How do you evaluate $sin (\frac{5 π}{9}) cos (\frac{7 π}{18}) - cos (\frac{5 π}{9}) sin (\frac{7 π}{18})$ $sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)$ ?

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How do you evaluate $sin (\frac{5 π}{9}) cos (\frac{7 π}{18}) - cos (\frac{5 π}{9}) sin (\frac{7 π}{18})$ $sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)$ ?

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Answer 1 · 2016-04-16T00:29:05

$\frac{1}{2}$ $1/2$

Explanation:

This equation can be solved using some knowledge about some trigonometric identities. In this case, the expansion of $sin (A - B)$ $sin(A-B)$ should be known:

$sin (A - B) = sin A cos B - cos A sin B$ $sin(A-B)=sinAcosB-cosAsinB$

You'll notice that this looks awfully similar to the equation in the question. Using the knowledge, we can solve it:
$sin (\frac{5 π}{9}) cos (\frac{7 π}{18}) - cos (\frac{5 π}{9}) sin (\frac{7 π}{18})$ $sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)$
$= sin (\frac{5 π}{9} - \frac{7 π}{18})$ $=sin((5pi)/9-(7pi)/18)$
$= sin (\frac{10 π}{18} - \frac{7 π}{18})$ $=sin((10pi)/18-(7pi)/18)$
$= sin (\frac{3 π}{18})$ $=sin((3pi)/18)$
$= sin (\frac{π}{6})$ $=sin((pi)/6)$ , and that has exact value of $\frac{1}{2}$ $1/2$

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How do you evaluate $sin (\frac{5 π}{9}) cos (\frac{7 π}{18}) - cos (\frac{5 π}{9}) sin (\frac{7 π}{18})$ $sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)$ ?

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Explanation:

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How do you evaluate sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)sin(5π9)cos(7π18)−cos(5π9)sin(7π18)sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)?

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Explanation:

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How do you evaluate $sin (\frac{5 π}{9}) cos (\frac{7 π}{18}) - cos (\frac{5 π}{9}) sin (\frac{7 π}{18})$ $sin((5pi)/9)cos((7pi)/18)-cos((5pi)/9)sin((7pi)/18)$ ?