Introduction

Let's take a look at the isosceles right triangle whose legs are both equal to 1, as shown below.

By considering either of the 45^\circ angles in this triangle, we find that \eqalign{ \sin 45^\circ &= \dfrac{\textrm{opp.}}{\textrm{hyp.}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}, \\[5pt] \cos 45^\circ &= \dfrac{\textrm{adj.}}{\textrm{hyp.}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}, \\[5pt] \tan 45^\circ &= \dfrac{\textrm{opp.}}{\textrm{adj.}} = \dfrac{1}{1} = 1. }

Isosceles right triangles are very common, so the values above occur a lot in trigonometry. It's a good idea to remember them.

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Example: Evaluating Trigonometric Expressions With 45 Degree Angles

Evaluate 2 \tan{45^\circ} - 4 \sin{45^\circ}.

EXPLANATION

The angle 45^\circ is special, and its trigonometric ratios have the values

\sin{45^\circ} = \dfrac{\sqrt 2}{2}, \qquad \tan{45^\circ} = 1.

Therefore,

\begin{align*} 2\tan{45^\circ} - 4 \sin{45^\circ} & = \\[5pt] 2(1) - 4\left(\dfrac{\sqrt 2}{2}\right) & = \\[5pt] 2 - 2\sqrt{2} & = \\[5pt] 2(1 - \sqrt{2}) &. \end{align*}

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Practice: Evaluating Trigonometric Expressions With 45 Degree Angles

$\sin{45^\circ} - \cos{45^\circ} = $

a
 $2\sqrt{2}$
b
 $5$
c
 $0$
d
 $1$
e
 $3\sqrt{2}$
Practice: Evaluating Trigonometric Expressions With 45 Degree Angles

$\csc{C} =$

a
b
c
d
e
Practice: Evaluating Trigonometric Expressions With 45 Degree Angles

$7\sin^2{45^\circ} = $

a
 $\dfrac 7 2$
b
 $7$
c
 $\dfrac {63} 4$
d
 $\dfrac {63} 2$
e
 $\dfrac 7 4$
Trigonometric Ratios for 30-60-90 Triangles

Let's now consider the 30 - 60 - 90 triangle whose shortest leg has a length equal to 1, shown below.

Now, we can calculate some trigonometric ratios without a calculator. Let's start with the 30^\circ -angle. By the definition of sine, cosine, and tangent, we have \begin{align*} \sin 30^\circ &= \dfrac{\textrm{opp.}}{\textrm{hyp.}} = \dfrac{1}{2}, \\[5pt] \cos 30^\circ &= \dfrac{\textrm{adj.}}{\textrm{hyp.}} = \dfrac{\sqrt{3}}{2}, \\[5pt] \tan 30^\circ &= \dfrac{\textrm{opp.}}{\textrm{adj.}} = \dfrac{1}{\sqrt{3}}= \dfrac{\sqrt{3}}{3}. \end{align*}

Similarly, by considering the 60^\circ -angle, we find that

\begin{align*} \sin 60^\circ &= \dfrac{\textrm{opp.}}{\textrm{hyp.}} = \dfrac{\sqrt 3}{2}, \\[5pt] \cos 60^\circ &= \dfrac{\textrm{adj.}}{\textrm{hyp.}} = \dfrac{1}{2}, \\[5pt] \tan 60^\circ &= \dfrac{\textrm{opp.}}{\textrm{adj.}} = \dfrac{\sqrt 3}{1} = \sqrt 3. \end{align*}

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Example: Evaluating Trigonometric Expressions With 30 and 60 Degree Angles

Evaluate \sin{30^\circ} + \cos{60^\circ} + \tan^2{30^\circ}.

EXPLANATION

The angles 30^\circ and 60^\circ are special angles, and their trigonometric ratios have the particular values

\sin{30^\circ} = \dfrac{1}{2}, \qquad \cos 60^\circ = \dfrac{1}{2} , \qquad \tan{30^\circ} = \dfrac{\sqrt{3}}{3}.

Also, note that \tan^2{30^\circ} simply means (\tan{30^\circ})^2.

Therefore,

\begin{align*} \sin{30^\circ} + \cos{60^\circ} + \tan^2{30^\circ} &= \\[5pt] \sin{30^\circ} + \cos{60^\circ} + \Bigl(\tan{30^\circ}\Bigr)^2 &= \\[5pt] \dfrac{1}{2} + \dfrac{1}{2} + \left(\dfrac{\sqrt{3}}{3}\right)^2 &= \\[5pt] 1 + \dfrac{3}{9} &= \\[5pt] 1 + \dfrac{1}{3} &= \\[5pt] \dfrac{4}{3} &. \end{align*}

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Practice: Evaluating Trigonometric Expressions With 30 and 60 Degree Angles

$\sin {30^\circ}+\cos {60^\circ}=$

a
 $2$
b
 $1$
c
 $3$
d
 $\sqrt{2}$
e
 $0$
Practice: Evaluating Trigonometric Expressions With 30 and 60 Degree Angles

$\cot A = $

a
b
c
d
e
Practice: Evaluating Trigonometric Expressions With 30 and 60 Degree Angles

$\cos^2{60^\circ} = $

a
 $2$
b
 $1$
c
 $\dfrac 1 2$
d
 $\dfrac 1 4$
e
 $\dfrac 3 4$
Special Values of the Zero Degree Angle

The values of the trigonometric ratios for 0^\circ angles are the following:





Trigonometric ratio Value
\sin \left(0^\circ\right) 0
\cos \left(0^\circ\right) 1
\tan \left(0^\circ\right) 0


Why is this so? If we look at the right triangle below, we can imagine making the angle \theta smaller by moving the vertex C towards A.

As the angle \theta gets closer and closer to zero, the opposite side AC also approaches zero. Also, the hypotenuse BC and the adjacent side AB become nearly the same length.

So, as \theta approaches zero (which we can write as \theta \rightarrow 0^\circ ) we have

AC \approx 0 \quad \text{and} \quad BC \approx AB.

Therefore, when \theta = 0^\circ we have:

\begin{align} \sin \left(0^\circ\right) &= \dfrac {AC} {BC} = 0, \\[5pt] \cos \left(0^\circ\right) &= \dfrac {AB} {BC} = 1, \\[5pt] \tan \left(0^\circ\right) &= \dfrac {AC} {AB} = 0. \end{align}

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Example: Evaluating Trigonometric Expressions With 0 Degree Angles

Find \sin {0^\circ} + 4 \cos {0^\circ}.

EXPLANATION

The angle 0^\circ is special, and its trigonometric ratios have the particular values

\sin{0^\circ} = 0, \qquad \cos{0^\circ} = 1.

Therefore,

\begin{align} \sin {0^\circ} + 4 \cos {0^\circ} & = \\ 0 + 4 (1) &= \\ 4. \end{align}

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Practice: Evaluating Trigonometric Expressions With 0 Degree Angles

$\tan{0^\circ} + \tan{60^\circ} =$

a
b
c
d
e
Practice: Evaluating Trigonometric Expressions With 0 Degree Angles

$\cos 0^{\circ}+\cos 30^{\circ}+\sin 0^{\circ}=$

a
 $\dfrac{3+\sqrt{3}}{7}$
b
 $\dfrac{3+\sqrt{3}}{2}$
c
 $\dfrac{2+\sqrt{3}}{5}$
d
 $\dfrac{2+\sqrt{3}}{2}$
e
 $\dfrac{2+\sqrt{5}}{2}$
Special Values of the 90 Degree Angle

The values of the trigonometric ratios for 90^\circ angles are the following:





Trigonometric ratio Value
\sin \left(90^\circ\right) 1
\cos \left(90^\circ\right) 0
\tan \left(90^\circ\right) Undefined


To understand where the values come from, this time, we can imagine increasing the value of the angle \theta by moving the vertex B towards A.

As B moves towards A, the adjacent side AB gets very close to zero, and the angle \theta gets very close to 90^\circ. Also, the hypotenuse BC and the opposite side AC become nearly the same length.

So, as \theta approaches 90^\circ (which we can write as \theta \rightarrow 90^\circ ) we have

AB \approx 0 \quad \text{and} \quad BC \approx AC.

Therefore, when \theta = 90^\circ, we have:

\begin{align} \sin \left(90^\circ\right) &= \dfrac {AC} {BC} = 1, \\ \cos \left(90^\circ\right) &= \dfrac {AB} {BC} = 0, \\ \tan \left(90^\circ\right) &= \dfrac {AC} {AB} = \dfrac{AC}{0}= \text{undefined.} \\ \end{align}

Note that when calculating the tangent, we get a 0 in the denominator. We know that we can't divide by zero. Therefore, the tangent of a right angle is said to be undefined.

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Example: Evaluating Trigonometric Expressions With 90 Degree Angles

Evaluate \cos {90^\circ} - 2 \sin {90^\circ}.

EXPLANATION

The angle 90^\circ is special, and its trigonometric ratios have the particular values

\cos{90^\circ} = 0, \qquad \sin{90^\circ} = 1.

Therefore,

\begin{align} \cos {90^\circ} - 2 \sin {90^\circ} & = \\ 0 - 2 (1) &= \\ -2. \end{align}

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Practice: Evaluating Trigonometric Expressions With 90 Degree Angles

$\cos{60^\circ} + \sin{90^\circ}=$

a
b
c
d
e
Practice: Evaluating Trigonometric Expressions With 90 Degree Angles

$\sin 45^{\circ} +\sin 90^{\circ}+ \sin 30^{\circ}=$

a
 $\dfrac{\sqrt{2}+3}{5}$
b
 $\dfrac{\sqrt{3}+5}{2}$
c
 $\dfrac{\sqrt{5}+3}{2}$
d
 $\dfrac{\sqrt{3}+2}{2}$
e
 $\dfrac{\sqrt{2}+3}{2}$
Practice: Evaluating Trigonometric Expressions With 90 Degree Angles

$\tan^2 45^{\circ}-\cos^230^{\circ}+\sin^290^{\circ}=$

a
 $\dfrac{3}{4}$
b
 $\dfrac{5}{7}$
c
 $\dfrac{5}{3}$
d
 $\dfrac{4}{5}$
e
 $\dfrac{5}{4}$
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