Trig Question Watch

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Martha
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#1
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1. Given that q cosec x = 17 and q cot x = 8, q>0, find q and the smallest positive value if x.

2. If 24 sin x = m and 7 cos x = m, m >0, find m and the possible values of x in the interval 0<x<360

3. prove that cos ^4 x - sin^4 x is identical to cos^2 x -sin^2 x


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misshn
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1. q cosecx=17, q cotx = 8 = q cosx cosecx =>cosx=8/17
As q>0 and qcosecx>0, sinx>0 =>sinx = rt(a-cos^2 x) = 15/17
Substitute into q cosecx=17 =>q = 17^2/15
cosx=8/17,sinx=15/17 => min x = 61.9 degree

2. 24sinx = m =>sinx = m/24 =>(sinx)^2=m^2/576
7cosx = m =>cosx = m/7 =>(cosx)^2=m^2/49
As (sinx)^2+(cosx)^2 =1 => m^2/576+m^2/49 = 1 => m = 6.72
So sinx=0.28 and cosx=0.96 => x = 16.3

3. (cosx)^4 - (sinx)^4 = [(cosx)^2 + (sinx)^2][(cosx)^2 - (sinx)^2]
= (cosx)^2 - (sinx)^2
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Nima
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#3
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(Original post by Martha)
1. Given that q cosec x = 17 and q cot x = 8, q>0, find q and the smallest positive value if x.
2. If 24 sin x = m and 7 cos x = m, m >0, find m and the possible values of x in the interval 0<x<360
3. prove that cos^4x - sin^4x is identical to cos^2 x - sin^2 x
1.) qcosecx = 17 ---> q = 17/cosecx
qcotx = 8 ---> q = 8/cotx
Equating the expressions for q:
---> 17/cosecx = 8/cotx
---> 17cotx = 8cosecx
---> cotx/cosecx = 8/17
---> sinx/tanx = 8/17
---> cosx = 8/17
---> Smallest +ve x = 1.08 Rads (3.S.F)

Hence: q = 17/cosecx = 17sinx = 17sin[cos^-1(8/17)]
---> q = 15

2.) 24sinx = m and 7cosx = m
Equating the expressions for m:
---> 24sinx = 7cosx
---> tanx = 7/24
---> x = 16.3, 196.3 Deg For 0 < x < 360 Deg

Hence: m = 7cosx = 7cos[tan^-1(7/24)]
---> m = 6.72

3.) You need to recognize that cos^4x - sin^4x is a difference of 2 squares as there is no middle term in the expression.

Hence: cos^4x - sin^4x = (cos^2x + sin^2x)(cos^2x - sin^2x) = 1(cos^2x - sin^2x) = cos^2x - sin^2x

QED.
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Gaz031
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Values of x for cosx=sin3x anyone?
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BCHL85
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(Original post by Gaz031)
Values of x for cosx=sin3x anyone?
cosx = sin3x = cos(pi/2 - 3x)
so x = pi/2 - 3x + k2pi
or x = pi-(pi/2 -3x) +k2pi (k E Z)
-> x = pi/8 + kpi/2
or x = -pi/4 + kpi (k E Z)
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Gaz031
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or x = pi-(pi/2 -3x) +k2pi (k E Z)
Why is this one present?

Is there a more natural way to do it?
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BCHL85
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(Original post by Gaz031)
Why is this one present?

Is there a more natural way to do it?
I'm sorry ... I mixed sine and cosine
It should be x = -(pi/2 - 3x) +k2pi
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Gaz031
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#8
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I see. That is presumably because cos is an even function so cosx = cos(-x) ?

Thanks for the assistance. Someone i know rung up and asked me a few questions and I was a bit worried that i couldn't do this quickly, given i'm starting P6.

Still, there's plenty of time left for me to revise.

I see where your logic is and will have to gather my thoughts before trying to explain it across the phone.
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BCHL85
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#9
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(Original post by Gaz031)
I see. That is presumably because cos is an even function so cosx = cos(-x) ?

Thanks for the assistance. Someone i know rung up and asked me a few questions and I was a bit worried that i couldn't do this quickly, given i'm starting P6.

Still, there's plenty of time left for me to revise.

I see where your logic is and will have to gather my thoughts before trying to explain it across the phone.
Yes, cos(x) = cos(-x) = cos(x+k2pi) where k E Z
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