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Find the least value for $sin x - cos^2 x -1$
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Find all the values of $x$ for which the function $y = sin x - cos^2 x -1$ assumes the least value. What is that value?
At first I found the first derivative to be $y' = cos x + 2 sin x cos x$.
Critical point $0$, $-À/6$ (principal)
$y'' = - sin x + 2(cos 2x)$
Then substitution of $x$ by critical points I found minima. But my answer is incorrect.
Correct minimum value is $-9/4$
real-analysis trigonometry derivatives optimization maxima-minima
edited Aug 2 at 21:30
Michael Rozenberg
87.1k1576178
asked Aug 2 at 21:17
user579305
363
add a comment |Â
up vote
3
down vote
favorite
Find all the values of $x$ for which the function $y = sin x - cos^2 x -1$ assumes the least value. What is that value?
At first I found the first derivative to be $y' = cos x + 2 sin x cos x$.
Critical point $0$, $-À/6$ (principal)
$y'' = - sin x + 2(cos 2x)$
Then substitution of $x$ by critical points I found minima. But my answer is incorrect.
Correct minimum value is $-9/4$
real-analysis trigonometry derivatives optimization maxima-minima
edited Aug 2 at 21:30
Michael Rozenberg
87.1k1576178
asked Aug 2 at 21:17
user579305
363
Use $cos^2(x)=1- sin^2(x)$ and complete the square ...
– Donald Splutterwit
Aug 2 at 21:21
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Find all the values of $x$ for which the function $y = sin x - cos^2 x -1$ assumes the least value. What is that value?
At first I found the first derivative to be $y' = cos x + 2 sin x cos x$.
Critical point $0$, $-À/6$ (principal)
$y'' = - sin x + 2(cos 2x)$
Then substitution of $x$ by critical points I found minima. But my answer is incorrect.
Correct minimum value is $-9/4$
real-analysis trigonometry derivatives optimization maxima-minima
edited Aug 2 at 21:30
Michael Rozenberg
87.1k1576178
asked Aug 2 at 21:17
user579305
363
Find all the values of $x$ for which the function $y = sin x - cos^2 x -1$ assumes the least value. What is that value?
At first I found the first derivative to be $y' = cos x + 2 sin x cos x$.
Critical point $0$, $-À/6$ (principal)
$y'' = - sin x + 2(cos 2x)$
Then substitution of $x$ by critical points I found minima. But my answer is incorrect.
Correct minimum value is $-9/4$
real-analysis trigonometry derivatives optimization maxima-minima
edited Aug 2 at 21:30
Michael Rozenberg
87.1k1576178
asked Aug 2 at 21:17
user579305
363
edited Aug 2 at 21:30
Michael Rozenberg
87.1k1576178
edited Aug 2 at 21:30
Michael Rozenberg
87.1k1576178
edited Aug 2 at 21:30
Michael Rozenberg
87.1k1576178
87.1k1576178
asked Aug 2 at 21:17
user579305
363
asked Aug 2 at 21:17
user579305
363
asked Aug 2 at 21:17
user579305
363
363
Use $cos^2(x)=1- sin^2(x)$ and complete the square ...
– Donald Splutterwit
Aug 2 at 21:21
add a comment |Â
Use $cos^2(x)=1- sin^2(x)$ and complete the square ...
– Donald Splutterwit
Aug 2 at 21:21
Use $cos^2(x)=1- sin^2(x)$ and complete the square ...
– Donald Splutterwit
Aug 2 at 21:21
Use $cos^2(x)=1- sin^2(x)$ and complete the square ...
– Donald Splutterwit
Aug 2 at 21:21
add a comment |Â
3 Answers
3
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oldest
votes
up vote
2
down vote
accepted
Yes your way is correct indeed for $x=-pi/6$ we have
$$f(x)=-frac12-frac34-1=-frac94$$
As an alternative, recall that $cos^2 x=1-sin^2 x$ and therefore we have
$$f(x)=sin x -cos^2 x-1=sin^2 x+sin x-2$$
then set $t=sin x$ and consider the parabola (concave up)
$$g(t)=t^2+t-2 implies g'(t)=2t+1=0 implies t=-frac 12$$
answered Aug 2 at 21:20
gimusi
63.8k73480
Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3sin x-cos^2 x-1$, this approach would have selected $t = -frac32$, which is not valid for real $x$.
– Brian Tung
Aug 2 at 21:25
@BrianTung Yes indeed in that case we need to find the max/min for the extrema values of $sin x$.
– gimusi
Aug 2 at 21:28
add a comment |Â
up vote
5
down vote
$$f(x)=sin^2x+sinx-2=left(sinx+frac12right)^2-2.25geq-2.25.$$
The equality occurs for $sinx=-frac12,$ which says that $-2.25$ is a minimal value.
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
1
Nice way Michael!
– gimusi
Aug 2 at 21:30
add a comment |Â
up vote
2
down vote
The critical points in $[-pi,pi]$ are those for which
$$
cos x(1+2sin x)=0
$$
which means $x=pi/2$, $x=-pi/2$, $x=-pi/6$ and $x=-5pi/6$.
Note that
$$
y''=-sin x+2cos^2x-2sin^2x
$$
Since
$$
y''(pi/2)=-1-2=-3,
quad
y''(-pi/2)=1-2=-1,
\
y''(-pi/6)=frac12+2frac34-2frac14=frac32=y''(-5pi/6)
$$
the points of local minimum are $-pi/6$ and $-5pi/6$.
Just compute
$$
y(-pi/6)=-frac12-frac34-1=-frac94
$$
The value at $-5pi/6$ is the same.
Simpler: write
$$
y=sin x-1+sin^2x-1=sin^2x+sin x-2=
left(sin x+frac12right)^!2-frac94
$$
Obviously, the minimum value is where $sin x=-frac12$ and is $-9/4$.
The maximum is where $sin x=1$.
answered Aug 2 at 21:30
egreg
164k1180187
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes your way is correct indeed for $x=-pi/6$ we have
$$f(x)=-frac12-frac34-1=-frac94$$
As an alternative, recall that $cos^2 x=1-sin^2 x$ and therefore we have
$$f(x)=sin x -cos^2 x-1=sin^2 x+sin x-2$$
then set $t=sin x$ and consider the parabola (concave up)
$$g(t)=t^2+t-2 implies g'(t)=2t+1=0 implies t=-frac 12$$
answered Aug 2 at 21:20
gimusi
63.8k73480
Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3sin x-cos^2 x-1$, this approach would have selected $t = -frac32$, which is not valid for real $x$.
– Brian Tung
Aug 2 at 21:25
@BrianTung Yes indeed in that case we need to find the max/min for the extrema values of $sin x$.
– gimusi
Aug 2 at 21:28
add a comment |Â
up vote
2
down vote
accepted
Yes your way is correct indeed for $x=-pi/6$ we have
$$f(x)=-frac12-frac34-1=-frac94$$
As an alternative, recall that $cos^2 x=1-sin^2 x$ and therefore we have
$$f(x)=sin x -cos^2 x-1=sin^2 x+sin x-2$$
then set $t=sin x$ and consider the parabola (concave up)
$$g(t)=t^2+t-2 implies g'(t)=2t+1=0 implies t=-frac 12$$
answered Aug 2 at 21:20
gimusi
63.8k73480
Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3sin x-cos^2 x-1$, this approach would have selected $t = -frac32$, which is not valid for real $x$.
– Brian Tung
Aug 2 at 21:25
@BrianTung Yes indeed in that case we need to find the max/min for the extrema values of $sin x$.
– gimusi
Aug 2 at 21:28
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes your way is correct indeed for $x=-pi/6$ we have
$$f(x)=-frac12-frac34-1=-frac94$$
As an alternative, recall that $cos^2 x=1-sin^2 x$ and therefore we have
$$f(x)=sin x -cos^2 x-1=sin^2 x+sin x-2$$
then set $t=sin x$ and consider the parabola (concave up)
$$g(t)=t^2+t-2 implies g'(t)=2t+1=0 implies t=-frac 12$$
answered Aug 2 at 21:20
gimusi
63.8k73480
Yes your way is correct indeed for $x=-pi/6$ we have
$$f(x)=-frac12-frac34-1=-frac94$$
As an alternative, recall that $cos^2 x=1-sin^2 x$ and therefore we have
$$f(x)=sin x -cos^2 x-1=sin^2 x+sin x-2$$
then set $t=sin x$ and consider the parabola (concave up)
$$g(t)=t^2+t-2 implies g'(t)=2t+1=0 implies t=-frac 12$$
answered Aug 2 at 21:20
gimusi
63.8k73480
edited Aug 2 at 21:27
answered Aug 2 at 21:20
gimusi
63.8k73480
answered Aug 2 at 21:20
gimusi
63.8k73480
answered Aug 2 at 21:20
gimusi
63.8k73480
63.8k73480
Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3sin x-cos^2 x-1$, this approach would have selected $t = -frac32$, which is not valid for real $x$.
– Brian Tung
Aug 2 at 21:25
@BrianTung Yes indeed in that case we need to find the max/min for the extrema values of $sin x$.
– gimusi
Aug 2 at 21:28
add a comment |Â
Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3sin x-cos^2 x-1$, this approach would have selected $t = -frac32$, which is not valid for real $x$.
– Brian Tung
Aug 2 at 21:25
@BrianTung Yes indeed in that case we need to find the max/min for the extrema values of $sin x$.
– gimusi
Aug 2 at 21:28
Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3sin x-cos^2 x-1$, this approach would have selected $t = -frac32$, which is not valid for real $x$.
– Brian Tung
Aug 2 at 21:25
Note to OP: Care should be taken in general when proceeding this way. Supposing the original function to have been $y = 3sin x-cos^2 x-1$, this approach would have selected $t = -frac32$, which is not valid for real $x$.
– Brian Tung
Aug 2 at 21:25
@BrianTung Yes indeed in that case we need to find the max/min for the extrema values of $sin x$.
– gimusi
Aug 2 at 21:28
@BrianTung Yes indeed in that case we need to find the max/min for the extrema values of $sin x$.
– gimusi
Aug 2 at 21:28
add a comment |Â
up vote
5
down vote
$$f(x)=sin^2x+sinx-2=left(sinx+frac12right)^2-2.25geq-2.25.$$
The equality occurs for $sinx=-frac12,$ which says that $-2.25$ is a minimal value.
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
1
Nice way Michael!
– gimusi
Aug 2 at 21:30
add a comment |Â
up vote
5
down vote
$$f(x)=sin^2x+sinx-2=left(sinx+frac12right)^2-2.25geq-2.25.$$
The equality occurs for $sinx=-frac12,$ which says that $-2.25$ is a minimal value.
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
1
Nice way Michael!
– gimusi
Aug 2 at 21:30
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$f(x)=sin^2x+sinx-2=left(sinx+frac12right)^2-2.25geq-2.25.$$
The equality occurs for $sinx=-frac12,$ which says that $-2.25$ is a minimal value.
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
$$f(x)=sin^2x+sinx-2=left(sinx+frac12right)^2-2.25geq-2.25.$$
The equality occurs for $sinx=-frac12,$ which says that $-2.25$ is a minimal value.
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
answered Aug 2 at 21:28
Michael Rozenberg
87.1k1576178
87.1k1576178
1
Nice way Michael!
– gimusi
Aug 2 at 21:30
add a comment |Â
1
Nice way Michael!
– gimusi
Aug 2 at 21:30
1
1
Nice way Michael!
– gimusi
Aug 2 at 21:30
Nice way Michael!
– gimusi
Aug 2 at 21:30
add a comment |Â
up vote
2
down vote
The critical points in $[-pi,pi]$ are those for which
$$
cos x(1+2sin x)=0
$$
which means $x=pi/2$, $x=-pi/2$, $x=-pi/6$ and $x=-5pi/6$.
Note that
$$
y''=-sin x+2cos^2x-2sin^2x
$$
Since
$$
y''(pi/2)=-1-2=-3,
quad
y''(-pi/2)=1-2=-1,
\
y''(-pi/6)=frac12+2frac34-2frac14=frac32=y''(-5pi/6)
$$
the points of local minimum are $-pi/6$ and $-5pi/6$.
Just compute
$$
y(-pi/6)=-frac12-frac34-1=-frac94
$$
The value at $-5pi/6$ is the same.
Simpler: write
$$
y=sin x-1+sin^2x-1=sin^2x+sin x-2=
left(sin x+frac12right)^!2-frac94
$$
Obviously, the minimum value is where $sin x=-frac12$ and is $-9/4$.
The maximum is where $sin x=1$.
answered Aug 2 at 21:30
egreg
164k1180187
add a comment |Â
up vote
2
down vote
The critical points in $[-pi,pi]$ are those for which
$$
cos x(1+2sin x)=0
$$
which means $x=pi/2$, $x=-pi/2$, $x=-pi/6$ and $x=-5pi/6$.
Note that
$$
y''=-sin x+2cos^2x-2sin^2x
$$
Since
$$
y''(pi/2)=-1-2=-3,
quad
y''(-pi/2)=1-2=-1,
\
y''(-pi/6)=frac12+2frac34-2frac14=frac32=y''(-5pi/6)
$$
the points of local minimum are $-pi/6$ and $-5pi/6$.
Just compute
$$
y(-pi/6)=-frac12-frac34-1=-frac94
$$
The value at $-5pi/6$ is the same.
Simpler: write
$$
y=sin x-1+sin^2x-1=sin^2x+sin x-2=
left(sin x+frac12right)^!2-frac94
$$
Obviously, the minimum value is where $sin x=-frac12$ and is $-9/4$.
The maximum is where $sin x=1$.
answered Aug 2 at 21:30
egreg
164k1180187
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The critical points in $[-pi,pi]$ are those for which
$$
cos x(1+2sin x)=0
$$
which means $x=pi/2$, $x=-pi/2$, $x=-pi/6$ and $x=-5pi/6$.
Note that
$$
y''=-sin x+2cos^2x-2sin^2x
$$
Since
$$
y''(pi/2)=-1-2=-3,
quad
y''(-pi/2)=1-2=-1,
\
y''(-pi/6)=frac12+2frac34-2frac14=frac32=y''(-5pi/6)
$$
the points of local minimum are $-pi/6$ and $-5pi/6$.
Just compute
$$
y(-pi/6)=-frac12-frac34-1=-frac94
$$
The value at $-5pi/6$ is the same.
Simpler: write
$$
y=sin x-1+sin^2x-1=sin^2x+sin x-2=
left(sin x+frac12right)^!2-frac94
$$
Obviously, the minimum value is where $sin x=-frac12$ and is $-9/4$.
The maximum is where $sin x=1$.
answered Aug 2 at 21:30
egreg
164k1180187
The critical points in $[-pi,pi]$ are those for which
$$
cos x(1+2sin x)=0
$$
which means $x=pi/2$, $x=-pi/2$, $x=-pi/6$ and $x=-5pi/6$.
Note that
$$
y''=-sin x+2cos^2x-2sin^2x
$$
Since
$$
y''(pi/2)=-1-2=-3,
quad
y''(-pi/2)=1-2=-1,
\
y''(-pi/6)=frac12+2frac34-2frac14=frac32=y''(-5pi/6)
$$
the points of local minimum are $-pi/6$ and $-5pi/6$.
Just compute
$$
y(-pi/6)=-frac12-frac34-1=-frac94
$$
The value at $-5pi/6$ is the same.
Simpler: write
$$
y=sin x-1+sin^2x-1=sin^2x+sin x-2=
left(sin x+frac12right)^!2-frac94
$$
Obviously, the minimum value is where $sin x=-frac12$ and is $-9/4$.
The maximum is where $sin x=1$.
answered Aug 2 at 21:30
egreg
164k1180187
answered Aug 2 at 21:30
egreg
164k1180187
answered Aug 2 at 21:30
egreg
164k1180187
answered Aug 2 at 21:30
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
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Use $cos^2(x)=1- sin^2(x)$ and complete the square ...
– Donald Splutterwit
Aug 2 at 21:21