// Numbas version: exam_results_page_options {"name": "Pythagorean Identity: find cos given sin", "extensions": [], "custom_part_types": [], "resources": [["question-resources/right_angled_triangle_4erLEm1.svg", "/srv/numbas/media/question-resources/right_angled_triangle_4erLEm1.svg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Pythagorean Identity: find cos given sin", "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

Use $\\cos^2\\theta+\\sin^2\\theta=1$ and/or an understanding on the unit circle definitions to determine $\\cos\\theta$ given $\\sin\\theta$ and the quadrant theta is in.

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You are told that $\\sin\\theta=\\simplify{{a}/sqrt({cc})}$.

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If $\\theta$ is in the first fourth quadrant, then the exact value of $\\cos\\theta$ is [[0]].

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Note: In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example, if your answer is $\\frac{\\sqrt{5}}{\\sqrt{17}}$, then enter sqrt(5)/sqrt(17)

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If $\\theta$ is in the second third quadrant, then the exact value of $\\cos\\theta$ is [[0]].

\n

Note: In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example, if your answer is $\\frac{\\sqrt{5}}{\\sqrt{17}}$, then enter sqrt(5)/sqrt(17)

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The equation of the unit circle is \\[x^2+y^2=1\\]

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This equation is a consequence of Pythagoras' Theorem, $a^2+b^2=c^2$, on a triangle in the unit circle with hypotenuse 1.

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Given that $\\cos\\theta$ is the $x$ coordinate of a point on the unit circle, and that $\\sin\\theta$ is the $y$ coordinate of the same point, by substitution we have the following Pythagorean identity \\[\\cos^2\\theta+\\sin^2\\theta=1\\]

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where $\\cos^2\\theta$ and $\\sin^2\\theta$ is the notation used to represent $(\\cos\\theta)^2$ and $(\\sin\\theta)^2$ respectively.

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Using the vaue of $\\sin\\theta$ given in the question we have that $\\cos^2\\theta+\\left(\\simplify{{a}/sqrt({cc})}\\right)^2=1$. We solve for $\\sin\\theta$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\cos^2\\theta+\\simplify{{aa}/{cc}}$$=$$1$
$\\cos^2\\theta$$=$$1-\\simplify{{aa}/{cc}}$
$\\cos^2\\theta$$=$$\\simplify[fractionNumbers]{{diff}}$
$\\cos\\theta$$=$$\\pm\\simplify[fractionNumbers]{sqrt({diff})}$
\n

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Because we are told that $\\sin\\theta$ is positive, we know that $\\theta$ is in the first or second quadrant (since $\\sin\\theta$ is the $y$ coordinate). If $\\theta$ is in the first quadrant, then (since $\\cos\\theta$ is the $x$ coordinate) $\\cos\\theta$ is positive, that is, $\\cos\\theta=\\simplify[fractionNumbers]{sqrt({diff})}$. However, if $\\theta$ is in the second quadrant, then (since $\\cos\\theta$ is the $x$ coordinate) $\\cos\\theta$ is negative, that is, $\\cos\\theta=-\\simplify[fractionNumbers]{sqrt({diff})}$.  negative, we know that $\\theta$ is in the third or fourth quadrant (since $\\sin\\theta$ is the $y$ coordinate). If $\\theta$ is in the fourth quadrant, then (since $\\cos\\theta$ is the $x$ coordinate) $\\cos\\theta$ is positive, that is, $\\cos\\theta=\\simplify[fractionNumbers]{sqrt({diff})}$. However, if $\\theta$ is in the third quadrant, then (since $\\cos\\theta$ is the $x$ coordinate) $\\cos\\theta$ is negative, that is, $\\cos\\theta=-\\simplify[fractionNumbers]{sqrt({diff})}$.  

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