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Mathematics: Post your doubts here!

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  1. PLZ HELP ME WITH THE QUESTION 5 !!
Let theta be x,
Part i)
sin^4x - cos^4x
= (sin^2x)(sin^2x) - (cos^2x)(cos^2x)
= (sin^2x)^2 - (cos^2x)^2
We know a^2 - b^2 = (a - b)(a + b)
= (sin^2x - cos^2x)(sin^2x + cos^2x)
We know sin^2x + cos^2x = 1
= (sin^2x - cos^2x)(1)
= sin^2x - cos^2x
sin^2x - cos^2x
= sin^2x - (1 - sin^2x)
= 2sin^2x - 1
Part ii)
sin^4x - cos^4x = 2sin^2x - 1
2sin^2x - 1 = 1/2
2sin^2x = 3/2
sin^2x = 3/4
sin(x) = (Roof of 3)/2
x =
60
As x lies between 0 to 360, there will be more 3 values of x.
x = 60 ; (180 - x) = (180 -60) =
120
x = 60 ; (x + 180) = (60 + 180) =
240
x = 120 ; (x +180) = (120 + 180) =
300
Therefore,
x = 60, 120, 240, 300.
 
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Thanks a lot !!!
Let theta be x,
Part i)
sin^4x - cos^4x
= (sin^2x)(sin^2x) - (cos^2x)(cos^2x)
= (sin^2x)^2 - (cos^2x)^2
We know a^2 - b^2 = (a - b)(a + b)
= (sin^2x - cos^2x)(sin^2x + cos^2x)
We know sin^2x + cos^2x = 1
= (sin^2x - cos^2x)(1)
= sin^2x - cos^2x
sin^2x - cos^2x
= sin^2x - (1 - sin^2x)
= 2sin^2x - 1
Part ii)
sin^4x - cos^4x = 2sin^2x - 1
2sin^2x - 1 = 1/2
2sin^2x = 3/2
sin^2x = 3/4
sin(x) = (Roof of 3)/2
x =
60
As x lies between 0 to 360, there will be more 3 values of x.
x = 60 ; (180 - x) = (180 -60) =
120
x = 60 ; (x + 180) = (60 + 180) = 240
x = 120 ; (x +180) = (120 + 180) = 300
Therefore, x = 60, 120, 240, 300.
ks
 
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PLZ SOME HELP TO ME TO SOLVE THIS QUESTION !!
 

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PART ( i ) : Differentiate y = ( 4x + 1 ) ^ 1/2
dy/dx = 2 ( 4x + 1 ) ^ -1/2
put x = 2 to get gradient of 1st tangent
gradient of tangent = 2 ( 4 * 2 + 1 )^ -1/2
gradient of tangent = 2/3

Differentiate y = ( 1/2) X^2 +1
dy/dx = x
put x = 2 in above equation to get gradient of 2nd tangent
gradient of 2nd tangent = 2

let theta be x
Tan x = ( M2 - M1 ) / 1 + M2M1
Tan x = ( 2 - 2/3 ) / 1 + 4/3
Tan x = 4/7
theta = 29.7 degree

Part ( ii ) : find area under the 1st curve using 2 and 0 as limits .
find area under 2nd curve using 2 and 0 as limits .
subtract both areas to get shaded area ... final answer is 1 unit square ..
 
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