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

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Answer 1 · 2016-07-26T12:02:56

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

Let $alpha = arccos(-2/3)$ and $beta = arcsin(1/4)$

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

$sin(alpha) = sqrt(1-cos^2(alpha)) = sqrt(1-(-2/3)^2) = sqrt(5/9) = sqrt(5)/3$

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

$cos(beta) = sqrt(1-sin^2(beta))= sqrt(1-(1/4)^2) = sqrt(15/16) = sqrt(15)/4$

Then:

$sin(alpha+beta)$

$= sin alpha cos beta + sin beta cos alpha$

$=(sqrt(5)/3)(sqrt(15)/4) + (1/4)(-2/3)$

$=(5sqrt(3))/12-1/6$

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