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May 25, 2019, 11:05:57 am

Author Topic: Maths problem - from 2006 vce exam 1  (Read 2364 times)  Share 

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azhtey

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Maths problem - from 2006 vce exam 1
« on: October 20, 2007, 12:24:24 pm »
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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.
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Ahmad

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Maths problem - from 2006 vce exam 1
« Reply #1 on: October 20, 2007, 12:27:12 pm »
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f(0) = 5 cos( 2/3 * pi ) = -5 cos (pi/3) = -5 * 1/2 = -5/2 :)
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azhtey

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Maths problem - from 2006 vce exam 1
« Reply #2 on: October 20, 2007, 12:46:52 pm »
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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??
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rustic_metal

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Maths problem - from 2006 vce exam 1
« Reply #3 on: October 20, 2007, 12:48:26 pm »
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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

Collin Li

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Maths problem - from 2006 vce exam 1
« Reply #4 on: October 20, 2007, 07:10:28 pm »
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5cos(2pi/3) = 5cos(pi - pi/3) = -5cos(pi/3)

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