How do I evaluate #cos48cos12#?

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Answer 1 · 2017-05-16T15:23:29

See below.

#cos48cos12# is similar to the sum-to-product formula for

#cos"P"+cos"Q"=2cos(("P"+"Q")/2)cos(("P"-"Q")/2)#

#"P"+"Q"=96#

#"P"-"Q"=24#

#2"P"=120#

#"P"=60#

#2"Q"=72#

#"Q"=36#

#cos48cos12=1/2(cos60+cos36)#

#=1/2(1/2+cos36)#

#cos36=cos2(18)=2cos^2 18-1#

From the above video, we know #sin18=(-1+sqrt5)/4#

#cos^2 18=1-sin^2 18=1-((-1+sqrt5)/4) ^2=(5+sqrt5)/8#

#cos36=2cos^2 18-1=2((5+sqrt5)/8)^2-1=(1+sqrt5)/4#

#1/2(1/2+cos36)#

#=1/2(1/2+(1+sqrt5)/4)#

#=1/2((3+sqrt5)/4)#

#=(3+sqrt5)/8#

#thereforecos48cos12=(3+sqrt5)/8# #sf(QED"/"OEDelta)#