ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: azhtey on October 20, 2007, 12:24:24 pm

Title: Maths problem - from 2006 vce exam 1
Post by: azhtey on October 20, 2007, 12:24:24 pm

f(x) = 5cos(2(x + pie/3))

The answer says the y-intercept is -2.5 but i don't know how they got that.

I bet you its some stupid thing im doing wrong. Can someone show me how they got -2.5 when x=0. Thanks.

Title: Maths problem - from 2006 vce exam 1
Post by: Ahmad on October 20, 2007, 12:27:12 pm

f(0) = 5 cos( 2/3 * pi ) = -5 cos (pi/3) = -5 * 1/2 = -5/2 :)

Title: Maths problem - from 2006 vce exam 1
Post by: azhtey on October 20, 2007, 12:46:52 pm

i still don't get it. Im sorry but im not very good at methods (maths isnt my thing).

I have no idea where you got the negative from. Is there a way to go through it step by step for a dumbass like me lol??

Title: Maths problem - from 2006 vce exam 1
Post by: rustic_metal on October 20, 2007, 12:48:26 pm

from what ahmad said:

f(0) = 5 cos( 2/3 * pi )             ----- sub in x=0

= -5 cos (pi/3)                        -----pi/3 is reference angle in 1st quadrant, therefore is negative in second (hence -5)

= -5 * 1/2 = -5/2                       ----cos pi/3 = 1/2

Title: Maths problem - from 2006 vce exam 1
Post by: Collin Li on October 20, 2007, 07:10:28 pm

5cos(2pi/3) = 5cos(pi - pi/3) = -5cos(pi/3)

(pi - pi/3) is your quadrant 2 angle for "pi/3"