The following questions require some familiarity with trigonometric identities.
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\n\\[\\simplify{(cos^2(x)+sin^2(x))^{m1}+{m2}}\\] \\[\\simplify{{m1}+{m2}(2-sin^2(y)-cos^2(y))}\\] \\[\\simplify{{m1^2}cos^4(z)+{2*m1^2}cos^2(z)sin^2(z)+{m1^2}sin^4(z)}\\]
\ncan be simplified to [[0]].
\nNote: For this question your answer should be a number.
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\n\\[\\simplify{{n}sin(theta)-{n}sqrt(1-cos^2(theta))}\\] \\[\\simplify{{n}sin(theta)+{n}sqrt(1-cos^2(theta))}\\]\\[\\simplify{{n}cos(theta)-{n}sqrt(1-sin^2(theta))}\\] \\[\\simplify{{n}cos(theta)+{n}sqrt(1-sin^2(theta))}\\]
\ncan be simplified to [[0]] for $\\theta$ in the first or second quadrant. third or fourth quadrant. first or fourth quadrant. second or third quadrant.
\nNote: For this question your answer should be a number.
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\nFor part a)
\nGiven $\\simplify{(cos^2(x)+sin^2(x))^{m1}+{m2}}$ we can replace $\\cos^2(x)+\\sin^2(x)$ with $1$. So our expression is $\\simplify{1^{m1}+{m2}}$. Therefore our expression simplifies to $\\var{m2+1}$.
\n| $\\simplify{{m1}+{m2}(2-sin^2(y)-cos^2(y))}$ | \n$=$ | \n$\\simplify{{m1}+{m2}(2-(sin^2(y)+cos^2(y)))}$ | \n
| \n | $=$ | \n$\\simplify[!collectNumbers]{{m1}+{m2}(2-1)}$ | \n
| \n | $=$ | \n$\\simplify[!collectNumbers]{{m1}+{m2}}$ | \n
| \n | $=$ | \n$\\var{m1+m2}$ | \n
| $\\simplify{{m1^2}cos^4(z)+{2*m1^2}cos^2(z)sin^2(z)+{m1^2}sin^4(z)}$ | \n$=$ | \n$\\simplify{{m1^2}(cos^4(z)+2cos^2(z)sin^2(z)+sin^4(z))}$ | \n
| \n | $=$ | \n$\\simplify{{m1^2}(cos^2(z)+sin^2(z))^2}$ | \n
| \n | $=$ | \n$\\var{m1^2}\\times 1^2$ | \n
| \n | $=$ | \n$\\var{m1^2}$ | \n
For part b)
\nRearranging the Pythagorean identity $\\cos^2\\theta+\\sin^2\\theta=1$ for $\\sin\\theta$ gives the equation \\[\\sin\\theta=\\pm\\sqrt{1-\\cos^2\\theta}\\]
\nRecall that $\\sin\\theta$ is the $y$ value of a point on the unit circle, whether $\\sin\\theta$ is taken as the postive square root or as the negative square root depends on the whether the point on the circle is on the top semicircle (positive $y$ value) or the bottom semicircle (negative $y$ value).
\nSince we are told $\\theta$ is in the first or second quadrant, the $y$ value must be postive, that is $\\sin\\theta=\\sqrt{1-\\cos^2\\theta}$. Therefore our expresson simplifies as follows
\n| $\\simplify{{n}sin(theta)-{n}sqrt(1-cos^2(theta))}$ | \n$=$ | \n$\\simplify{{n}sin(theta)-{n}sin(theta)}$ | \n
| \n | $=$ | \n$0$ | \n
Since we are told $\\theta$ is in the third or fourth quadrant, the $y$ value must be negative, that is $\\sin\\theta=-\\sqrt{1-\\cos^2\\theta}$ or equivalently $-\\sin\\theta=\\sqrt{1-\\cos^2\\theta}$. Therefore our expresson simplifies as follows
\n\n| $\\simplify{{n}sin(theta)-{n}sqrt(1-cos^2(theta))}$ | \n$=$ | \n$\\simplify{{n}sin(theta)-{n}sin(theta)}$ | \n
| \n | $=$ | \n$0$ | \n
Rearranging the Pythagorean identity $\\cos^2\\theta+\\sin^2\\theta=1$ for $\\cos\\theta$ gives the equation \\[\\cos\\theta=\\pm\\sqrt{1-\\sin^2\\theta}\\]
\nRecall that $\\cos\\theta$ is the $x$ value of a point on the unit circle, whether $\\cos\\theta$ is taken as the postive square root or as the negative square root depends on the whether the point on the circle is on the right semicircle (positive $x$ value) or the left semicircle (negative $x$ value).
\nSince we are told $\\theta$ is in the first or fourth quadrant, the $x$ value must be postive, that is $\\cos\\theta=\\sqrt{1-\\sin^2\\theta}$. Therefore our expresson simplifies as follows
\n| $\\simplify{{n}cos(theta)-{n}sqrt(1-sin^2(theta))}$ | \n$=$ | \n$\\simplify{{n}cos(theta)-{n}cos(theta)}$ | \n
| \n | $=$ | \n$0$ | \n
Since we are told $\\theta$ is in the second or third quadrant, the $x$ value must be negative, that is $\\cos\\theta=-\\sqrt{1-\\sin^2\\theta}$ or equivalently $-\\cos\\theta=\\sqrt{1-\\sin^2\\theta}$. Therefore our expresson simplifies as follows
\n\n| $\\simplify{{n}cos(theta)-{n}sqrt(1-sin^2(theta))}$ | \n$=$ | \n$\\simplify{{n}cos(theta)-{n}cos(theta)}$ | \n
| \n | $=$ | \n$0$ | \n