#1
Need help in maths. I got given three sets of questions for homework, if anyone can figure out one of each to get me on my way (I've forgotten some basics) with full explanations I should be okay and be very happy )

1. For 0 < Θ < 360, solve

SinΘ(2SinΘ-1)=0

2. For -180 < Θ < 180, solve

(2sinΘ-1)(3sinΘ-1)=0

3. For 0 < Θ < 2pi, solve

8cos^3Θ=1
#2
Quote by Craigo
Need help in maths. I got given three sets of questions for homework, if anyone can figure out one of each to get me on my way (I've forgotten some basics) with full explanations I should be okay and be very happy )

1. For 0 < Θ < 360, solve

SinΘ(2SinΘ-1)=0

2. For -180 < Θ < 180, solve

(2sinΘ-1)(3sinΘ-1)=0

3. For 0 < Θ < 2pi, solve

8cos^3Θ=1



I can't help you, but I have a question, who actually uses these formulas irl?
#3
i'll do the first one

EDIT: hope i helped, ive got a NAB on this tomorrow

1. For 0 < Θ < 360, solve

SinΘ(2SinΘ-1)=0


SinΘ = 0	or	2SinΘ-1 = 0		find the 0's

		SinΘ	= 1/2		angle cant = 0 so 1/2 is only option

Sin +ve =>  1st or 2nd quadrant

=> Θ = 30	or	180 - 30
=> Θ = 30	or	150

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Last edited by EMGs_rule at Feb 24, 2008,
#4
Quote by Guitarfreak777
I can't help you, but I have a question, who actually uses these formulas irl?

Architects/physicists often use them. TS, I'd help but we haven't done trigonometric equations yet.
#5
Quote by Wratheh
Architects/physicists often use them. TS, I'd help but we haven't done trigonometric equations yet.


Thank you, NONE of the math teachers in my school can tell me what they( or other high level math equations) are used for, and it ticks me off. Thanks.
#7
For the first one:

SinΘ(2SinΘ-1)=0
SinΘ=0 2SinΘ-1=0
Θ= 0, 180, 360 SinΘ=1/2
Θ=30, 150


Basically, Θ= 0, 30, 150, 360
What have we learned.....

Words are weightless here on earth,
because they're free


.....from this wee exercise?
Last edited by con job at Feb 24, 2008,
#8
I'll do the last one.. i'll use Ω instead of thèta 'cause I can't find it on my keyboard..
here we go..
8cos^3(Ω = 1 <->
cos^3(Ω = 1/8 <->
cos(Ω = (1/8)^(1/3) <->
Ω = arccos((1/8)^(1/3))

now you have 1 solution, but between 0 and 2∏ there are 2 solutions,
just look at the goneometric circle (if you know what it is..) the cos of an angle is the same as the cos of the negative angle, so your two solutions are:

arccos((1/8)^(1/3)) and -arccos((1/8)^(1/3)), or between 0 and 2∏: -arccos((1/8)^(1/3)) = 2∏-arccos((1/8)^(1/3))

so, there you go...
Last edited by Future at Feb 24, 2008,
#9
^^^^^

this guys does the sin-1 0 bit right :P

con job i mean

,--.-'-,--.
\ /-~-\ /
/ )' a a `( \
( ( ,---. ) )
THIS WAS MEANT TO BE A PIG
\ `(_o_o_)' /
\ `-' /
| |---| |
[_] [_]
#10
Quote by Guitarfreak777
I can't help you, but I have a question, who actually uses these formulas irl?



Math teachers.
<Han> I love Hitler