- First quadrant (A)⇒ Angles = 0
^{o}– 90^{o}

⇒ All = all values are positive. - Second quadrant (S)⇒ Angles = 90
^{o}< θ ≤ 180^{o}

⇒ Sin = the positive values only for sine and cosecant. - Third Quadrant (T)⇒ Angles = 180
^{o}< θ ≤ 270^{o}

⇒ Tan = the positive values only for tangent and cotangent. - Fourth quadrant (C)⇒ Angles = 270
^{o}< θ ≤ 360^{o}

⇒ Cos = the positive values only for cosine and secant.

From the picture above, we can clearly see the relationship between sine and cosecant, cosine and secant, tangent and cotangent. Since they are in reverse relationship, then we just need to memorize the values of sine, cosine, and tangent. The values of cosecant, secant, and cotangent can be calculated by their related trigonometry. For example, we can calculate the values of csc 45^{o} by using the values of sin 45^{o}.

### Table for Trigonomeric Values of Special Angles

From the table below, we can see that the value of sines are started from 0 to 1 to 0 to 1 to 0. And the opposite, the values of cosine are started from 1 to 0 to 1 to 0 to 1. The pattern must be 0−½−½√2−½√3−1−½√3−½√2−½−0.

– | ^{o} | 30^{o} | 45^{o} | 60^{o} | 90^{o} | 120^{o} | 135^{o} | 150^{o} | 180^{o} |

sin | ½ | ½√2 | ½√3 | 1 | ½√3 | ½√2 | ½ | ||

cos | 1 | ½√3 | ½√2 | ½ | -½ | -½√2 | -½√3 | -1 | |

tan | 1/3√3 | 1 | √3 | – | -√3 | -1 | -1/3√3 |

– | 210^{o} | 225^{o} | 240^{o} | 270^{o} | 300^{o} | 315^{o} | 330^{o} | 360^{o} |

sin | -½ | -½√2 | -½√3 | -1 | -½√3 | -½√2 | -½ | |

cos | -½√3 | -½√2 | -½ | ½ | ½√2 | ½√3 | 1 | |

tan | 1/3√3 | 1 | √3 | – | -√3 | -1 | -1/3√3 |

^{o}– 90

^{o}). For the other angles, we can use the pattern or use trigonometric identities. For more detail, see the following example :

We suppose that you have memorized the trigonometric values for the angles of (0o – 90o). Then you’re asked for calculate sin 150o and cos 135o. For answer this question, there are two methode :

- You must remember the table above and the pattern.If you dont not completely memorize the values, you can create auxiliary table like the tabel below.
^{o}30 ^{o}45 ^{o}60 ^{o}90 ^{o}120 ^{o}135 ^{o}150 ^{o}Next, you should memorize the pattern of trigonometric values.

⇒ For sine = 0 − ½ − ½√2 − ½√3 − 1 − ½√3 − ½√2 − ½ − 0.

⇒ For cosine = 1 − ½√3 − ½√2 − ½ − 0 − ½ − ½√2 − ½√3 − 1.– ^{o}30 ^{o}45 ^{o}60 ^{o}90 ^{o}120 ^{o}135 ^{o}150 ^{o}sin ½ ½√2 ½√3 1 ½√3 ½√2 ½ cos 1 ½√3 ½√2 ½ -½ -½√2 -½√3 Well, according to the table we made, we can see that :

sin 150^{o}= ½

cos 135^{o}= -½√2The first try may seem still complicated, but trust me if you are already familiar with the pattern then you will immediately know the value without having to make the auxiliary table.

- You should understand the concept of related angles.⇒ For angles of (90 ± a) and (270 ± a) remember this :

sin = cos, cos = sin, tan = cot, cot = tan, sec = cosec, cosec = sec ;

The mark (positive and negative) is adjusted according ASTC.

⇒ For ngles of (180 ± a) and (360 ± a) remember this :

sin = sin, cos = cos, tan = tan, cot = cot, sec = sec, cosec = cosec ;

The mark (positive and negative) is adjusted according ASTC.Now, back to the problem.

sin 150^{o}= sin (90 + 60)

⇒ sin 150^{o}= cos 60

⇒ sin 150^{o}= ½Note :

The angle of 150^{o }is in second quadrant (sine and cosecant are positive), then sin 150^{o}has positive value.

“sin” changes into “cos” because we use the formula for angles of (90 + a).cos 135

^{o}= cos (180 – 45)

⇒ cos 135^{o}= – cos 45

⇒ cos 135^{o}= -½√2.Note :

The angle of 135^{o }is in second quadrant (sine and cosecant are positive), then cos 135^{o}has negative value.

“cos” still be “cos” because we use the formula for angles of (180 – a).For alternative solutions, we can use the formula for angles of (90 + a) to answer the second problem.

cos 135^{o}= cos (90 + 45)

⇒ cos 135^{o}= – sin 45

⇒ cos 135^{o}= -½√2.Note :

the angle of 135^{o }is in second quadrant (sine and cosecant are positive), then cos 135^{o}has negative value. “cos” changes into “sin” because we use the formula for angles of (90 + a).