How do you express sin(pi/ 8 ) * cos(( ( 5 pi) / 3 ) sin(π8)cos((5π3) without using products of trigonometric functions?

2 Answers
Dec 9, 2017

The answer is =1/2(sin(1/3pi)-sin(5/24pi))=12(sin(13π)sin(524π))

Explanation:

Apply the formula

sinacosb=1/2(sin(a+b)+sin(a-b))sinacosb=12(sin(a+b)+sin(ab))

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

Here,

a=1/8pia=18π and b=5/3pib=53π

Therefore,

sin(1/8pi)cos(5/3pi)=1/2(sin(1/8pi+5/3pi)+sin(1/8pi-5/3pi))sin(18π)cos(53π)=12(sin(18π+53π)+sin(18π53π))

=1/2(sin(43/24pi)+sin(-37/24pi))=12(sin(4324π)+sin(3724π))

=1/2(sin(-5/24pi)+sin(8/24pi))=12(sin(524π)+sin(824π))

=1/2(sin(1/3pi)-sin(5/24pi))=12(sin(13π)sin(524π))

Dec 9, 2017

Simply calculate the given terms.
= 0.00682=0.00682

Explanation:

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