How do you evaluate $sin [arccos (- \frac{2}{3}) + arcsin (\frac{1}{4})]$ $sin[arccos(-2/3)+arcsin(1/4)]$ ?

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Answer 1 · 2015-07-05T05:13:36

Find sin

Use calculator.

a. $cos x = \frac{2}{3} - \to x = 48.19$ $cos x = 2/3 --> x = 48.19$ deg

b. $sin y = \frac{1}{4} \to y = 14.48$ $sin y = 1/4 -> y = 14.48$ deg

sin (x + y) = sin (48.19 + 14.48) = sin (62.67) = 0.89

How do you evaluate sin[arccos(−23)+arcsin(14)]sin[arccos(-2/3)+arcsin(1/4)]?