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

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Answer 1 · 2015-06-13T18:31:35

Find value of $sin [arccos (- \frac{4}{5}) + arcsin (- \frac{3}{5})]$ $sin[ arccos (-4/5) + arcsin (-3/5)]$

On the trig unit circle.

$cos x = - \frac{4}{5} = - 0.80 - \to x = - 143.13$ $cos x = -4/5 = -0.80 --> x = - 143.13$

$sin y = - \frac{3}{5} = - 0.60 - \to y = - 36.87$ $sin y = -3/5 = -0.60 --> y = -36.87$

$sin (x + y) = sin (- 143.13 - 36.87) = sin (- 180) = 0$ $sin (x + y) = sin (-143.13 - 36.87) = sin (-180) = 0$

How do you evaluate sin[arccos(−45)+arcsin(−35)]sin[arccos(-4/5)+arcsin(-3/5)]?