4.3ValuesofTrigonometricFunctions.notebook. January 27, 2019

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1 St. Peter Catholic High School Algonquin College MCT4C Mr. M. Couturier Jan 27 10:05 AM 4.3 Values of Trigonometric Functions In this class, we will look at special angle measurements notably, the 0 o, 30 o,, 60 o, 90 o triangles. By the end of this class, you should know how to derive: sin 0 o = cos 0 o = tan 0 o = sin 30 o = cos 30 o = tan 30 o = sin = cos = tan = sin 60 o = cos 60 o = tan 60 o = sin 90 o = cos 90 o = tan 90 o = Jan 27 10:04 AM Let's start with an equilateral triangle of length : 60 o Now let's cut it in half! 30 o 3 cm 60 o 90 o 60 o 60 o Let's find the height of this triangle using Pythagorean Theorem: a = (c 2 - b 2 ) a = ( ) a = 3 Feb 10 9:07 AM 1

2 Now let's use SOH CAH TOA to determine our trigonometric ratios: 30 o sin 30 o = ½ 2 cos 30 o = 3/2 tan 30 o = 3/3* 3 sin 60 o = 3/2 cos 60 o = ½ tan 60 o = 3 60 o 90 o 1 * tan 30 o is written in this way since a radical cannot appear as a denominator Jan 27 10:08 AM Let's start with a right triangle of length : Let's find the hypotenuse of this triangle using Pythagorean Theorem: c = (a 2 + b 2 ) c = ( ) c = 2 90 o Jan 27 10:48 AM Now let's use SOH CAH TOA to determine our trigonometric ratios: sin = 2/2* cos = 2/2* tan = 1 90 o * sin and cos are written in this way since a radical cannot appear as a denominator Jan 27 10:45 AM 2

3 Using your calculators: First let's make sure that you are in the proper mode. Calculators can be in three modes for trignometry: degree (DEG), radians (RAD) and gradians (GRAD). Look at the display on your calculator. What does it say? We are currently using degrees, therefore make sure there is a D or a DEG written on the display. If not, find a bottom that reads DRG and push it until you get Degrees. Otherwise, you may have to go into your mode or settings to change it. Test: Find the cos(23.93 o ): Answer in DEG mode: Jan 27 10:48 AM When given the angle, simply type in the angle into the trigonometric operator. You are finding the RATIO! sin(23.3 o )= cos(12.9 o )= tan(81.6 o )= Feb 1 12:17 PM sin(23.3 o )= cos(12.9 o )= tan(81.6 o )= Jan 27 11:51 AM 3

4 When given the ratio, take the inverse function of the trigonometric operator. You are finding the ANGLE! sin θ = cos θ = tan θ = Jan 27 11:11 AM sin θ = θ = sin -1 (0.5632) θ = 34.3 o cos θ = θ = cos -1 (0.1298) θ = 82.5 o tan θ = (2.1334) θ = tan -1 (2.1334) θ = 64.9 o Jan 27 11:12 AM When given the angle for a reciprocal function, simply convert the reciprocal function into its reciprocal function and type in the angle. You are finding the RATIO! csc(23.3 o )= sec(12.9 o )= cot(81.6 o )= Feb 1 10:02 AM 4

5 csc(23.3 o )= 1/sin(23.3 o ) csc(23.3 o )= 1/ csc(23.3 o )= sec(12.9 o )= 1/cos(12.9 o ) sec(12.9 o )= 1/ sec(12.9 o )= cot(81.6 o )= 1/tan(81.6 o ) cot(81.6 o )= 1/ cot(81.6 o )= Feb 6 9:07 AM When given the ratio of an inverse trigonometric function, first convert to the reciprocal and then, take the inverse function of the trigonometric operator. You are again finding the ANGLE! csc θ = sec θ = Feb 1 10:04 AM csc θ = sin θ = 1/csc θ sin θ = 1/ θ = sin -1 (1/1.5632) θ = o sec θ = cos θ = 1/sec θ cos θ = 1/ θ = cos -1 (1/1.1298) θ = o Feb 1 10:07 AM 5

6 Example: Given sec θ = , find tanθ. Feb 1 10:11 AM Solution: Given sec θ = , find tanθ. Find θ first: sec θ = cos θ = 1/sec θ cos θ = 1/ θ = cos -1 ( ) θ = o Now plug θ into tanθ. tanθ = tan(27.73 o ) tan(27.73 o ) = Feb 1 10:15 AM When a rocket is launched, its horizontal velocity v x, is related to the velocity v with which it is fired by the equation v x = vcosθ. Here, θ is the angle between the horizontal and the direction in which it is fired. Find v x if v = 1250m/s and θ = 36.0 o. Feb 10 11:21 AM 6

7 Solution: v x = vcosθ v x = (1250m/s)cos(36.0 o ) v x = 1010m/s The horizontal velocity is 1010m/s. Feb 10 11:44 AM Exit Ticket: In class/group activity: Complete & hand-in. 4.3 #58 Homework: 4.3 #9-39 (every other odd) Feb 1 12:10 PM 7