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How do you evaluate #cos((3pi)/5)cos((4pi)/15)+sin((3pi)/5)sin((4pi)/15)#?

1 Answer

Feb 13, 2015

Use the standard trig. formula:
#cos( a - b ) = cos(a)*cos(b) + sin(a)*sin(b)#

#cos( (3Pi)/5) cos( (4Pi)/15) + sin( (3Pi)/5) sin( (4Pi)/15)# becomes

#cos ( (3Pi)/5 - (4Pi)/15)#

# = cos( (5Pi)/15) = cos( Pi/3)#

#Pi/3# (#60^o#) is one of the standard triangles

#cos(Pi/3) = 1/2

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