Section 1.2 Trigonometric Ratios

Objectives: 1. To state and apply the Pythagorean theorem. 2. To define the six trigonometric ratios.

Pythagorean Theorem In right ABC, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. a2 + b2 = c2 B A C

Trigonometric Ratios leg opposite A sine of A = hypotenuse leg adjacent A cosine of A = leg adjacent A leg opposite A tangent of A =

SOHCAHTOA ine pposite ypotenuse osine djacent ypotenuse angent pposite

Trigonometric Ratios h o A sin = h a A cos = a o A tan = hypotenuse opposite adjacent A h o A sin = h a A cos = a o A tan =

Reciprocal Ratios 1 cosecant of A = cscA = sinA 1 cosA 1 secant of A = secA = tanA 1 cotangent of A = cotA =

Reciprocal Ratios o h A csc = a h A sec = o a A cot = A opposite adjacent hypotenuse o h A csc = a h A sec = o a A cot =

EXAMPLE 1 Find the six trigonometric ratios for G in right EFG. g2 + e2 = f2 g2 + 62 = 82 g2 + 36 = 64 g2 = 28 g = 2 7 8 6 E F

Practice Question: Find the six trigonometric ratios for E in right EFG. 9 11 G E F sin E = 1. 2. 3. 4. 9 11 2 10 11 9 10 20 2 10 9

P(x,y) r y  x

Trigonometric Ratios r y hyp opp sin = q y r opp hyp csc = q r x hyp adj cos = q x r adj hyp sec = q x y adj opp tan = q y x opp adj cot = q

EXAMPLE 2 Find the six trigonometric ratios for a 90º angle. x=0, y=1, r=1 cos  = 0 sin  = 1 tan  = und. 90° P 

EXAMPLE 2 Find the six trigonometric ratios for a 90º angle. x=0, y=1, r=1 sec  = und. csc  = 1 cot  = 0 90° P 

Practice Question: Find the six trigonometric ratios for a 180º angle. sin  = 1. -1 2. 0 3. 1 4. und. 180°  P

Special Triangles a2 + b2 = c2 12 + 12 = c2 1 + 1 = c2 c2 = 2 45 1 2 c

Special Triangles 2 ÷ ø ö ç è æ 2 1 sin 45° = 2 c = 45 1

Special Triangles 2 sin 45° = 2 c = 45 1 2 cos 45° = tan 45° = 1

Special Triangles csc 45° = 2 2 c = 45 1 sec 45° = 2 cot 45° = 1

2 sin 45° = cos 45° = tan 45° = 1 csc 45° = sec 45° = cot 45° =

Special Triangles a2 + b2 = c2 12 + b2 = 22 1 + b2 = 4 b2 = 3 60 2 1 30 3 b =

Special Triangles 3 ÷ ø ö ç è æ 3 1 60 2 1 30 tan 30° = 3 b =

Special Triangles 60 2 1 30 3 b = tan 30° = 3 1 sin 30° = 2 3 cos 30° =

Special Triangles 60 2 1 30 3 b = cot 30° = 3 csc 30° = 2 3 2 sec 30° =

3 tan 30° = sin 30° = cos 30° = 2 1 cot 30° = csc 30° = sec 30° =

Special Triangles 60 2 1 30 3 b = 2 3 sin 60° = 2 1 cos 60° = tan 60° = 3

Special Triangles 60 2 1 30 3 b = 3 2 csc 60° = sec 60° = 2 3 cot 60° =

sin 60° = cos 60° = tan 60° = 2 3 1 csc 60° = sec 60° = cot 60° =

Homework: pp. 12-13

►A. Exercises 1. Find the six trig. ratios for both acute angles in each triangle. A 5 2 B C 21

1st-solve for side n n2 = 122-22 = 144-4 = 140 22 + n2 = 122 35 2 140 ►A. Exercises 3. Find the six trig. ratios for both acute angles in each triangle. n M L 2 12 N 1st-solve for side n n2 = 122-22 = 144-4 = 140 22 + n2 = 122 35 2 140 n =

#3. 35 2 M L 2 12 N

►A. Exercises 7. Find the six trig. functions for an angle in standard pos. whose terminal ray passes through the point (-6, -1). -6  -1 37

-6 -1 37  sin  = csc  = cos  = sec  = tan  = cot  =

►B. Exercises 15. Find the six trig. ratios for the quadrantal angle measuring 180°.

►B. Exercises 15. Find the six trig. ratios for the quadrantal angle measuring 180°. (-1, 0)

■ Cumulative Review 30. Give the distance between (2, 7) and (-3, -1).

■ Cumulative Review 31. Give the midpoint of the segment joining (2, 7) and (-3, -1).

■ Cumulative Review 32. Give the angle  coterminal with 835 if 0    360.

■ Cumulative Review 33. Convert 88 to radians.

■ Cumulative Review 34. If sec  = 7, find cos .