Given that cos (x) = 1/3, find sin (90 - x)

2 answers:

3 0

$\sin{(A - B)} \equiv \sin{A}\cos{B} - \cos{A}\sin{B} \\\\
\therefore \sin{(90 - x)}\\
 = \sin{(90)}\cos{(x)} - \cos{(90)}\sin{(x)} \\
= (1)\cos{(x)} - (0)\sin{(x)} \\
= \cos{x} - 0 \\
= \cos{x} \\
= \frac{1}{3}
$

denis23 [38]3 years ago

3 0

$90а= \frac{ \pi }{2} \\ sin(90а-x)=sin (\frac{ \pi }{2}-x)=cos (x) \\ cos(x)= \frac{1}{3} \\ sin(x)= \frac{1}{3} \approx0.33333 \\ x=19а$

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