How do you express $cos (\frac{π}{3}) \cdot sin (\frac{7 π}{8})$ $cos(pi/ 3 ) * sin( ( 7 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{7 π}{8})$ $cos(pi/ 3 ) * sin( ( 7 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-02-13T02:31:08

Use Sum or Difference
$cos (\frac{π}{3}) \cdot sin (\frac{7 π}{8}) = \frac{1}{2} \cdot sin (\frac{13 π}{24}) - \frac{1}{2} \cdot sin (\frac{5 π}{24})$ $cos(pi/3)*sin((7pi)/8)=1/2*sin((13pi)/24)-1/2*sin ((5pi)/24)$

Explanation:

This is the derivation

$sin (x + y) = sin x cos y + cos x sin y$ $sin (x+y)=sin x cos y+ cos x sin y" " " "$ 1st equation
$sin (x - y) = sin x cos y - cos x sin y$ $sin (x-y)=sin x cos y- cos x sin y" " " "$ 2nd equation

subtract 2nd from the 1st

$sin (x + y) - sin (x - y) = 2 \cdot cos x sin y$ $sin (x+y)-sin(x-y)=2*cos x sin y$
it follows

$cos x sin y = \frac{1}{2} [sin (x + y) - sin (x - y)]$ $cos x sin y=1/2[sin (x+y)-sin(x-y)]$

Now, let $x = \frac{π}{3}$ $x=pi/3$ and $y = \frac{7 π}{8}$ $y=(7pi)/8$

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) = \frac{1}{2} [sin (\frac{π}{3} + \frac{7 π}{8}) - sin (\frac{π}{3} - \frac{7 π}{8})]$ $cos (pi/3) sin ((7pi)/8)=1/2[sin (pi/3+(7pi)/8)-sin(pi/3-(7pi)/8)]$

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) = \frac{1}{2} [sin (\frac{8 π + 21 π}{24}) - sin (\frac{8 π - 21 π}{24})]$ $cos (pi/3) sin ((7pi)/8)=1/2[sin ((8pi+21pi)/24)-sin ((8pi-21pi)/24)]$

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) = \frac{1}{2} [sin (\frac{29 π}{24}) - sin (\frac{- 13 π}{24})]$ $cos (pi/3) sin ((7pi)/8)=1/2[sin ((29pi)/24)-sin ((-13pi)/24)]$

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) = \frac{1}{2} [sin (π + \frac{5 π}{24}) - sin (\frac{- 13 π}{24})]$ $cos (pi/3) sin ((7pi)/8)=1/2[sin (pi+(5pi)/24)-sin ((-13pi)/24)]$

Note: $sin (\frac{- 13 π}{24}) = - sin (\frac{13 π}{24})$ $sin ((-13pi)/24)=-sin ((13pi)/24)$

so that

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) = \frac{1}{2} [sin (π + \frac{5 π}{24}) - (- sin (\frac{13 π}{24}))]$ $cos (pi/3) sin ((7pi)/8)=1/2[sin (pi+(5pi)/24)-(-sin ((13pi)/24))]$

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) = \frac{1}{2} [sin (π + \frac{5 π}{24}) + sin (\frac{13 π}{24})]$ $cos (pi/3) sin ((7pi)/8)=1/2[sin (pi+(5pi)/24)+sin ((13pi)/24)]$

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) =$ $cos (pi/3) sin ((7pi)/8)=$
$\frac{1}{2} [sin π cos (\frac{5 π}{24}) + cos π sin (\frac{5 π}{24}) + sin (\frac{13 π}{24})]$ $1/2[sin pi cos((5pi)/24)+cos pi sin ((5pi)/24)+sin ((13pi)/24)]$

$cos (\frac{π}{3}) sin (\frac{7 π}{8}) = \frac{1}{2} [sin (\frac{13 π}{24}) - sin (\frac{5 π}{24})]$ $cos (pi/3) sin ((7pi)/8)=1/2[sin((13pi)/24)-sin ((5pi)/24)]$

final answer

$\frac{1}{2} \cdot sin (\frac{13 π}{24}) - \frac{1}{2} \cdot sin (\frac{5 π}{24})$ $1/2*sin((13pi)/24)-1/2*sin ((5pi)/24)$

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How do you express $cos (\frac{π}{3}) \cdot sin (\frac{7 π}{8})$ $cos(pi/ 3 ) * sin( ( 7 pi) / 8 )$ without using products of trigonometric functions?

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