos(58^\circ) = \sqrt{\frac{1 + \cos(116^\circ)}{2}} $$

Then you need to find the value of $ \cos(116^\circ) $ using either the sum and difference formulas or the unit circle. After finding the value, plug it into the equation to calculate $ \cos(58^\circ) $.