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 $sin18=(-1+sqrt5)/4$

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

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

$1/2(1/2+cos36)$

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

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

$=(3+sqrt5)/8$

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

How do I evaluate cos48cos12cos48cos12?