# How do you express the value of #cos 2^o15' #, in mathematical exactitude?

##### 1 Answer

#### Explanation:

If you know that

#cos(theta/2) = +-sqrt(1/2+1/2cos theta)#

We need to apply this

So:

#cos 2^@15' = sqrt(1/2 + 1/2cos 4^@30')#

#color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2cos 9^@))#

#color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2cos 18^@)))#

#color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2sqrt(1/2+1/2cos 36^@))))#

#color(white)(cos 2^@15') = sqrt(1/2 + 1/2sqrt(1/2 + 1/2sqrt(1/2+1/2sqrt(1/2+1/2(1/4(sqrt(5)+1)))))#

This can be simplified a little, but has a pleasing form as is, which indicates the derivation.

**Derivation of #cos 36^@#**

Here's a derivation of the formula for

#-1 = cos 180^@ + i sin 180^@#

#color(white)(-1) = (cos 36^@ + i sin 36^@)^5#

#color(white)(-1) = (cos^5 36^@ - 10 cos^3 36^@ sin^2 36^@ + 5 cos 36^@ sin^4 36^@) + i (5cos^4 36^@ sin 36^@ - 10cos^2 36^@ sin^3 36^@ + sin^5 36^@)#

Equating Real parts:

#-1 = cos^5 36^@ - 10 cos^3 36^@ sin^2 36^@ + 5 cos 36^@ sin^4 36^@#

#color(white)(-1) = cos^5 36^@ - 10 cos^3 36^@ (1-cos^2 36^@) + 5 cos 36^@ (1-cos^2 36^@)^2#

#color(white)(-1) = 16 cos^5 36^@ - 20 cos^3 36^@ + 5 cos 36^@#

Hence

#16x^5-20x^3+5x+1 = (x+1)(16x^4-16x^3-4x^2+4x+1)#

#color(white)(16x^5-20x^3+5x+1) = (x+1)(4x^2-2x-1)^2#

We can discard

#0 = 4(4x^2-2x-1)#

#color(white)(0) = 16x^2-8x+1-5#

#color(white)(0) = (4x-1)^2-sqrt(5)^2#

#color(white)(0) = ((4x-1)-sqrt(5))((4x-1)+sqrt(5))#

#color(white)(0) = (4x-1-sqrt(5))(4x-1+sqrt(5))#

Hence:

#x = 1/4(1+-sqrt(5))#

Then since

#cos 36^@ = 1/4(sqrt(5)+1)#