How do you express $cos (\frac{π}{3}) \cdot sin (\frac{5 π}{8})$ $cos(pi/ 3 ) * sin( ( 5 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{5 π}{8})$ $cos(pi/ 3 ) * sin( ( 5 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-08-01T19:10:07

It may be "cheating", but I would just substitute $\frac{1}{2}$ $1/2$ for $cos (\frac{π}{3})$ $cos(\pi/3)$ .

Explanation:

You're probably supposed to use the identity

$cos a sin b = (\frac{1}{2}) (sin (a + b) - sin (a - b))$ $cos a sin b = (1/2)(sin(a+b)-sin(a-b))$ .

Put in $a = \frac{π}{3} = \frac{8 π}{24}, b = \frac{5 π}{8} = \frac{15 π}{24}$ $a=\pi/3={8\pi}/24, b={5\pi}/8={15\pi}/24$ .

Then

$cos (\frac{π}{3}) sin (\frac{5 \cdot π}{8}) = (\frac{1}{2}) (sin (\frac{23 \cdot π}{24}) - sin (\frac{- 7 \cdot π}{24}))$ $cos(\pi/3)sin({5*pi}/8)=(1/2)(sin({23*\pi}/24)-sin({-7*pi}/24))$

$= (\frac{1}{2}) (sin (\frac{π}{24}) + sin (\frac{7 \cdot π}{24}))$ $=(1/2)(sin({\pi}/24)+sin({7*\pi}/24))$

where in the last line we use $sin (π - x) = sin (x)$ $sin(\pi-x)=sin(x)$ and $sin (- x) = - sin (x)$ $sin(-x)=-sin(x)$ .

As you can see, this is unwieldy compared with just putting in $cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3)=1/2$ . The trigonometric product-sum and product-difference relations are more useful when you can't evaluate either factor in the product.

Answer 2 · 2016-08-02T05:01:36

$- (\frac{1}{2}) cos (\frac{π}{8})$ $- (1/2)cos (pi/8)$

Explanation:

$P = cos (\frac{π}{3}) . sin (\frac{5 π}{8})$ $P = cos (pi/3).sin ((5pi)/8)$
Trig table --> $cos (\frac{π}{3}) = \frac{1}{2}$ $cos (pi/3) = 1/2$
Trig unit circle and property of complementary arcs -->
$sin (\frac{5 π}{8}) = sin (\frac{π}{8} + \frac{4 π}{8}) = sin (\frac{π}{8} + \frac{π}{2}) =$ $sin ((5pi)/8) = sin (pi/8 + (4pi)/8) = sin (pi/8 + pi/2) =$
$= - cos (\frac{π}{8}) .$ $= - cos (pi/8).$
P can be expressed as:
$P = - (\frac{1}{2}) cos (\frac{π}{8})$ $P = - (1/2)cos (pi/8)$
NOTE. We can evaluate $cos (\frac{π}{8})$ $cos (pi/8)$ by using the trig identity:
$1 + cos (\frac{π}{4}) = 2 {cos}^{2} (\frac{π}{8})$ $1 + cos (pi/4) = 2cos^2 (pi/8)$

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{5 π}{8})$ $cos(pi/ 3 ) * sin( ( 5 pi) / 8 )$ without using products of trigonometric functions?

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