8. Triangular & Circular Functions

Common Core Math 3 - 2020 Edition

8.01 Trigonometric functions as right triangle ratios

Lesson

Trigonometric ratios describe the relationships between the different sides of a right triangle. These sides are described in terms of where they are compared to one of the two acute angles.

So a trigonometric ratio tells us which two sides are in a given ratio, as well as the angle the ratio is being measured from.

In the triangle below we’ve chosen one of the acute angles, labeled $x$`x`. The sides of the triangle can then be referred to as the side **opposite** $x$`x`, the side **adjacent** to $x$`x` and the **hypotenuse**.

Three trigonometric ratios are:

The sine ratio: $\sin\left(x\right)=\frac{\text{Opposite}}{\text{Hypotenuse}}$`s``i``n`(`x`)=OppositeHypotenuse

The cosine ratio: $\cos\left(x\right)=\frac{\text{Adjacent}}{\text{Hypotenuse}}$`c``o``s`(`x`)=AdjacentHypotenuse

The tangent ratio: $\tan\left(x\right)=\frac{\text{Opposite}}{\text{Adjacent}}$`t``a``n`(`x`)=OppositeAdjacent

There are also what we call the reciprocal trigonometric ratios, each of which are a reciprocal of one of the three ratios above.

The cosecant ratio: $\csc\left(x\right)=\frac{1}{\sin\left(x\right)}=\frac{\text{Hypotenuse}}{\text{Opposite}}$`c``s``c`(`x`)=1`s``i``n`(`x`)=HypotenuseOpposite

The secant ratio: $\sec\left(x\right)=\frac{1}{\cos\left(x\right)}=\frac{\text{Hypotenuse}}{\text{Adjacent}}$`s``e``c`(`x`)=1`c``o``s`(`x`)=HypotenuseAdjacent

The cotangent ratio: $\cot\left(x\right)=\frac{1}{\tan\left(x\right)}=\frac{\text{Adjacent}}{\text{Opposite}}$`c``o``t`(`x`)=1`t``a``n`(`x`)=AdjacentOpposite

Notice similar triangles will have the same the trigonometric ratio of a particular angle, because enlarging or shrinking a triangle doesn't change the size of its angles or the ratio of its sides.

For example, if $\tan x=\frac{2}{1}$`t``a``n``x`=21 we can tell that the side opposite to $x$`x` is twice as big as the side adjacent to $x$`x`. The sides might have lengths $6$6 and $3$3, not necessarily $2$2 and $1$1.

Even though we don’t know the exact length of the sides just from knowing these ratios, we can still use one ratio to calculate another ratio.

Given one trigonometric ratio, we can construct a right triangle that has an angle and sides that match the information described by the ratio.

We can then use the Pythagorean Theorem to find the length of the other side.

Once we have all three side lengths, we can find any other trigonometric ratio of the acute angle by using the appropriate formula.

A right triangle has $\sin\theta=\frac{13}{85}$`s``i``n``θ`=1385, find the value of $\csc\theta$`c``s``c``θ` and $\sec\theta$`s``e``c``θ`.

**Think:** If $\sin\theta=\frac{13}{85}$`s``i``n``θ`=1385, then this means we can make the length of the hypotenuse is $85$85 and the length of the opposite side $13$13. We need to find the length of the adjacent side for $\sec\theta$`s``e``c``θ`.

**Do:**

Finding $\csc\theta$`c``s``c``θ`:

$\csc\theta$cscθ |
$=$= | $\frac{1}{\sin\theta}$1sinθ |

$=$= | $\frac{1}{\frac{13}{85}}$11385 | |

$=$= | $\frac{85}{13}$8513 |

Finding $\sec\theta$`s``e``c``θ`, by first finding the adjacent side:

$a^2+b^2$a2+b2 |
$=$= | $c^2$c2 |
State the Pythagorean Theorem |

$a^2+13^2$a2+132 |
$=$= | $85^2$852 |
Fill in the given information |

$a^2+169$a2+169 |
$=$= | $7225$7225 |
Evaluate the squares |

$a^2$a2 |
$=$= | $7056$7056 |
Subtract $169$169 from both sides |

$a$a |
$=$= | $\sqrt{7056}$√7056 |
Take the square root of both sides |

$a$a |
$=$= | $84$84 |
Simplify the square root |

We get the triangle below

Now to find $\sec\theta$`s``e``c``θ`

$\sec\theta$secθ |
$=$= | $\frac{1}{\cos\theta}$1cosθ |

$=$= | $\frac{\text{hypotenuse }}{\text{adjacent }}$hypotenuse adjacent | |

$=$= | $\frac{85}{84}$8584 |

Evaluate $\cos\theta$`c``o``s``θ` within $\triangle ABC$△`A``B``C`.

Consider the following triangle.

Find the value of $x$

`x`.Hence find the value of $\sin\theta$

`s``i``n``θ`.Hence find the value of $\cos\theta$

`c``o``s``θ`.

Consider the triangle below. Given that $\sin x=\frac{4}{5}$`s``i``n``x`=45 , find the value of $\csc x$`c``s``c``x`.

Consider the triangle below.

Calculate the length of the missing side.

Find each of the following trigonometric ratios.

$\sin x$ `s``i``n``x`$=$= $\editable{}$ $\cos x$ `c``o``s``x`$=$= $\editable{}$ $\tan x$ `t``a``n``x`$=$= $\editable{}$ $\sec x$ `s``e``c``x`$=$= $\editable{}$ $\operatorname{cosec}x$ `c``o``s``e``c``x`$=$= $\editable{}$ $\cot x$ `c``o``t``x`$=$= $\editable{}$