How do you evaluate $[cos (arccos (\frac{3}{5}) - arcsin (\frac{4}{5}))]$ $[cos(arccos (3/5) - arcsin (4/5))]$ ?

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Answer 1 · 2015-10-27T16:25:56

$arccos (\frac{3}{5}) = arcsin (\frac{4}{5})$ $arccos(3/5) = arcsin(4/5)$ so this is just $cos (0) = 1$ $cos(0) = 1$

$3^{2} + 4^{2} = 5^{2}$ $3^2+4^2=5^2$

So a $3, 4, 5$ $3,4,5$ triangle is a right angled triangle.

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$arccos (\frac{3}{5}) = B = arcsin (\frac{4}{5})$ $arccos(3/5) = B = arcsin(4/5)$

So: $cos (arccos (\frac{3}{5}) - arcsin (\frac{4}{5})) = cos (0) = 1$ $cos(arccos(3/5)-arcsin(4/5)) = cos(0) = 1$

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