Maximum of $sqrtfrac14cdot sin^2(t)+sin^2(t+fracpi3)$ [closed]

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Maximum of $sqrtfrac14cdotsin^2(t)+sin^2(t+fracpi3)$

In my opinion, the maximum of the sinus is 1, so I calculated $sqrtfrac14cdot1+1$

This is wrong, why?

periodic-functions

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edited Jul 24 at 12:54

gimusi

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asked Jul 24 at 12:36

WinstonCherf

49816

closed as off-topic by amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson Jul 25 at 0:14

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson

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up vote
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Maximum of $sqrtfrac14cdotsin^2(t)+sin^2(t+fracpi3)$

In my opinion, the maximum of the sinus is 1, so I calculated $sqrtfrac14cdot1+1$

This is wrong, why?

periodic-functions

share|cite|improve this question

edited Jul 24 at 12:54

gimusi

65.2k73583

asked Jul 24 at 12:36

WinstonCherf

49816

closed as off-topic by amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson Jul 25 at 0:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
1
down vote

favorite

2

up vote
1
down vote

favorite

2

2

Maximum of $sqrtfrac14cdotsin^2(t)+sin^2(t+fracpi3)$

In my opinion, the maximum of the sinus is 1, so I calculated $sqrtfrac14cdot1+1$

This is wrong, why?

periodic-functions

share|cite|improve this question

edited Jul 24 at 12:54

gimusi

65.2k73583

asked Jul 24 at 12:36

WinstonCherf

49816

Maximum of $sqrtfrac14cdotsin^2(t)+sin^2(t+fracpi3)$

In my opinion, the maximum of the sinus is 1, so I calculated $sqrtfrac14cdot1+1$

This is wrong, why?

periodic-functions

share|cite|improve this question

edited Jul 24 at 12:54

gimusi

65.2k73583

asked Jul 24 at 12:36

WinstonCherf

49816

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 12:54

gimusi

65.2k73583

edited Jul 24 at 12:54

gimusi

65.2k73583

edited Jul 24 at 12:54

gimusi

65.2k73583

65.2k73583

asked Jul 24 at 12:36

WinstonCherf

49816

asked Jul 24 at 12:36

WinstonCherf

49816

asked Jul 24 at 12:36

WinstonCherf

49816

49816

closed as off-topic by amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson Jul 25 at 0:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson Jul 25 at 0:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Claude Leibovici, Mostafa Ayaz, Taroccoesbrocco, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

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add a comment | 

4 Answers
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accepted

You are right about the sine. However, the max for $sin t$ and the max for $sin(t+pi/3)$ do not occur at the same value of $t$, so these expressions can't be maximized separately.

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answered Jul 24 at 12:39

MPW

28.4k11853

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Other people have pointed out the OP's error. Now, I am supplying a solution.

Note that
$$sinleft(t+fracpi3right)=frac12sin(t)+fracsqrt32cos(t),.$$
That is,
$$beginalignfrac14sin^2(t)+sin^2left(t+fracpi3right)&=fracsin^2(t)+big(sin(t)+sqrt3cos(t)big)^24\&=frac58+frac2sqrt3sin(2t)+cos(2t)8,.endalign$$
We then apply the Cauchy-Schwarz Inequality to get
$$-fracsqrt138leq frac2sqrt3sin(2t)+cos(2t)8leq +fracsqrt138,.$$
Consequently,
$$sqrtfrac5-sqrt138leq sqrtfrac14sin^2(t)+sin^2left(t+fracpi3right)leq sqrtfrac5+sqrt138,.$$
It is not difficult to see that both the inequality on the left-hand side and the inequality on the right-hand side are sharp.

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answered Jul 24 at 12:49

Batominovski

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You've definitely calculated an upper bound for the function. The issue is that $ sin(t) $ and $ sin(t + fracpi 3 ) $ cannot both equal $ 1 $ at the same time (i.e. for the same value of $ t $).

To find the exact maximum, you'll need to differentiate the function (using the chain rule a couple of times).

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answered Jul 24 at 12:43

Malkin

1,268523

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Note that the two values for $sin^2$ don't occur for the same value of $t$, indeed

$$sin^2(t)=1implies t=fracpi2+kpi implies sin^2(t+fracpi3)=frac14$$

To evaluate the maximum let observe that

$$left(sqrt f(x)right)'=fracf'(x)2sqrt f(x)$$

then it suffices to evaluate

$$f(t)=frac14sin^2(t)+sin^2(t+fracpi3)implies f'(t)=frac12sin t cos t+2sin(t+fracpi3)cos(t+fracpi3)=0$$

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edited Jul 24 at 12:48

answered Jul 24 at 12:38

gimusi

65.2k73583

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

You are right about the sine. However, the max for $sin t$ and the max for $sin(t+pi/3)$ do not occur at the same value of $t$, so these expressions can't be maximized separately.

share|cite|improve this answer

answered Jul 24 at 12:39

MPW

28.4k11853

add a comment | 

up vote
3
down vote



accepted

You are right about the sine. However, the max for $sin t$ and the max for $sin(t+pi/3)$ do not occur at the same value of $t$, so these expressions can't be maximized separately.

share|cite|improve this answer

answered Jul 24 at 12:39

MPW

28.4k11853

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

You are right about the sine. However, the max for $sin t$ and the max for $sin(t+pi/3)$ do not occur at the same value of $t$, so these expressions can't be maximized separately.

share|cite|improve this answer

answered Jul 24 at 12:39

MPW

28.4k11853

You are right about the sine. However, the max for $sin t$ and the max for $sin(t+pi/3)$ do not occur at the same value of $t$, so these expressions can't be maximized separately.

share|cite|improve this answer

answered Jul 24 at 12:39

MPW

28.4k11853

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 12:39

MPW

28.4k11853

answered Jul 24 at 12:39

MPW

28.4k11853

answered Jul 24 at 12:39

MPW

28.4k11853

28.4k11853

add a comment | 

add a comment | 

up vote
2
down vote

Other people have pointed out the OP's error. Now, I am supplying a solution.

Note that
$$sinleft(t+fracpi3right)=frac12sin(t)+fracsqrt32cos(t),.$$
That is,
$$beginalignfrac14sin^2(t)+sin^2left(t+fracpi3right)&=fracsin^2(t)+big(sin(t)+sqrt3cos(t)big)^24\&=frac58+frac2sqrt3sin(2t)+cos(2t)8,.endalign$$
We then apply the Cauchy-Schwarz Inequality to get
$$-fracsqrt138leq frac2sqrt3sin(2t)+cos(2t)8leq +fracsqrt138,.$$
Consequently,
$$sqrtfrac5-sqrt138leq sqrtfrac14sin^2(t)+sin^2left(t+fracpi3right)leq sqrtfrac5+sqrt138,.$$
It is not difficult to see that both the inequality on the left-hand side and the inequality on the right-hand side are sharp.

share|cite|improve this answer

answered Jul 24 at 12:49

Batominovski

23.2k22777

add a comment | 

up vote
2
down vote

Other people have pointed out the OP's error. Now, I am supplying a solution.

Note that
$$sinleft(t+fracpi3right)=frac12sin(t)+fracsqrt32cos(t),.$$
That is,
$$beginalignfrac14sin^2(t)+sin^2left(t+fracpi3right)&=fracsin^2(t)+big(sin(t)+sqrt3cos(t)big)^24\&=frac58+frac2sqrt3sin(2t)+cos(2t)8,.endalign$$
We then apply the Cauchy-Schwarz Inequality to get
$$-fracsqrt138leq frac2sqrt3sin(2t)+cos(2t)8leq +fracsqrt138,.$$
Consequently,
$$sqrtfrac5-sqrt138leq sqrtfrac14sin^2(t)+sin^2left(t+fracpi3right)leq sqrtfrac5+sqrt138,.$$
It is not difficult to see that both the inequality on the left-hand side and the inequality on the right-hand side are sharp.

share|cite|improve this answer

answered Jul 24 at 12:49

Batominovski

23.2k22777

add a comment | 

up vote
2
down vote

up vote
2
down vote

Other people have pointed out the OP's error. Now, I am supplying a solution.

Note that
$$sinleft(t+fracpi3right)=frac12sin(t)+fracsqrt32cos(t),.$$
That is,
$$beginalignfrac14sin^2(t)+sin^2left(t+fracpi3right)&=fracsin^2(t)+big(sin(t)+sqrt3cos(t)big)^24\&=frac58+frac2sqrt3sin(2t)+cos(2t)8,.endalign$$
We then apply the Cauchy-Schwarz Inequality to get
$$-fracsqrt138leq frac2sqrt3sin(2t)+cos(2t)8leq +fracsqrt138,.$$
Consequently,
$$sqrtfrac5-sqrt138leq sqrtfrac14sin^2(t)+sin^2left(t+fracpi3right)leq sqrtfrac5+sqrt138,.$$
It is not difficult to see that both the inequality on the left-hand side and the inequality on the right-hand side are sharp.

share|cite|improve this answer

answered Jul 24 at 12:49

Batominovski

23.2k22777

Other people have pointed out the OP's error. Now, I am supplying a solution.

Note that
$$sinleft(t+fracpi3right)=frac12sin(t)+fracsqrt32cos(t),.$$
That is,
$$beginalignfrac14sin^2(t)+sin^2left(t+fracpi3right)&=fracsin^2(t)+big(sin(t)+sqrt3cos(t)big)^24\&=frac58+frac2sqrt3sin(2t)+cos(2t)8,.endalign$$
We then apply the Cauchy-Schwarz Inequality to get
$$-fracsqrt138leq frac2sqrt3sin(2t)+cos(2t)8leq +fracsqrt138,.$$
Consequently,
$$sqrtfrac5-sqrt138leq sqrtfrac14sin^2(t)+sin^2left(t+fracpi3right)leq sqrtfrac5+sqrt138,.$$
It is not difficult to see that both the inequality on the left-hand side and the inequality on the right-hand side are sharp.

share|cite|improve this answer

answered Jul 24 at 12:49

Batominovski

23.2k22777

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 12:49

Batominovski

23.2k22777

answered Jul 24 at 12:49

Batominovski

23.2k22777

answered Jul 24 at 12:49

Batominovski

23.2k22777

23.2k22777

add a comment | 

add a comment | 

up vote
1
down vote

You've definitely calculated an upper bound for the function. The issue is that $ sin(t) $ and $ sin(t + fracpi 3 ) $ cannot both equal $ 1 $ at the same time (i.e. for the same value of $ t $).

To find the exact maximum, you'll need to differentiate the function (using the chain rule a couple of times).

share|cite|improve this answer

answered Jul 24 at 12:43

Malkin

1,268523

add a comment | 

up vote
1
down vote

You've definitely calculated an upper bound for the function. The issue is that $ sin(t) $ and $ sin(t + fracpi 3 ) $ cannot both equal $ 1 $ at the same time (i.e. for the same value of $ t $).

To find the exact maximum, you'll need to differentiate the function (using the chain rule a couple of times).

share|cite|improve this answer

answered Jul 24 at 12:43

Malkin

1,268523

add a comment | 

up vote
1
down vote

up vote
1
down vote

You've definitely calculated an upper bound for the function. The issue is that $ sin(t) $ and $ sin(t + fracpi 3 ) $ cannot both equal $ 1 $ at the same time (i.e. for the same value of $ t $).

To find the exact maximum, you'll need to differentiate the function (using the chain rule a couple of times).

share|cite|improve this answer

answered Jul 24 at 12:43

Malkin

1,268523

You've definitely calculated an upper bound for the function. The issue is that $ sin(t) $ and $ sin(t + fracpi 3 ) $ cannot both equal $ 1 $ at the same time (i.e. for the same value of $ t $).

To find the exact maximum, you'll need to differentiate the function (using the chain rule a couple of times).

share|cite|improve this answer

answered Jul 24 at 12:43

Malkin

1,268523

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 12:43

Malkin

1,268523

answered Jul 24 at 12:43

Malkin

1,268523

answered Jul 24 at 12:43

Malkin

1,268523

1,268523

add a comment | 

add a comment | 

up vote
1
down vote

Note that the two values for $sin^2$ don't occur for the same value of $t$, indeed

$$sin^2(t)=1implies t=fracpi2+kpi implies sin^2(t+fracpi3)=frac14$$

To evaluate the maximum let observe that

$$left(sqrt f(x)right)'=fracf'(x)2sqrt f(x)$$

then it suffices to evaluate

$$f(t)=frac14sin^2(t)+sin^2(t+fracpi3)implies f'(t)=frac12sin t cos t+2sin(t+fracpi3)cos(t+fracpi3)=0$$

share|cite|improve this answer

edited Jul 24 at 12:48

answered Jul 24 at 12:38

gimusi

65.2k73583

add a comment | 

up vote
1
down vote

Note that the two values for $sin^2$ don't occur for the same value of $t$, indeed

$$sin^2(t)=1implies t=fracpi2+kpi implies sin^2(t+fracpi3)=frac14$$

To evaluate the maximum let observe that

$$left(sqrt f(x)right)'=fracf'(x)2sqrt f(x)$$

then it suffices to evaluate

$$f(t)=frac14sin^2(t)+sin^2(t+fracpi3)implies f'(t)=frac12sin t cos t+2sin(t+fracpi3)cos(t+fracpi3)=0$$

share|cite|improve this answer

edited Jul 24 at 12:48

answered Jul 24 at 12:38

gimusi

65.2k73583

add a comment | 

up vote
1
down vote

up vote
1
down vote

Note that the two values for $sin^2$ don't occur for the same value of $t$, indeed

$$sin^2(t)=1implies t=fracpi2+kpi implies sin^2(t+fracpi3)=frac14$$

To evaluate the maximum let observe that

$$left(sqrt f(x)right)'=fracf'(x)2sqrt f(x)$$

then it suffices to evaluate

$$f(t)=frac14sin^2(t)+sin^2(t+fracpi3)implies f'(t)=frac12sin t cos t+2sin(t+fracpi3)cos(t+fracpi3)=0$$

share|cite|improve this answer

edited Jul 24 at 12:48

answered Jul 24 at 12:38

gimusi

65.2k73583

Note that the two values for $sin^2$ don't occur for the same value of $t$, indeed

$$sin^2(t)=1implies t=fracpi2+kpi implies sin^2(t+fracpi3)=frac14$$

To evaluate the maximum let observe that

$$left(sqrt f(x)right)'=fracf'(x)2sqrt f(x)$$

then it suffices to evaluate

$$f(t)=frac14sin^2(t)+sin^2(t+fracpi3)implies f'(t)=frac12sin t cos t+2sin(t+fracpi3)cos(t+fracpi3)=0$$

share|cite|improve this answer

edited Jul 24 at 12:48

answered Jul 24 at 12:38

gimusi

65.2k73583

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 12:48

edited Jul 24 at 12:48

edited Jul 24 at 12:48

answered Jul 24 at 12:38

gimusi

65.2k73583

answered Jul 24 at 12:38

gimusi

65.2k73583

answered Jul 24 at 12:38

gimusi

65.2k73583

65.2k73583

add a comment | 

add a comment |