How do you simplify $cos [{cos}^{- 1} (\frac{3}{5}) - {sin}^{- 1} (- \frac{4}{5})]$ $Cos[cos^-1 (3/5) - sin^-1 (-4/5)]$ ?

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Answer 1 · 2016-06-21T17:10:19

Let $a = {cos}^{- 1} (\frac{3}{5})$ $a = cos^(-1)(3/5)$ . Then, $cos a = \frac{3}{5} > 0$ $cos a = 3/5>0$ .a is in either 1st

quadrant or in the 4th. Accordingly, $sin a = \pm \frac{4}{5}$ $sin a = +-4/5$ .

Let $b = {sin}^{- 1} (- \frac{4}{5})$ $b = sin^(-1)(-4/5)$ . Then, $sin b = - \frac{4}{5} > 0$ $sin b = -4/5>0$ .b is in either 3rd

quadrant or in the 4th. Accordingly, $cos b = \pm \frac{3}{5}$ $cos b = +-3/5$ .

Now, the given expression is

$cos (a - b) = cos a cos b + sin a sin b$ $cos(a-b)=cos a cos b+sin a sin b$

$= ((\frac{3}{5}) (\pm \frac{3}{5}) - (\pm \frac{4}{5}) ((- \frac{4}{5}))$ $=((3/5)(+-3/5)-(+-4/5)((-4/5))$

$= \frac{\pm 9 \pm 16}{25}$ $=(+-9+-16)/25$

$= \pm 1, \pm \frac{7}{25}$ $=+-1, +-7/25$ .

How do you simplify Cos[cos^-1 (3/5) - sin^-1 (-4/5)] cos[cos−1(35)−sin−1(−45)]Cos[cos^-1 (3/5) - sin^-1 (-4/5)] ?