cant use lhopital rule :s
a = 1/2
Can you use Taylor's expansion?
Otherwise (1-cosx)cos^2x = a sin^2x, but sin^2x = 1-cos^2x = (1+cosx)(1-cosx), so that
(1-cosx)cos^2x = a (1+cosx)(1-cosx) .
Rewrite as [1-cosx)(cos^2x -a(1+cosx)]=0
For (cos^2x -a(1+cosx) to be = 0 in the limit x->0, 1-2a=0 => a = 1/2
Hello,
For any real x, we have:
cos²(x) + sin²(x) = 1 âââ Most famous trigonometric identity
1 + tan²(x) = 1/cos²(x) âââ Divide all by cos²(x)
tan²(x) = [1/cos²(x)] - 1 âââ Subtract 1 from both side.
tan²(x) = [1² - cos²(x)] / cos²(x) âââ Set to same denominator
tan²(x) = [1 - cos(x)][1 + cos(x)] / cos²(x) âââ Because a²-b²=(a-b)(a+b)
a.tan²(x) = a[1 - cos(x)][1 + cos(x)] / cos²(x) âââ Multiply everything by a
a.tan²(x) / [1 - cos(x)] = a[1 + cos(x)] / cos²(x) âââ Divide everything by 1-cos(x)
[1 - cos(x)] / [a.tan²(x)] = cos²(x) / {a[1 + cos(x)]} âââ Equality of the reverse
And obviously:
Lim (xâ0) [1 - cos(x)] / [a.tan²(x)]
   = Lim (xâ0) cos²(x) / {a[1 + cos(x)]}
   = cos²(0) / {a[1 + cos(0)]}
   = 1² / [a(1 + 1)]
   = 1 / (2a)
Since we are told
Lim (xâ0) [1 - cos(x)] / [a.tan²(x)] = 1
Then:
Lim (xâ0) [1 - cos(x)] / [a.tan²(x)] = 1/(2a) = 1
Thus:
1 / (2a) = 1
a = ½
Regards,
Dragon.Jade :-)
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Answers & Comments
Verified answer
a = 1/2
Can you use Taylor's expansion?
Otherwise (1-cosx)cos^2x = a sin^2x, but sin^2x = 1-cos^2x = (1+cosx)(1-cosx), so that
(1-cosx)cos^2x = a (1+cosx)(1-cosx) .
Rewrite as [1-cosx)(cos^2x -a(1+cosx)]=0
For (cos^2x -a(1+cosx) to be = 0 in the limit x->0, 1-2a=0 => a = 1/2
Hello,
For any real x, we have:
cos²(x) + sin²(x) = 1 âââ Most famous trigonometric identity
1 + tan²(x) = 1/cos²(x) âââ Divide all by cos²(x)
tan²(x) = [1/cos²(x)] - 1 âââ Subtract 1 from both side.
tan²(x) = [1² - cos²(x)] / cos²(x) âââ Set to same denominator
tan²(x) = [1 - cos(x)][1 + cos(x)] / cos²(x) âââ Because a²-b²=(a-b)(a+b)
a.tan²(x) = a[1 - cos(x)][1 + cos(x)] / cos²(x) âââ Multiply everything by a
a.tan²(x) / [1 - cos(x)] = a[1 + cos(x)] / cos²(x) âââ Divide everything by 1-cos(x)
[1 - cos(x)] / [a.tan²(x)] = cos²(x) / {a[1 + cos(x)]} âââ Equality of the reverse
And obviously:
Lim (xâ0) [1 - cos(x)] / [a.tan²(x)]
   = Lim (xâ0) cos²(x) / {a[1 + cos(x)]}
   = cos²(0) / {a[1 + cos(0)]}
   = 1² / [a(1 + 1)]
   = 1 / (2a)
Since we are told
Lim (xâ0) [1 - cos(x)] / [a.tan²(x)] = 1
Then:
Lim (xâ0) [1 - cos(x)] / [a.tan²(x)] = 1/(2a) = 1
Thus:
1 / (2a) = 1
a = ½
Regards,
Dragon.Jade :-)