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JEE Main 2019
Sequences & Series
Sequences and Series
Medium

Question

If sin 4 α\alpha + 4 cos 4 β\beta + 2 = 42\sqrt 2 sin α\alpha cos β\beta ; α\alpha , β\beta \in [0, π\pi ], then cos(α\alpha + β\beta ) - cos(α\alpha - β\beta ) is equal to :

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Solution

Key Concepts and Formulas

  • Arithmetic Mean - Geometric Mean (AM-GM) Inequality: For non-negative real numbers a1,a2,,ana_1, a_2, \dots, a_n, a1+a2++anna1a2ann\frac{a_1 + a_2 + \dots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \dots a_n}. Equality holds if and only if a1=a2==ana_1 = a_2 = \dots = a_n.
  • Trigonometric Identities:
    • sin2x+cos2x=1\sin^2 x + \cos^2 x = 1
    • cos(A+B)cos(AB)=2sinAsinB\cos(A+B) - \cos(A-B) = -2 \sin A \sin B
  • Properties of Trigonometric Functions: Understanding the signs and values of sinx\sin x and cosx\cos x in the interval [0,π][0, \pi].

Step-by-Step Solution

Step 1: Analyze the Given Equation and Prepare for AM-GM The given equation is sin4α+4cos4β+2=42sinαcosβ\sin^4 \alpha + 4 \cos^4 \beta + 2 = 4\sqrt{2} \sin \alpha \cos \beta, with α,β[0,π]\alpha, \beta \in [0, \pi]. We observe that the left side is a sum of terms, and the right side involves a product. This suggests the application of the AM-GM inequality. To apply AM-GM, we need non-negative terms. sin4α0\sin^4 \alpha \ge 0, 4cos4β04 \cos^4 \beta \ge 0, and 2>02 > 0. The constant term '2' can be split into 1+11+1 to facilitate the application of AM-GM with four terms. Let the four non-negative terms be a1=sin4αa_1 = \sin^4 \alpha, a2=4cos4βa_2 = 4 \cos^4 \beta, a3=1a_3 = 1, and a4=1a_4 = 1.

Step 2: Apply the AM-GM Inequality Applying the AM-GM inequality to these four terms: sin4α+4cos4β+1+14sin4α4cos4β114\frac{\sin^4 \alpha + 4 \cos^4 \beta + 1 + 1}{4} \ge \sqrt[4]{\sin^4 \alpha \cdot 4 \cos^4 \beta \cdot 1 \cdot 1} Simplifying the inequality: sin4α+4cos4β+244sin4αcos4β4\frac{\sin^4 \alpha + 4 \cos^4 \beta + 2}{4} \ge \sqrt[4]{4 \sin^4 \alpha \cos^4 \beta} sin4α+4cos4β+242sin4αcos4β4\frac{\sin^4 \alpha + 4 \cos^4 \beta + 2}{4} \ge \sqrt{2} \sqrt[4]{\sin^4 \alpha \cos^4 \beta} Since x44=x\sqrt[4]{x^4} = |x|, we have: sin4α+4cos4β+242sinαcosβ\frac{\sin^4 \alpha + 4 \cos^4 \beta + 2}{4} \ge \sqrt{2} |\sin \alpha \cos \beta| Multiplying both sides by 4: sin4α+4cos4β+242sinαcosβ\sin^4 \alpha + 4 \cos^4 \beta + 2 \ge 4\sqrt{2} |\sin \alpha \cos \beta|

Step 3: Utilize the Equality Condition of AM-GM We are given that sin4α+4cos4β+2=42sinαcosβ\sin^4 \alpha + 4 \cos^4 \beta + 2 = 4\sqrt{2} \sin \alpha \cos \beta. Comparing this with the result from AM-GM, sin4α+4cos4β+242sinαcosβ\sin^4 \alpha + 4 \cos^4 \beta + 2 \ge 4\sqrt{2} |\sin \alpha \cos \beta|, we see that the equality must hold in the AM-GM inequality. For equality to hold, two conditions must be met:

  1. All the terms in the AM-GM inequality must be equal: sin4α=4cos4β=1\sin^4 \alpha = 4 \cos^4 \beta = 1.
  2. The term sinαcosβ\sin \alpha \cos \beta must be non-negative, so that sinαcosβ=sinαcosβ|\sin \alpha \cos \beta| = \sin \alpha \cos \beta. This is because the left side of the given equation is a sum of non-negative terms and thus non-negative, implying the right side 42sinαcosβ4\sqrt{2} \sin \alpha \cos \beta must also be non-negative.

Step 4: Solve for α\alpha and β\beta From the equality condition sin4α=1\sin^4 \alpha = 1: Since α[0,π]\alpha \in [0, \pi], sinα0\sin \alpha \ge 0. Thus, sinα=1\sin \alpha = 1. For sinα=1\sin \alpha = 1 in the interval [0,π][0, \pi], we get α=π2\alpha = \frac{\pi}{2}.

From the equality condition 4cos4β=14 \cos^4 \beta = 1: cos4β=14\cos^4 \beta = \frac{1}{4}. Taking the square root of both sides, cos2β=14=12\cos^2 \beta = \sqrt{\frac{1}{4}} = \frac{1}{2} (since cos2β0\cos^2 \beta \ge 0). So, cosβ=±12\cos \beta = \pm \frac{1}{\sqrt{2}}.

Now, we use the condition sinαcosβ0\sin \alpha \cos \beta \ge 0. Since sinα=1\sin \alpha = 1 (which is positive), we must have cosβ0\cos \beta \ge 0. Therefore, cosβ=12\cos \beta = \frac{1}{\sqrt{2}}. For cosβ=12\cos \beta = \frac{1}{\sqrt{2}} in the interval [0,π][0, \pi], we get β=π4\beta = \frac{\pi}{4}.

Step 5: Calculate the Required Expression We need to find the value of cos(α+β)cos(αβ)\cos(\alpha + \beta) - \cos(\alpha - \beta). Using the trigonometric identity cos(A+B)cos(AB)=2sinAsinB\cos(A+B) - \cos(A-B) = -2 \sin A \sin B: Let A=αA = \alpha and B=βB = \beta. Then, cos(α+β)cos(αβ)=2sinαsinβ\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2 \sin \alpha \sin \beta.

Substitute the values of α\alpha and β\beta we found: α=π2    sinα=sinπ2=1\alpha = \frac{\pi}{2} \implies \sin \alpha = \sin \frac{\pi}{2} = 1. β=π4    sinβ=sinπ4=12\beta = \frac{\pi}{4} \implies \sin \beta = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}.

Therefore, cos(α+β)cos(αβ)=2(1)(12)=22=2\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2 \cdot (1) \cdot \left(\frac{1}{\sqrt{2}}\right) = -\frac{2}{\sqrt{2}} = -\sqrt{2}.

Common Mistakes & Tips

  • Incorrect AM-GM Term Selection: Splitting the constant term '2' into '1+1' is crucial for the geometric mean to simplify correctly.
  • Ignoring Equality Conditions: The equality condition of AM-GM is not just about terms being equal; it also implies that the expression on the right side of the given equation must be consistent with the inequality (i.e., non-negative in this case).
  • Domain Issues: Carefully use the given domain [0,π][0, \pi] for α\alpha and β\beta to determine the correct signs and values of their trigonometric functions. For example, sinα=±1\sin \alpha = \pm 1 becomes sinα=1\sin \alpha = 1 due to α[0,π]\alpha \in [0, \pi].
  • Absolute Value in Geometric Mean: Remember that x44=x\sqrt[4]{x^4} = |x|. The context of the problem (given equality) helps resolve the absolute value.

Summary

This problem is elegantly solved using the AM-GM inequality. By applying AM-GM to carefully chosen terms, we establish an inequality that, when compared to the given equation, forces the equality condition of AM-GM to hold. This equality condition allows us to determine the specific values of α\alpha and β\beta. Finally, a standard trigonometric identity is used to compute the required expression with the found values of α\alpha and β\beta.

The final answer is 2\boxed{-\sqrt{2}}.

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