August 20, 2019, 12:58:51 pm

### AuthorTopic: Maths problem - from 2006 vce exam 1  (Read 2480 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### azhtey

• Victorian
• Trendsetter
• Posts: 169
• Respect: +1
##### Maths problem - from 2006 vce exam 1
« on: October 20, 2007, 12:24:24 pm »
0
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.
"BROWS"

• Victorian
• Part of the furniture
• Posts: 1296
• *dreamy sigh*
• Respect: +15
##### Maths problem - from 2006 vce exam 1
« Reply #1 on: October 20, 2007, 12:27:12 pm »
0
f(0) = 5 cos( 2/3 * pi ) = -5 cos (pi/3) = -5 * 1/2 = -5/2
Mandark: Please, oh please, set me up on a date with that golden-haired angel who graces our undeserving school with her infinite beauty!

The collage of ideas. The music of reason. The poetry of thought. The canvas of logic.

$-=\mathbb{A}\mathfrak{hmad Issa}=-$

#### azhtey

• Victorian
• Trendsetter
• Posts: 169
• Respect: +1
##### Maths problem - from 2006 vce exam 1
« Reply #2 on: October 20, 2007, 12:46:52 pm »
0
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??
"BROWS"

#### rustic_metal

• Guest
##### Maths problem - from 2006 vce exam 1
« Reply #3 on: October 20, 2007, 12:48:26 pm »
0

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

• VCE Tutor
• Victorian
• ATAR Notes Legend
• Posts: 4966
• Respect: +17
##### Maths problem - from 2006 vce exam 1
« Reply #4 on: October 20, 2007, 07:10:28 pm »
0
5cos(2pi/3) = 5cos(pi - pi/3) = -5cos(pi/3)