Solve $cos(sin^-1(x))= 1/9$ for $x$ ?

Question

15824 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-10T08:44:08

$x=+-sqrt80/9$

To solve $cos(sin^-1(x))= 1/9$ for $x$ , let us assume

$x=sintheta$ and then $sin^-1x=theta$

and hence $cos(sin^-1(x))$

= $costheta=sqrt(1-sin^2theta)$

= $sqrt(1-x^2)$

As such $sqrt(1-x^2)=1/9$

$:.1-x^2=1/81$ or $x^2-80/81=0$

or $(x-sqrt80/9)(x+sqrt80/9)=0$

Hence, $x=+-sqrt80/9$

Answer 2 · 2017-02-10T15:37:22

$+- 83^@62$

sin^-1 x --> arcsin x
cos (arcsin x) --> $cos x = 1/9$
Use calculator and unit circle:
$cos x = 1/9$ --> $x == +- 83^@62$

Solve cos(sin^-1(x))= 1/9cos(sin^-1(x))= 1/9 for xx?