How do you find the exact value of #cos48# using the sum and difference, double angle or half angle formulas?

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Answer 1 · 2017-01-28T17:52:21

#cos48^@=1/8(sqrt(30+6sqrt5)+1-sqrt5)#

Let #A=18^@#, hence #5A=90^@# and

#2A=90^@-3A# and #sin2A=sin(90^@-3A)=cos3A#

or #2sinAcosA=4cos^3A-3cosA#

i.e. #cosA(2sinA-4cos^2A+3)=0# and as #cosA=cos18^@!=0#

we have #2sinA-4cos^2A+3=0#

or #2sinA-4(1-sin^2A)+3=0#

or #4sin^2A+2sinA-1=0# and hence

#sinA=(-2+-sqrt(2^2-4*4*(-1)))/8# or #(-2+-2sqrt5)/8# i.e. #(sqrt5-1)/4#

(as #m/_A=18^@#, #sinA# and #cosA# cannot be negative)

and #cosA=sqrt(1-(sqrt5-1)^2/16)=1/4sqrt(16-(5+1-2sqrt5)#

= #sqrt(10+2sqrt5)/4#

And #cos48^@=cos(30^@+18^@)#

= #cos30^@cos18^@-sin30^@sin18^@#

= #sqrt3/2xxsqrt(10+2sqrt5)/4-1/2xx(sqrt5-1)/4#

= #1/8(sqrt(30+6sqrt5)+1-sqrt5)#