How do you express #sin(pi/ 8 ) * cos(( ( 5 pi) / 3 ) # without using products of trigonometric functions?

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Answer 1 · 2017-12-09T07:33:46

The answer is #=1/2(sin(1/3pi)-sin(5/24pi))#

Apply the formula

#sinacosb=1/2(sin(a+b)+sin(a-b))#

You can easily prove this formula by using #sin(a+b)# and #sin(a-b)#

Here,

#a=1/8pi# and #b=5/3pi#

Therefore,

#sin(1/8pi)cos(5/3pi)=1/2(sin(1/8pi+5/3pi)+sin(1/8pi-5/3pi))#

#=1/2(sin(43/24pi)+sin(-37/24pi))#

#=1/2(sin(-5/24pi)+sin(8/24pi))#

#=1/2(sin(1/3pi)-sin(5/24pi))#

Answer 2 · 2017-12-09T07:34:57

Simply calculate the given terms.
# = 0.00682#

You "using products of trig functions" statement is not clear. I took it to mean revising the expression to form a single trigonometric identity.

Thus, to avoid that, I simply evaluated the expression by each term.
#sin(π/8) xx cos(5π/3) = 0.00685 xx 0.9958 = 0.00682#