Proving the Pythagorean Identity - Geometry
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Find $\tan(\theta)$ if $\sin(\theta)=\frac{3}{5}$ and $\theta$ is in Quadrant II.
Find $\tan(\theta)$ if $\sin(\theta)=\frac{3}{5}$ and $\theta$ is in Quadrant II.
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$\tan(\theta)=-\frac{3}{4}$. Find $\cos(\theta)=-\frac{4}{5}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
$\tan(\theta)=-\frac{3}{4}$. Find $\cos(\theta)=-\frac{4}{5}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
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State the quotient identity defining $\cot(\theta)$ in terms of $\cos(\theta)$ and $\sin(\theta)$.
State the quotient identity defining $\cot(\theta)$ in terms of $\cos(\theta)$ and $\sin(\theta)$.
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$\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$. The basic quotient identity defining cotangent.
$\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$. The basic quotient identity defining cotangent.
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What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant I?
What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant I?
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$\sin(\theta)>0$ and $\cos(\theta)>0$. In Quadrant I, both coordinates are positive.
$\sin(\theta)>0$ and $\cos(\theta)>0$. In Quadrant I, both coordinates are positive.
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What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant II?
What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant II?
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$\sin(\theta)>0$ and $\cos(\theta)<0$. In Quadrant II, $y>0$ but $x<0$.
$\sin(\theta)>0$ and $\cos(\theta)<0$. In Quadrant II, $y>0$ but $x<0$.
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What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant III?
What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant III?
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$\sin(\theta)<0$ and $\cos(\theta)<0$. In Quadrant III, both coordinates are negative.
$\sin(\theta)<0$ and $\cos(\theta)<0$. In Quadrant III, both coordinates are negative.
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What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant IV?
What are the signs of $\sin(\theta)$ and $\cos(\theta)$ in Quadrant IV?
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$\sin(\theta)<0$ and $\cos(\theta)>0$. In Quadrant IV, $x>0$ but $y<0$.
$\sin(\theta)<0$ and $\cos(\theta)>0$. In Quadrant IV, $x>0$ but $y<0$.
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What is the sign of $\tan(\theta)$ in Quadrant II?
What is the sign of $\tan(\theta)$ in Quadrant II?
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$\tan(\theta)<0$. In Quadrant II, $\sin>0$ and $\cos<0$, so $\tan<0$.
$\tan(\theta)<0$. In Quadrant II, $\sin>0$ and $\cos<0$, so $\tan<0$.
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What is the sign of $\tan(\theta)$ in Quadrant III?
What is the sign of $\tan(\theta)$ in Quadrant III?
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$\tan(\theta)>0$. In Quadrant III, both sine and cosine are negative, so $\tan>0$.
$\tan(\theta)>0$. In Quadrant III, both sine and cosine are negative, so $\tan>0$.
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What is the sign of $\tan(\theta)$ in Quadrant IV?
What is the sign of $\tan(\theta)$ in Quadrant IV?
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$\tan(\theta)<0$. In Quadrant IV, $\sin<0$ and $\cos>0$, so $\tan<0$.
$\tan(\theta)<0$. In Quadrant IV, $\sin<0$ and $\cos>0$, so $\tan<0$.
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Identify the general method to find $\cos(\theta)$ given $\sin(\theta)$ and the quadrant.
Identify the general method to find $\cos(\theta)$ given $\sin(\theta)$ and the quadrant.
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$\cos(\theta)=\pm\sqrt{1-\sin^2(\theta)}$ with sign from quadrant. Solve for cosine using the identity, then apply the quadrant sign.
$\cos(\theta)=\pm\sqrt{1-\sin^2(\theta)}$ with sign from quadrant. Solve for cosine using the identity, then apply the quadrant sign.
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Identify the general method to find $\sin(\theta)$ given $\cos(\theta)$ and the quadrant.
Identify the general method to find $\sin(\theta)$ given $\cos(\theta)$ and the quadrant.
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$\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}$ with sign from quadrant. Solve for sine using the identity, then apply the quadrant sign.
$\sin(\theta)=\pm\sqrt{1-\cos^2(\theta)}$ with sign from quadrant. Solve for sine using the identity, then apply the quadrant sign.
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What is $\cos(\theta)$ if $\sin(\theta)=\frac{3}{5}$ and $\theta$ is in Quadrant I?
What is $\cos(\theta)$ if $\sin(\theta)=\frac{3}{5}$ and $\theta$ is in Quadrant I?
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$\cos(\theta)=\frac{4}{5}$. Using $\cos^2(\theta)=1-\sin^2(\theta)=1-\frac{9}{25}=\frac{16}{25}$, so $\cos(\theta)=\frac{4}{5}$ in Q1.
$\cos(\theta)=\frac{4}{5}$. Using $\cos^2(\theta)=1-\sin^2(\theta)=1-\frac{9}{25}=\frac{16}{25}$, so $\cos(\theta)=\frac{4}{5}$ in Q1.
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Find $\sin(\theta)$ if $\cos(\theta)=\frac{24}{25}$ and $\theta$ is in Quadrant IV.
Find $\sin(\theta)$ if $\cos(\theta)=\frac{24}{25}$ and $\theta$ is in Quadrant IV.
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$\sin(\theta)=-\frac{7}{25}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{576}{625}=\frac{49}{625}$, negative in Q4.
$\sin(\theta)=-\frac{7}{25}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{576}{625}=\frac{49}{625}$, negative in Q4.
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Find $\cos(\theta)$ if $\sin(\theta)=\frac{7}{25}$ and $\theta$ is in Quadrant II.
Find $\cos(\theta)$ if $\sin(\theta)=\frac{7}{25}$ and $\theta$ is in Quadrant II.
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$\cos(\theta)=-\frac{24}{25}$. Same calculation as Q1, but cosine is negative in Quadrant II.
$\cos(\theta)=-\frac{24}{25}$. Same calculation as Q1, but cosine is negative in Quadrant II.
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Find $\cos(\theta)$ if $\sin(\theta)=\frac{7}{25}$ and $\theta$ is in Quadrant I.
Find $\cos(\theta)$ if $\sin(\theta)=\frac{7}{25}$ and $\theta$ is in Quadrant I.
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$\cos(\theta)=\frac{24}{25}$. Using $\cos^2(\theta)=1-\sin^2(\theta)=1-\frac{49}{625}=\frac{576}{625}$, positive in Q1.
$\cos(\theta)=\frac{24}{25}$. Using $\cos^2(\theta)=1-\sin^2(\theta)=1-\frac{49}{625}=\frac{576}{625}$, positive in Q1.
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Find $\cos(\theta)$ if $\tan(\theta)=-\frac{5}{12}$ and $\theta$ is in Quadrant IV.
Find $\cos(\theta)$ if $\tan(\theta)=-\frac{5}{12}$ and $\theta$ is in Quadrant IV.
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$\cos(\theta)=\frac{12}{13}$. From $\tan(\theta)=-\frac{5}{12}$, find cosine using $1+\tan^2(\theta)=\sec^2(\theta)$.
$\cos(\theta)=\frac{12}{13}$. From $\tan(\theta)=-\frac{5}{12}$, find cosine using $1+\tan^2(\theta)=\sec^2(\theta)$.
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Find $\sin(\theta)$ if $\tan(\theta)=-\frac{5}{12}$ and $\theta$ is in Quadrant IV.
Find $\sin(\theta)$ if $\tan(\theta)=-\frac{5}{12}$ and $\theta$ is in Quadrant IV.
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$\sin(\theta)=-\frac{5}{13}$. From $\tan(\theta)=-\frac{5}{12}$, find sine using $1+\tan^2(\theta)=\sec^2(\theta)$.
$\sin(\theta)=-\frac{5}{13}$. From $\tan(\theta)=-\frac{5}{12}$, find sine using $1+\tan^2(\theta)=\sec^2(\theta)$.
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Find $\cos(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant III.
Find $\cos(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant III.
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$\cos(\theta)=-\frac{4}{5}$. Same calculation as Q1, but cosine is negative in Quadrant III.
$\cos(\theta)=-\frac{4}{5}$. Same calculation as Q1, but cosine is negative in Quadrant III.
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Find $\sin(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant III.
Find $\sin(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant III.
Tap to reveal answer
$\sin(\theta)=-\frac{3}{5}$. Same calculation as Q1, but sine is negative in Quadrant III.
$\sin(\theta)=-\frac{3}{5}$. Same calculation as Q1, but sine is negative in Quadrant III.
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Find $\cos(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant I.
Find $\cos(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant I.
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$\cos(\theta)=\frac{4}{5}$. From $\tan(\theta)=\frac{3}{4}$, use identity $1+\tan^2(\theta)=\sec^2(\theta)$ to find cosine.
$\cos(\theta)=\frac{4}{5}$. From $\tan(\theta)=\frac{3}{4}$, use identity $1+\tan^2(\theta)=\sec^2(\theta)$ to find cosine.
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Find $\sin(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant I.
Find $\sin(\theta)$ if $\tan(\theta)=\frac{3}{4}$ and $\theta$ is in Quadrant I.
Tap to reveal answer
$\sin(\theta)=\frac{3}{5}$. From $\tan(\theta)=\frac{3}{4}$, use identity $1+\tan^2(\theta)=\sec^2(\theta)$ to find sine.
$\sin(\theta)=\frac{3}{5}$. From $\tan(\theta)=\frac{3}{4}$, use identity $1+\tan^2(\theta)=\sec^2(\theta)$ to find sine.
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Find $\tan(\theta)$ if $\cos(\theta)=-\frac{8}{17}$ and $\theta$ is in Quadrant III.
Find $\tan(\theta)$ if $\cos(\theta)=-\frac{8}{17}$ and $\theta$ is in Quadrant III.
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$\tan(\theta)=\frac{15}{8}$. Find $\sin(\theta)=-\frac{15}{17}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
$\tan(\theta)=\frac{15}{8}$. Find $\sin(\theta)=-\frac{15}{17}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
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Find $\tan(\theta)$ if $\cos(\theta)=\frac{12}{13}$ and $\theta$ is in Quadrant IV.
Find $\tan(\theta)$ if $\cos(\theta)=\frac{12}{13}$ and $\theta$ is in Quadrant IV.
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$\tan(\theta)=-\frac{5}{12}$. Find $\sin(\theta)=-\frac{5}{13}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
$\tan(\theta)=-\frac{5}{12}$. Find $\sin(\theta)=-\frac{5}{13}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
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Find $\tan(\theta)$ if $\sin(\theta)=\frac{3}{5}$ and $\theta$ is in Quadrant I.
Find $\tan(\theta)$ if $\sin(\theta)=\frac{3}{5}$ and $\theta$ is in Quadrant I.
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$\tan(\theta)=\frac{3}{4}$. Find $\cos(\theta)=\frac{4}{5}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
$\tan(\theta)=\frac{3}{4}$. Find $\cos(\theta)=\frac{4}{5}$ first, then $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$.
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What is $\sin(\theta)$ if $\cos(\theta)=-\frac{8}{17}$ and $\theta$ is in Quadrant III?
What is $\sin(\theta)$ if $\cos(\theta)=-\frac{8}{17}$ and $\theta$ is in Quadrant III?
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$\sin(\theta)=-\frac{15}{17}$. Same calculation as Q2, but sine is negative in Quadrant III.
$\sin(\theta)=-\frac{15}{17}$. Same calculation as Q2, but sine is negative in Quadrant III.
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What is $\sin(\theta)$ if $\cos(\theta)=-\frac{8}{17}$ and $\theta$ is in Quadrant II?
What is $\sin(\theta)$ if $\cos(\theta)=-\frac{8}{17}$ and $\theta$ is in Quadrant II?
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$\sin(\theta)=\frac{15}{17}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{64}{289}=\frac{225}{289}$, sine is positive in Q2.
$\sin(\theta)=\frac{15}{17}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{64}{289}=\frac{225}{289}$, sine is positive in Q2.
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What is $\cos(\theta)$ if $\sin(\theta)=-\frac{5}{13}$ and $\theta$ is in Quadrant IV?
What is $\cos(\theta)$ if $\sin(\theta)=-\frac{5}{13}$ and $\theta$ is in Quadrant IV?
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$\cos(\theta)=\frac{12}{13}$. Same sine calculation, but cosine is positive in Quadrant IV.
$\cos(\theta)=\frac{12}{13}$. Same sine calculation, but cosine is positive in Quadrant IV.
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What is $\cos(\theta)$ if $\sin(\theta)=-\frac{5}{13}$ and $\theta$ is in Quadrant III?
What is $\cos(\theta)$ if $\sin(\theta)=-\frac{5}{13}$ and $\theta$ is in Quadrant III?
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$\cos(\theta)=-\frac{12}{13}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=\frac{25}{169}$, cosine is negative in Q3.
$\cos(\theta)=-\frac{12}{13}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=\frac{25}{169}$, cosine is negative in Q3.
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What is $\sin(\theta)$ if $\cos(\theta)=\frac{12}{13}$ and $\theta$ is in Quadrant IV?
What is $\sin(\theta)$ if $\cos(\theta)=\frac{12}{13}$ and $\theta$ is in Quadrant IV?
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$\sin(\theta)=-\frac{5}{13}$. Same calculation as Q1, but sine is negative in Quadrant IV.
$\sin(\theta)=-\frac{5}{13}$. Same calculation as Q1, but sine is negative in Quadrant IV.
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What is $\sin(\theta)$ if $\cos(\theta)=\frac{12}{13}$ and $\theta$ is in Quadrant I?
What is $\sin(\theta)$ if $\cos(\theta)=\frac{12}{13}$ and $\theta$ is in Quadrant I?
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$\sin(\theta)=\frac{5}{13}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{144}{169}=\frac{25}{169}$, so $\sin(\theta)=\frac{5}{13}$ in Q1.
$\sin(\theta)=\frac{5}{13}$. Using $\sin^2(\theta)=1-\cos^2(\theta)=1-\frac{144}{169}=\frac{25}{169}$, so $\sin(\theta)=\frac{5}{13}$ in Q1.
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