How do I know the maximum value of $\sin x + \cos x$ is $\sqrt{2}$?

This is a standard result in trigonometry! Any expression of the form $a \sin x + b \cos x$ can be rewritten as $R \sin(x + \phi)$, where $R = \sqrt{a^2 + b^2}$. In your case, $a=1$ and $b=1$, so $R = \sqrt{1^2 + 1^2} = \sqrt{2}$. Since the maximum value of the sine function is $1$, the maximum value of the entire expression is $\sqrt{2} \times 1 = \sqrt{2}$.

JEE Advanced 1995

LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: Find the smallest positive number $p$ for which the equation $cos (p sin x) = sin (p cos x)$ has a solution $x \in [0, 2 π]$ .

Q: Why do we set $n=0$ in the general solution?

We are looking for the *smallest positive* value of $p$. Our expression for $p$ is $p = \frac{2n\pi + \frac{\pi}{2}}{\sin x + \cos x}$. To keep $p$ positive and as small as possible, we need the smallest positive numerator. Setting $n=0$ gives us the numerator $\frac{\pi}{2}$. If we chose $n=1$, the numerator would be $\frac{5\pi}{2}$, which is significantly larger, and negative values of $n$ would make $p$ negative or zero, which we must avoid.

Progressive Hints

Hint 1

Hint 1: The Spark. You are looking at a trigonometric equation where the functions are mismatched—a cosine on one side and a sine on the other. To make any progress, you must bridge this gap using the complementary angle identity:

sin θ = cos (\frac{π}{2} - θ)

. This will allow you to compare the arguments of the cosine functions directly.

Hint 2

Hint 2: The Arsenal. Once you have

cos (A) = cos (B)

, remember that the general solution is not simply

A = B

. You must account for the periodicity of the cosine function, which gives you

A = 2 nπ \pm B

, where

n

is any integer. This is the crucial step to capture all possible values of

x

Hint 3

Hint 3: The Execution. After substituting

A = p sin x

and

B = \frac{π}{2} - p cos x

, you will arrive at an equation involving

p (sin x + cos x)

. To find the smallest positive

p

, set

n = 0

to minimize the numerator and maximize the denominator,

sin x + cos x

, using the identity for the maximum value of

a sin x + b cos x

Visualized Solution

The Equation

Given: $cos (p sin x) = sin (p cos x)$
We need the smallest positive $p$ for $x \in [0, 2 π]$ .

Equating Trigonometric Ratios

To solve, we need the same trigonometric ratio on both sides.
Identity: $sin θ = cos (\frac{π}{2} - θ)$

Applying the Identity

Rewrite RHS: $sin (p cos x) = cos (\frac{π}{2} - p cos x)$
Equation becomes: $cos (p sin x) = cos (\frac{π}{2} - p cos x)$

General Solution for Cosine

If $cos A = cos B$ , then the general solution is:
$A = 2 nπ \pm B$ , where $n \in Z$

Substituting into General Solution

Let $A = p sin x$ and $B = \frac{π}{2} - p cos x$
$p sin x = 2 nπ \pm (\frac{π}{2} - p cos x)$

Case 1: Positive Sign

Taking the positive sign:
$p sin x = 2 nπ + \frac{π}{2} - p cos x$

Rearranging Case 1

Move $- p cos x$ to the left side:
$p sin x + p cos x = 2 nπ + \frac{π}{2}$
Factor out $p$ : $p (sin x + cos x) = 2 nπ + \frac{π}{2}$

Isolating $p$

$p = \frac{2 nπ + \frac{π}{2}}{s i n x + c o s x}$
To minimize positive $p$ , we need:
- Minimum positive numerator
- Maximum positive denominator

Minimizing the Numerator

Numerator: $2 nπ + \frac{π}{2}$
For smallest positive value, set $n = 0$ .
Numerator becomes $\frac{π}{2}$ .

Maximizing the Denominator

Denominator: $f (x) = sin x + cos x$
The maximum value of $a sin x + b cos x$ is $a^{2} + b^{2}$ .
Max value $= 1^{2} + 1^{2} = 2$ .

Visualizing the Maximum

The graph of $y = sin x + cos x$ reaches its peak at $x = \frac{π}{4}$ .
At this point, the value is exactly $2$ .

Calculating Minimum $p$

Substitute the values back:
$p = \frac{\frac{π}{2}}{2}$
$p = \frac{π}{2 2}$

Rationalizing the Result

Multiply numerator and denominator by $2$ :
$p = \frac{π}{2 2} \times \frac{2}{2}$
$p = \frac{π 2}{4}$

What About Case 2?

Case 2 (Negative sign): $p (sin x - cos x) = 2 nπ - \frac{π}{2}$
For $n = 1$ , numerator is $\frac{3 π}{2}$ (larger).
For $n = 0$ , numerator is $- \frac{π}{2}$ , requiring negative denominator.
Minimum positive $p$ remains $\frac{π 2}{4}$ .

00:00 / 00:00

The Sigma Insight: General Solution of Trigonometric Equations

Solution Diagram

Analyzing the Setup

The given equation is

cos (p sin x) = sin (p cos x)

. This represents a transcendental equation where we seek the smallest positive parameter

p

that allows for a solution in

x

The Identity Bridge

To solve this, we must align the trigonometric functions. We utilize the complementary angle identity,

sin θ = cos (\frac{π}{2} - θ)

, to rewrite the right-hand side.

The equation transforms into:

cos (p sin x) = cos (\frac{π}{2} - p cos x)

The General Solution Trap

Because the cosine function is periodic, the equality

cos A = cos B

implies the general solution

A = 2 nπ \pm B

, where

n

is any integer. Applying this to our equation, we obtain:

p sin x = 2 nπ \pm (\frac{π}{2} - p cos x)

The Optimization

To find the smallest positive

p

, we examine the case where

n = 0

and the positive sign is chosen:

p sin x = \frac{π}{2} - p cos x

Rearranging the terms to isolate

p

, we get:

p (sin x + cos x) = \frac{π}{2}

This yields the expression for

p

p = \frac{π /2}{sin x + cos x}

To minimize

p

, we must maximize the denominator

f (x) = sin x + cos x

. Using the harmonic addition theorem, the maximum value of

a sin x + b cos x

a^{2} + b^{2}

For

a = 1

and

b = 1

, the maximum value is

1^{2} + 1^{2} = 2

Final Calculation

Substituting the maximum value of the denominator into our expression for

p

p = \frac{π /2}{2} = \frac{π}{2 2}

Rationalizing the denominator by multiplying by

\frac{2}{2}

, we arrive at the final result:

$p = \frac{π 2}{4}$

Common Traps & FAQs

Wait, why can't we just cancel the cosines and say

p sin x = \frac{π}{2} - p cos x

That is a classic trap! If you simply equate the arguments, you are only finding one specific set of solutions. Trigonometric functions are periodic, and the equation

cos A = cos B

is satisfied whenever

A = 2 nπ \pm B

. By ignoring the

2 nπ

and the

\pm

sign, you miss an infinite number of potential solutions, which could lead you to a value of

p

that is not the smallest one.

Why do we set

n = 0

in the general solution?

We are looking for the *smallest positive* value of

p

. Our expression for

p

p = \frac{2 nπ + \frac{π}{2}}{s i n x + c o s x}

. To keep

p

positive and as small as possible, we need the smallest positive numerator. Setting

n = 0

gives us the numerator

\frac{π}{2}

. If we chose

n = 1

, the numerator would be

\frac{5 π}{2}

, which is significantly larger, and negative values of

n

would make

p

negative or zero, which we must avoid.

How do I know the maximum value of

sin x + cos x

2

This is a standard result in trigonometry! Any expression of the form

a sin x + b cos x

can be rewritten as

R sin (x + ϕ)

, where

R = a^{2} + b^{2}

. In your case,

a = 1

and

b = 1

, so

R = 1^{2} + 1^{2} = 2

. Since the maximum value of the sine function is

1

, the maximum value of the entire expression is

2 \times 1 = 2

Animated Solution for Mathematics - Trigonometry: Find the smallest positive number $p$ for which the equation $cos (p sin x) = sin (p cos x)$ has a solution $x \in [0, 2 π]$ .

Visualized Solution

The Equation

Equating Trigonometric Ratios

Applying the Identity

General Solution for Cosine

Substituting into General Solution

Case 1: Positive Sign

Rearranging Case 1

Isolating $p$

Minimizing the Numerator

Maximizing the Denominator

Visualizing the Maximum

Calculating Minimum $p$

Rationalizing the Result

What About Case 2?

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Identity Bridge

The General Solution Trap

The Optimization

Final Calculation

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the smallest positive number p for which the equation cos(psinx)=sin(pcosx) has a solution x∈[0,2π].

Visualized Solution

The Equation

Equating Trigonometric Ratios

Applying the Identity

General Solution for Cosine

Substituting into General Solution

Case 1: Positive Sign

Rearranging Case 1

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Minimizing the Numerator

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Visualizing the Maximum

Calculating Minimum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Rationalizing the Result

What About Case 2?

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Identity Bridge

The General Solution Trap

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