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March 29, 2024, 06:24:20 pm

Author Topic: 3U Trig  (Read 5376 times)  Share 

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RuiAce

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Re: 3U Trig
« Reply #15 on: January 24, 2017, 06:10:38 pm »
0
Let me just add something here.

This sort of stuff is GENERALLY considered "Harder 3U" material for the 4U course. It is very unlikely that it should be examinable.


The more common thing to do is what ellipse said; apply a sum-to-product formula. But 3U students are, in general, without HELP, NOT expected to know the sum-to-product formulae off by heart. Only 4U students are 'expected' to memorise these procedures.

(Even then, I've only seen it appear once, which was in my trial HSC for MX2)

prabhleenkaur~

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Re: 3U Trig
« Reply #16 on: January 27, 2017, 09:47:19 pm »
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sinx + 2sinxcosx + 3sinx - 4sin3x = 0
2sinx(2 + cosx - 2sin2x) = 0
2sinx[-2(1-cos2x) + cosx + 2) = 0
sinx(2cos2x + cosx) = 0
sinxcosx(2cosx +1) = 0
therefore sinx=0, cosx=0 or cosx=-1/2

sinx = 0 = sin0
∴ x = πn

cosx = 0 = cos(π/2)
∴ x = 2πn ± π/2

cosx = -1/2 = cos(2π/3)
∴ x = 2πn ± 2π/3

Thank youuu!! Appreciate it!

prabhleenkaur~

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Re: 3U Trig
« Reply #17 on: January 27, 2017, 09:50:05 pm »
+1
Thanks kiwiberry, you are a legend!! ;D prabhleenkaur~, welcome to the forums! ;D

Glad to be a part of this ATAR Notes journey :D

armtistic

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Re: 3U Trig
« Reply #18 on: February 04, 2017, 12:28:38 pm »
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Hi
i have a similar question,
would i be able to get some help on this,

sin2x + sin3x + sin4x = 0

much appreciated thank you!


Rewrite it as

sin(3x-x) + sin3x + sin(3x+x) =0

Expand those and you get

2sin3xcosx+sin3x=0

sin3x(2cosx+1)=0

Solve sin3x=0 and 2cosx+1=0

x=kπ/3
ATAR: 99.70