A maths questions
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Integration
integrate sin^2 x cos^2x between pi/2 and 0. My answer is pi/16, while that given by answer is pi/4.
Circles
It is given that P(0,3), Q(7,5), R(9,-2) and S are the vertices of a square. A circle A passes through the points, P,Q ,R and S. Find the equation of the circle. Hence/otherwise, find the equation of circle B which is the image of A under the reflection about hte y-axis. Equation of circle A is x^2 + y^2 - 9x - y - 6 =0, while for B my answer is x^2 + y^2 + 9x - y - 6 =0. But the answer given is x^2 + y^2 +x- y - 6 =0.
Please help me check if I am right or not. Thanks!
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eh. TS. for qn 1. How did you do it.??
From what i think, ..
change sin^2 x into 1-cos^x.. then
expand out, and solve by the reduction method.
But since you dont learn reduction, how did you do it?
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Originally posted by anpanman:
Integration
integrate sin^2 x cos^2x between pi/2 and 0. My answer is pi/16, while that given by answer is pi/4.
Circles
It is given that P(0,3), Q(7,5), R(9,-2) and S are the vertices of a square. A circle A passes through the points, P,Q ,R and S. Find the equation of the circle. Hence/otherwise, find the equation of circle B which is the image of A under the reflection about hte y-axis. Equation of circle A is x^2 + y^2 - 9x - y - 6 =0, while for B my answer is x^2 + y^2 + 9x - y - 6 =0. But the answer given is x^2 + y^2 +x- y - 6 =0.
Please help me check if I am right or not. Thanks!
You are correct, provided for A) the integration is between 0 and pi/2, not the other way round as you stated.
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Originally posted by Forbiddensinner:
You are correct, provided for A) the integration is between 0 and pi/2, not the other way round as you stated.
How did you manage to integrate sin^2x cos^2x without reduction? Cause TS i think is Olev right?
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i might have misinterpreted it, but its
integrate [ sine squared x, (times,) cosine squared x ] ??
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Originally posted by OHSheet:
How did you manage to integrate sin^2x cos^2x without reduction? Cause TS i think is Olev right?
They used...
sin 2A = 2sinAcosA 1st
followed by cos 2A = 1 - 2sin^2 A
to make the expression linear trigo.
Both questions are a waste of time. Come out hor I cut.
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Originally posted by Mikethm:
They used...
sin 2A = 2sinAcosA 1st
followed by cos 2A = 1 - 2sin^2 A
to make the expression linear trigo.
Both questions are a waste of time. Come out hor I cut.
eh??? but the question is Sine squared x. NOT sin2x.
and its cos squared x. NOT cos2x
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Originally posted by OHSheet:
eh??? but the question is Sine squared x. NOT sin2x.
and its cos squared x. NOT cos2x
cos 2 x = 1 - 2sin^2x = 2cos^2x - 1
sin^2x cos^2x
= 1/2 X ( 1 - cos2x ) X 1/2 X ( 1 + cos2x )
= 1/4 X ( 1 - cos^2 (2x) )
= 1/4 - [ 1/4 X 1/2 X ( cos4x + 1 ) ]
= 1/8 - ( 1/8 X cos4x )
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Originally posted by OHSheet:
eh??? but the question is Sine squared x. NOT sin2x.
and its cos squared x. NOT cos2x
sin^2 x cos^2 x
= (1/4)( 2sinxcosx)^2
= (1/4)(sin^2 2x)---------------------- ( from sin 2A = 2sinAcosA)
=(1/4)(1/2)(1-cos 4x)------------- ( From cos 2A = 1-2sin^2 A aka sin^2 A = (1/2)(1-cos2A)
=(1/8)(1- cos4x)
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hahaha! thanks!
