Trigonometric inequalities with substitution
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Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$
I tried with substitution $tan frac{x}{2} = t$
$sin x = frac{2t}{t^2+1}$
$t>frac{x}{2}$
$2t>x$
$t^2+1>1+frac{x^2}{4}$
After this I am unable to do it
Please help me
Thank you
trigonometry
edited Nov 13 at 9:39
MathFun123
459216
asked Nov 13 at 9:34
Ashwini
11
add a comment |
up vote
0
down vote
favorite
Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$
I tried with substitution $tan frac{x}{2} = t$
$sin x = frac{2t}{t^2+1}$
$t>frac{x}{2}$
$2t>x$
$t^2+1>1+frac{x^2}{4}$
After this I am unable to do it
Please help me
Thank you
trigonometry
edited Nov 13 at 9:39
MathFun123
459216
asked Nov 13 at 9:34
Ashwini
11
NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$
I tried with substitution $tan frac{x}{2} = t$
$sin x = frac{2t}{t^2+1}$
$t>frac{x}{2}$
$2t>x$
$t^2+1>1+frac{x^2}{4}$
After this I am unable to do it
Please help me
Thank you
trigonometry
edited Nov 13 at 9:39
MathFun123
459216
asked Nov 13 at 9:34
Ashwini
11
Using inequality $tan frac{x}{2} > frac{x}{2}$ prove that $sin x > x- frac{x^3}{4}$
I tried with substitution $tan frac{x}{2} = t$
$sin x = frac{2t}{t^2+1}$
$t>frac{x}{2}$
$2t>x$
$t^2+1>1+frac{x^2}{4}$
After this I am unable to do it
Please help me
Thank you
trigonometry
trigonometry
edited Nov 13 at 9:39
MathFun123
459216
asked Nov 13 at 9:34
Ashwini
11
edited Nov 13 at 9:39
MathFun123
459216
asked Nov 13 at 9:34
Ashwini
11
edited Nov 13 at 9:39
MathFun123
459216
edited Nov 13 at 9:39
MathFun123
459216
edited Nov 13 at 9:39
MathFun123
459216
459216
asked Nov 13 at 9:34
Ashwini
11
asked Nov 13 at 9:34
Ashwini
11
asked Nov 13 at 9:34
Ashwini
11
11
NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55
add a comment |
NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55
NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55
NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55
add a comment |
1 Answer
1
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0
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We know that, $$1-t^4<1$$
$Longrightarrow frac{1}{1+t^2}>1-t^2$
$Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$
$Longrightarrow sin(x)>x-frac{x^3}{4}$
Hence proved.
Hope it helps:)
answered Nov 13 at 9:45
Crazy for maths
4948
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We know that, $$1-t^4<1$$
$Longrightarrow frac{1}{1+t^2}>1-t^2$
$Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$
$Longrightarrow sin(x)>x-frac{x^3}{4}$
Hence proved.
Hope it helps:)
answered Nov 13 at 9:45
Crazy for maths
4948
add a comment |
up vote
0
down vote
We know that, $$1-t^4<1$$
$Longrightarrow frac{1}{1+t^2}>1-t^2$
$Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$
$Longrightarrow sin(x)>x-frac{x^3}{4}$
Hence proved.
Hope it helps:)
answered Nov 13 at 9:45
Crazy for maths
4948
add a comment |
up vote
0
down vote
up vote
0
down vote
We know that, $$1-t^4<1$$
$Longrightarrow frac{1}{1+t^2}>1-t^2$
$Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$
$Longrightarrow sin(x)>x-frac{x^3}{4}$
Hence proved.
Hope it helps:)
answered Nov 13 at 9:45
Crazy for maths
4948
We know that, $$1-t^4<1$$
$Longrightarrow frac{1}{1+t^2}>1-t^2$
$Longrightarrow frac{2t}{1+t^2}>2t-2t^3>2(frac{x}{2})-2(frac{x}{2})^3$
$Longrightarrow sin(x)>x-frac{x^3}{4}$
Hence proved.
Hope it helps:)
answered Nov 13 at 9:45
Crazy for maths
4948
answered Nov 13 at 9:45
Crazy for maths
4948
answered Nov 13 at 9:45
Crazy for maths
4948
answered Nov 13 at 9:45
Crazy for maths
4948
4948
add a comment |
add a comment |
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NB both the original and derived identities are only valid for $x > 0$.
– Travis
Nov 13 at 9:55