Trig expression help?

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Answer 1 · 2018-08-06T09:18:45

The answer is $= o p t i o n (4)$ $=option (4)$

Let

$x = {sin}^{- 1} (\frac{2}{3})$ $x=sin^-1(2/3)$

And

$y = {sin}^{- 1} (- \frac{1}{3})$ $y=sin^-1(-1/3)$

Then,

$cos (x + 2 y) = cos x cos 2 y - sin x sin 2 y$ $cos(x+2y)=cosxcos2y-sinxsin2y$

$sin x = \frac{2}{3}$ $sinx=2/3$

$cos x = \sqrt{1 - {sin}^{2} x} = \sqrt{1 - \frac{4}{9}}$ $cosx=sqrt(1-sin^2x)=sqrt(1-4/9)$

$= \frac{\sqrt{5}}{3}$ $=sqrt5/3$

$sin y = - \frac{1}{3}$ $siny=-1/3$

$cos y = \sqrt{1 - {sin}^{2} y} = \sqrt{1 - \frac{1}{9}} = \frac{2 \sqrt{2}}{3}$ $cosy=sqrt(1-sin^2y)=sqrt(1-1/9)=(2sqrt2)/3$

$sin (2 y) = 2 sin y cos y = 2 \cdot - \frac{1}{3} \cdot \frac{2 \sqrt{2}}{3}$ $sin(2y)=2sinycosy=2*-1/3*(2sqrt2)/3$

$= - \frac{4 \sqrt{2}}{9}$ $=-(4sqrt2)/9$

$cos (2 y) = 1 - 2 {sin}^{2} y = 1 - 2 \cdot \frac{1}{9} = \frac{7}{9}$ $cos(2y)=1-2sin^2y=1-2*1/9=7/9$

Finally,

$cos (x + 2 y) = \frac{\sqrt{5}}{3} \cdot \frac{7}{9} - \frac{2}{3} \cdot - \frac{4 \sqrt{2}}{9}$ $cos(x+2y)=sqrt5/3*7/9-2/3*-(4sqrt2)/9$

$= \frac{7 \sqrt{5}}{27} + \frac{8 \sqrt{2}}{27}$ $=(7sqrt5)/27+(8sqrt2)/27$

$= \frac{7 \sqrt{5} + 8 \sqrt{2}}{27}$ $=(7sqrt5+8sqrt2)/27$

The answer is $= o p t i o n (4)$ $=option (4)$