How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{2})$ $sin(pi/12) * cos(( pi)/2 )$ without products of trigonometric functions?

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{2})$ $sin(pi/12) * cos(( pi)/2 )$ without products of trigonometric functions?

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Answer 1 · 2016-06-14T15:33:21

$sin (\frac{π}{12}) \cdot cos (\frac{π}{2}) = \frac{1}{2} sin (\frac{7 π}{12}) - \frac{1}{2} sin (\frac{5 π}{12})$ $color(blue)(sin (pi/12)*cos (pi/2)=1/2sin ((7pi)/12)-1/2sin ((5pi)/12))$

Explanation:

Let us use the sum and difference formulas

$sin (x + y) = sin x \cdot cos y + cos x \cdot sin y$ $sin (x+y)=sin x*cos y +cos x* sin y$

$sin (x - y) = sin x \cdot cos y - cos x \cdot sin y$ $sin (x-y)=sin x*cos y -cos x* sin y$

After addition of equal values, we have

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

and we have the formula

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

We can now use the given

$sin (\frac{π}{12}) \cdot cos (\frac{π}{2}) = \frac{1}{2} sin (\frac{π}{12} + \frac{π}{2}) + \frac{1}{2} sin (\frac{π}{12} - \frac{π}{2})$ $sin (pi/12)*cos (pi/2)=1/2sin (pi/12+pi/2)+1/2sin (pi/12-pi/2)$

$sin (\frac{π}{12}) \cdot cos (\frac{π}{2}) = \frac{1}{2} sin (\frac{7 π}{12}) + \frac{1}{2} sin (\frac{- 5 π}{12})$ $sin (pi/12)*cos (pi/2)=1/2sin ((7pi)/12)+1/2sin ((-5pi)/12)$

$sin (\frac{π}{12}) \cdot cos (\frac{π}{2}) = \frac{1}{2} sin (\frac{7 π}{12}) - \frac{1}{2} sin (\frac{5 π}{12})$ $sin (pi/12)*cos (pi/2)=1/2sin ((7pi)/12)-1/2sin ((5pi)/12)$

God bless....I hope the explanation is useful.

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{2})$ $sin(pi/12) * cos(( pi)/2 )$ without products of trigonometric functions?

1 Answer

Explanation:

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How do you express $sin (\frac{π}{12}) \cdot cos (\frac{π}{2})$ $sin(pi/12) * cos(( pi)/2 )$ without products of trigonometric functions?