How do you express $sin (\frac{π}{12}) \cdot cos (\frac{13 π}{12})$ $sin(pi/12) * cos(( 13 pi)/12 )$ without products of trigonometric functions?

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{13 π}{12})$ $sin(pi/12) * cos(( 13 pi)/12 )$ without products of trigonometric functions?

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Answer 1 · 2018-07-03T09:12:07

$- sin (\frac{π}{6}) = - \frac{1}{2}$ $color(red)(-sin (pi/6)= -1/2$

Explanation:

![https://study.com/academy/lesson/http://product-to-sum-identities-uses-applications.html](https://useruploads.socratic.org/2xBiCxzSmmCr1qNXSs2e_trigonometric%20identities.png)

$sin (\frac{π}{12}) \cdot cos (\frac{13 π}{12})$ $sin (pi/12) * cos ((13pi)/12)$

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

$\Rightarrow (\frac{1}{2}) [sin (\frac{π}{12} + \frac{13 π}{12}) + sin (\frac{π}{12} - \frac{13 π}{12})]$ $=> (1/2)[sin (pi/12 + (13pi)/12) + sin (pi/12 - (13pi)/12)]$

$\Rightarrow \frac{sin (\frac{14 π}{12}) + sin (- π)}{2}$ $=> [sin ((14pi)/12) + sin (-pi)]/2$

$\Rightarrow \frac{sin (π + \frac{π}{6}) - sin (π)}{2}$ $=> [sin (pi + pi/6) - sin (pi)]/2$

$- sin (\frac{π}{6}) = - \frac{1}{2}$ $color(red)(-sin (pi/6)= -1/2$

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{13 π}{12})$ $sin(pi/12) * cos(( 13 pi)/12 )$ without products of trigonometric functions?

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How do you express sin(π12)⋅cos(13π12)sin(pi/12) * cos(( 13 pi)/12 ) without products of trigonometric functions?

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{13 π}{12})$ $sin(pi/12) * cos(( 13 pi)/12 )$ without products of trigonometric functions?