Unit Circle: Trig Ratios

Question 1 of 3

Find

\tan θ

$tan theta$ given the following

\sin θ = - \frac{21}{29}

$sin theta=-21/29$

\cos θ

$cos theta$

>

$>$

0

$0$

Write fractions in the format “a/b”

$\tan θ =$ $tan theta=$ (-21/20)

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Trigonometric Functions

\sin θ = \frac{opposite}{hypotenuse}

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

\cos θ = \frac{adjacent}{hypotenuse}

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

\tan θ = \frac{opposite}{adjacent}

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

Pythagoras’ Theorem Formula

$a^{2}$ $a^2$

+

$+$ $b^{2}$ $b^2$

=

$=$ $c^{2}$ $c^2$

a

$a$ and

b

$b$ are the two sides, and

c

$c$ is the hypotenuse

First, draw a triangle that satisfies

\sin θ = \frac{21}{29}

$sin theta=21/29$ (the sign is disregarded in this process)

Using the triangle above, use the Pythagoras’ Theorem to solve for

x

$x$

a = 21

$a=21$

b = x

$b=x$

c = 29

$c=29$

$a^{2}$ $a^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	$c^{2}$ $c^2$	Pythagoras’ Theorem
$21^{2}$ $21^2$ $+$ $+$ $x^{2}$ $x^2$	$=$ $=$	$29^{2}$ $29^2$	Substitute values
$441 + x^{2}$ $441+x^2$	$=$ $=$	$841$ $841$
$441 + x^{2}$ $441+x^2$ $- 411$ $-411$	$=$ $=$	$841$ $841$ $- 411$ $-411$	Subtract $441$ $441$ from both sides
$x^{2}$ $x^2$	$=$ $=$	$400$ $400$
$\sqrt{x^{2}}$ $sqrt(x^2)$	$=$ $=$	$\sqrt{400}$ $sqrt400$	Get the square root of both sides
$x$ $x$	$=$ $=$	$\pm 20$ $+-20$

Recall the right triangle to find the value of

\tan θ

$tan theta$

$\tan θ$ $tan theta$	$=$ $=$	$\frac{\text{opposite}}{\text{adjacent}}$

	$=$ $=$	$\frac{21}{20}$ $21/20$

Remember that

\tan θ = \frac{\sin θ}{\cos θ}

$tan theta=(sin theta)/(cos theta)$

\sin θ = - \frac{21}{29}

$sin theta=-21/29$

\cos θ

$cos theta$

>

$>$

0

$0$

Since

\sin θ

$sin theta$ is negative and

\cos θ

$cos theta$ is positive,

\tan θ

$tan theta$ is negative

Therefore,

\tan θ = - \frac{21}{20}

$tan theta=-21/20$

\tan θ = - \frac{21}{20}

$tan theta=-21/20$

Question 2 of 3

Find

\cos θ

$cos theta$ given the following

\sin θ = - \frac{3}{5}

$sin theta=-3/5$

\frac{3 π}{2}

$(3pi)/2$

<

$<$

θ

$theta$

<

$<$

2 π

$2pi$

Write fractions in the format “a/b”

$\cos θ =$ $cos theta=$ (4/5)

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Trigonometric Functions

\sin θ = \frac{opposite}{hypotenuse}

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

\cos θ = \frac{adjacent}{hypotenuse}

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

\tan θ = \frac{opposite}{adjacent}

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

Pythagoras’ Theorem Formula

$a^{2}$ $a^2$

+

$+$ $b^{2}$ $b^2$

=

$=$ $c^{2}$ $c^2$

a

$a$ and

b

$b$ are the two sides, and

c

$c$ is the hypotenuse

First, draw a triangle that satisfies

\sin θ = \frac{3}{5}

$sin theta=3/5$ (the sign is disregarded in this process)

Using the triangle above, use the Pythagoras’ Theorem to solve for

x

$x$

a = 3

$a=3$

b = x

$b=x$

c = 5

$c=5$

$a^{2}$ $a^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	$c^{2}$ $c^2$	Pythagoras’ Theorem
$3^{2}$ $3^2$ $+$ $+$ $x^{2}$ $x^2$	$=$ $=$	$5^{2}$ $5^2$	Substitute values
$9 + x^{2}$ $9+x^2$	$=$ $=$	$25$ $25$
$9 + x^{2}$ $9+x^2$ $- 9$ $-9$	$=$ $=$	$25$ $25$ $- 9$ $-9$	Subtract $9$ $9$ from both sides
$x^{2}$ $x^2$	$=$ $=$	$16$ $16$
$\sqrt{x^{2}}$ $sqrt(x^2)$	$=$ $=$	$\sqrt{16}$ $sqrt16$	Get the square root of both sides
$x$ $x$	$=$ $=$	$\pm 4$ $+-4$

Recall the right triangle to find the value of

\cos θ

$cos theta$

$\cos θ$ $cos theta$	$=$ $=$	$\frac{\text{adjacent}}{\text{hypotenuse}}$

	$=$ $=$	$\frac{4}{5}$ $4/5$

Identify the appropriate quadrants in the unit circle to find the sign of

\cos θ

$cos theta$

\frac{3 π}{2}

$(3pi)/2$

<

$<$

θ

$theta$

<

$<$

2 π

$2pi$

In the fourth quadrant,

\cos θ

$cos theta$ is positive

Therefore,

\cos θ = \frac{4}{5}

$cos theta=4/5$

\cos θ = \frac{4}{5}

$cos theta=4/5$

Question 3 of 3

Find

\sin θ

$sin theta$ given the following

\cos θ = - \frac{5}{13}

$cos theta=-5/13$

π

$pi$

<

$<$

θ

$theta$

<

$<$

\frac{3 π}{2}

$(3pi)/2$

Write fractions in the format “a/b”

$\sin θ =$ $sin theta=$ (-12/13)

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Trigonometric Functions

\sin θ = \frac{opposite}{hypotenuse}

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$

\cos θ = \frac{adjacent}{hypotenuse}

$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$

\tan θ = \frac{opposite}{adjacent}

$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$

Pythagoras’ Theorem Formula

$a^{2}$ $a^2$

+

$+$ $b^{2}$ $b^2$

=

$=$ $c^{2}$ $c^2$

a

$a$ and

b

$b$ are the two sides, and

c

$c$ is the hypotenuse

First, draw a triangle that satisfies

\cos θ = \frac{5}{13}

$cos theta=5/13$ (the sign is disregarded in this process)

Using the triangle above, use the Pythagoras’ Theorem to solve for

x

$x$

a = y

$a=y$

b = 5

$b=5$

c = 13

$c=13$

$a^{2}$ $a^2$ $+$ $+$ $b^{2}$ $b^2$	$=$ $=$	$c^{2}$ $c^2$	Pythagoras’ Theorem
$y^{2}$ $y^2$ $+$ $+$ $5^{2}$ $5^2$	$=$ $=$	$13^{2}$ $13^2$	Substitute values
$y^{2} + 25$ $y^2+25$	$=$ $=$	$169$ $169$
$y^{2} + 25$ $y^2+25$ $- 25$ $-25$	$=$ $=$	$169$ $169$ $- 25$ $-25$	Subtract $25$ $25$ from both sides
$y^{2}$ $y^2$	$=$ $=$	$144$ $144$
$\sqrt{y^{2}}$ $sqrt(y^2)$	$=$ $=$	$\sqrt{144}$ $sqrt144$	Get the square root of both sides
$y$ $y$	$=$ $=$	$\pm 12$ $+-12$

Recall the right triangle to find the value of

\sin θ

$sin theta$

$\sin θ$ $sin theta$	$=$ $=$	$\frac{\text{opposite}}{\text{hypotenuse}}$

	$=$ $=$	$\frac{12}{13}$ $12/13$

Identify the appropriate quadrants in the unit circle to find the sign of

\cos θ

$cos theta$

π

$pi$

<

$<$

θ

$theta$

<

$<$

\frac{3 π}{2}

$(3pi)/2$

In the third quadrant,

\sin θ

$sin theta$ is negative

Therefore,

\sin θ = - \frac{12}{13}

$sin theta=-12/13$

\sin θ = - \frac{12}{13}

$sin theta=-12/13$

Unit Circle: Trig Ratios

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Quiz summary

Information

1. Question

Hint

Fantastic!

Trigonometric Functions

Pythagoras’ Theorem Formula

2. Question

Hint

Excellent!

Trigonometric Functions

Pythagoras’ Theorem Formula

3. Question

Hint

Nice Job!

Trigonometric Functions

Pythagoras’ Theorem Formula

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