Predict the shape of the graph of y = 1 2 sin x . Graph the function on your GDC to check your prediction.

The graph of y = sin x is If the given function is f x and the new function is 1 a f x , the points c , d in graph of f x becomes c , 1 a d in the new graph of 1 a f x Therefore, the coordinate point π / 2 , 1 in the graph of y = sin x changes to π / 2 , 0 . 5 in the graph of y = 1 2 sin x Now, the graph of y = 1 2 sin x is

Find the amplitude of the graph of y = 1 2 sin x

The standard form of the sine function is y = a sin ( b x + c ) + d , where amplitude = a Given function is y = 1 2 sin x Hence, amplitude of the graph of y = 1 2 sin x is 1 2

Draw the graph of y = 2 cos x and y = 1 2 cos x

The graph of y = 2 cos x is y 2 cos x - 2 - 0 . 83229367 - 1 1 . 0806046 0 2 1 1 . 0806046 2 - 0 . 83229367 Now, the graph of y = 1 2 cos x is y 1 2 cos x - 2 - 0 . 20807342 - 1 0 . 27015115 0 0 . 5 1 0 . 27015115 2 - 0 . 20807342

Find the amplitude of the graph of y = 1 2 cos x

The standard form of the sine function is y = a cos ( b x + c ) + d , where amplitude = a Now, the graph of y = 1 2 cos x is y 1 2 cos x - 2 - 0 . 20807342 - 1 0 . 27015115 0 0 . 5 1 0 . 27015115 2 - 0 . 20807342 Hence, the amplitude of y = 1 2 cos x is 1 2

E M B I B E

Extended MYP Mathematics A Concept based approach Years 4 & 5>How Do They Measure Up?>Chapter 8 - Trigonometric Functions>Exploration 2

Rose Harrison, Clara Huizink, Aidan Sproat Clements and, Marlene Torres Skoumal Solutions for Exercise 5: Exploration 2

Author:Rose Harrison, Clara Huizink, Aidan Sproat Clements & Marlene Torres Skoumal

Rose Harrison Mathematics Solutions for Exercise - Rose Harrison, Clara Huizink, Aidan Sproat Clements and, Marlene Torres Skoumal Solutions for Exercise 5: Exploration 2

Attempt the free practice questions from Exercise 5: Exploration 2 with hints and solutions to strengthen your understanding. Extended MYP Mathematics A Concept based approach Years 4 & 5 solutions are prepared by Experienced Embibe Experts.