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Let ${⨍ (x)=e}^{\cos^{- 1} \sin (x + π / 3)}$ then

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a

$f (\frac{8 π}{9}) = e^{5 π / 18}$

b

$f (\frac{8 π}{9}) = e^{13 π / 18}$

c

$f (- \frac{7 π}{4}) = e^{π / 12}$

d

$f (- \frac{7 π}{4}) = e^{11 π / 12}$

answer is B.

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Detailed Solution

$⨍ (\frac{8 π}{9}) {=e}^{\cos^{- 1 (\sin (\frac{11 π}{9}))}} {=e}^{\cos^{- 1} cos \frac{13 π}{18}} {=e}^{\frac{13 π}{18}} ⇒⨍ (\frac{- 7 π}{4}) {=e}^{\cos^{- 1} sin (\frac{- 7 π}{4} + \frac{π}{3})} {=e}^{\cos^{- 1} cos \frac{π}{12}} {=e}^{\frac{π}{12}}$

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