Question

Sin(theta)=2/3 and theta is in quadrant II, find cos theta

Answer 1

Answer: Choice C.
`\displaystyle \boldsymbol{-(√(5))/(3)}`

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Work Shown:

Part 1

`\sin^2(\theta)+\cos^2(\theta) = 1\\\\\cos^2(\theta) = 1-\sin^2(\theta)\\\\\cos(\theta) = -√(1-\sin^2(\theta)) \ \ \text{ ..... cosine is negative in Q2}\\\\\cos(\theta) = -\sqrt{1-\left((2)/(3)\right)^2}\\\\\\cos(\theta) = -\sqrt{1-(4)/(9)}\\\\`

Part 2

`\cos(\theta) = -\sqrt{(9)/(9)-(4)/(9)}\\\\\cos(\theta) = -\sqrt{(9-4)/(9)}\\\\\cos(\theta) = -\sqrt{(5)/(9)}\\\\\cos(\theta) = -(√(5))/(√(9))\\\\\cos(\theta) = -(√(5))/(3)\\\\`