Common Core Algebra 2 - 2020 Edition

10.02 Cofunction identities

Lesson

Recall that in a right triangle, the two acute angles together make a right-angle. We say the acute angles are **complementary*** *(to one another). If the two acute angles have measures $\alpha$`α` and $\beta$`β`, then $\alpha+\beta=90^\circ$`α`+`β`=90° and so, $\beta=90^\circ-\alpha$`β`=90°−`α`.

The diagram below illustrates the following relationships.

$\cos\alpha=\frac{b}{h}=\sin\beta=\sin\left(90^\circ-\alpha\right)$`c``o``s``α`=`b``h`=`s``i``n``β`=`s``i``n`(90°−`α`)

$\sin\alpha=\frac{a}{h}=\cos\beta=\cos\left(90^\circ-\alpha\right)$`s``i``n``α`=`a``h`=`c``o``s``β`=`c``o``s`(90°−`α`)

$\cot\alpha=\frac{b}{a}=\tan\beta=\tan\left(90^\circ-\alpha\right)$`c``o``t``α`=`b``a`=`t``a``n``β`=`t``a``n`(90°−`α`)

Thus, the 'co' in complementary explains the meaning of *co*sine in relation to sine, and to *co*tangent in relation to tangent.

The relationships above are **identities** because they are true whatever the value of the angle $\alpha$`α`. We can extend the relationships to reciprocal trig functions to obtain the following set of cofunction identities:

Cofunction identities

$\sin\alpha\equiv\cos\left(90^\circ-\alpha\right)$sinα≡cos(90°−α) |
$\tan\alpha\equiv\cot\left(90^\circ-\alpha\right)$tanα≡cot(90°−α) |
$\sec\alpha\equiv\csc\left(90^\circ-\alpha\right)$secα≡csc(90°−α) |

$\cos\alpha\equiv\sin\left(90^\circ-\alpha\right)$cosα≡sin(90°−α) |
$\cot\alpha\equiv\tan\left(90^\circ-\alpha\right)$cotα≡tan(90°−α) |
$\csc\alpha\equiv\sec\left(90^\circ-\alpha\right)$cscα≡sec(90°−α) |

These identities are true not only in right triangle trigonometry, but they also hold for angles of any size. This can be confirmed by thinking about the geometry in the unit circle diagram that is used for defining the trigonometric functions of angles of any magnitude.

**Simplify** the relation $\sin\left(90^\circ-\theta\right)=\sqrt{3}\sin\theta$`s``i``n`(90°−`θ`)=√3`s``i``n``θ` where $0^\circ<\theta<360^\circ$0°<`θ`<360° is an acute angle.

**Think:** Rearrange the equation so that the trigonometric functions are on one side and the coefficients are on the other.

**Do:**

$\sin\left(90^\circ-\theta\right)$sin(90°−θ) |
$=$= | $\sqrt{3}\sin\theta$√3sinθ |
Given statement |

$\cos\theta$cosθ |
$=$= | $\sqrt{3}\sin\theta$√3sinθ |
Using the cofunction relationships |

$1$1 | $=$= | $\frac{\sqrt{3}\sin\theta}{\cos\theta}$√3sinθcosθ |
Dividing both sides by $\cos\theta$ |

$1$1 | $=$= | $\sqrt{3}\tan\theta$√3tanθ |
Using the fact that $\tan\theta=\frac{\sin\theta}{\cos\theta}$ |

$\tan\theta$tanθ |
$=$= | $\frac{1}{\sqrt{3}}$1√3 |
Dividing both sides by $\sqrt{3}$√3 |

We recognize an exact value for $\tan$`t``a``n` and conclude that $\theta=30^\circ$`θ`=30° if $\theta$`θ` is acute.

**Reflect: **We should check that there is also a third quadrant solution, $\theta=210^\circ$`θ`=210°.

Simplify the following expression using complementary angles:

$\frac{\sin51^\circ}{\cos39^\circ}$`s``i``n`51°`c``o``s`39°

Rewrite $\sec\frac{\pi}{9}$`s``e``c`π9 in terms of its cofunction.

Simplify $\sin\left(90^\circ-y\right)\times\tan y$`s``i``n`(90°−`y`)×`t``a``n``y`.

Prove the pythagorean identity sin^2(Î¸) cos^2(Î¸) = 1 and use it to find sin(Î¸), cos(Î¸), or tan(Î¸) given sin(Î¸), cos(Î¸), or tan(Î¸) and the quadrant of the angle.