How do you find the exact value of $sin [arccos (- \frac{2}{3}) + arcsin (\frac{1}{4})]$ $sin[arccos(-2/3)+arcsin(1/4)]$ ?

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How do you find the exact value of $sin [arccos (- \frac{2}{3}) + arcsin (\frac{1}{4})]$ $sin[arccos(-2/3)+arcsin(1/4)]$ ?

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Answer 1 · 2016-07-26T12:02:56

$sin (arccos (- \frac{2}{3}) + arcsin (\frac{1}{4})) = \frac{5 \sqrt{3}}{12} - \frac{1}{6}$ $sin(arccos(-2/3)+arcsin(1/4))=(5sqrt(3))/12-1/6$

Explanation:

Let $α = arccos (- \frac{2}{3})$ $alpha = arccos(-2/3)$ and $β = arcsin (\frac{1}{4})$ $beta = arcsin(1/4)$

$α$ $alpha$ is in Q2, so $sin (α) > 0$ $sin(alpha) > 0$ and we find:

$sin (α) = \sqrt{1 - {cos}^{2} (α)} = \sqrt{1 - {(- \frac{2}{3})}^{2}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$ $sin(alpha) = sqrt(1-cos^2(alpha)) = sqrt(1-(-2/3)^2) = sqrt(5/9) = sqrt(5)/3$

$β$ $beta$ is in Q1, so $cos (β) > 0$ $cos(beta) > 0$ and we find:

$cos (β) = \sqrt{1 - {sin}^{2} (β)} = \sqrt{1 - {(\frac{1}{4})}^{2}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4}$ $cos(beta) = sqrt(1-sin^2(beta))= sqrt(1-(1/4)^2) = sqrt(15/16) = sqrt(15)/4$

Then:

$sin (α + β)$ $sin(alpha+beta)$

$= sin α cos β + sin β cos α$ $= sin alpha cos beta + sin beta cos alpha$

$= (\frac{\sqrt{5}}{3}) (\frac{\sqrt{15}}{4}) + (\frac{1}{4}) (- \frac{2}{3})$ $=(sqrt(5)/3)(sqrt(15)/4) + (1/4)(-2/3)$

$= \frac{5 \sqrt{3}}{12} - \frac{1}{6}$ $=(5sqrt(3))/12-1/6$

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How do you find the exact value of $sin [arccos (- \frac{2}{3}) + arcsin (\frac{1}{4})]$ $sin[arccos(-2/3)+arcsin(1/4)]$ ?

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