If sin 4 α + 4 cos 4 β + 2 = 42 sin α cos β; α, β∈ [0, π], then cos(α + β) − cos(α−β) 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,…,an, na1+a2+⋯+an≥na1a2…an. Equality holds if and only if a1=a2=⋯=an.
Trigonometric Identities:
sin2x+cos2x=1
cos(A+B)−cos(A−B)=−2sinAsinB
Properties of Trigonometric Functions: Understanding the signs and values of sinx and cosx in the interval [0,π].
Step-by-Step Solution
Step 1: Analyze the Given Equation and Prepare for AM-GM
The given equation is sin4α+4cos4β+2=42sinαcosβ, with α,β∈[0,π].
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. sin4α≥0, 4cos4β≥0, and 2>0. The constant term '2' can be split into 1+1 to facilitate the application of AM-GM with four terms.
Let the four non-negative terms be a1=sin4α, a2=4cos4β, a3=1, and a4=1.
Step 2: Apply the AM-GM Inequality
Applying the AM-GM inequality to these four terms:
4sin4α+4cos4β+1+1≥4sin4α⋅4cos4β⋅1⋅1
Simplifying the inequality:
4sin4α+4cos4β+2≥44sin4αcos4β4sin4α+4cos4β+2≥24sin4αcos4β
Since 4x4=∣x∣, we have:
4sin4α+4cos4β+2≥2∣sinαcosβ∣
Multiplying both sides by 4:
sin4α+4cos4β+2≥42∣sinαcosβ∣
Step 3: Utilize the Equality Condition of AM-GM
We are given that sin4α+4cos4β+2=42sinαcosβ.
Comparing this with the result from AM-GM, sin4α+4cos4β+2≥42∣sinαcosβ∣, we see that the equality must hold in the AM-GM inequality.
For equality to hold, two conditions must be met:
All the terms in the AM-GM inequality must be equal: sin4α=4cos4β=1.
The term sinαcosβ must be non-negative, so that ∣sinαcosβ∣=sinαcosβ. 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 42sinαcosβ must also be non-negative.
Step 4: Solve for α and β
From the equality condition sin4α=1:
Since α∈[0,π], sinα≥0. Thus, sinα=1.
For sinα=1 in the interval [0,π], we get α=2π.
From the equality condition 4cos4β=1:
cos4β=41.
Taking the square root of both sides, cos2β=41=21 (since cos2β≥0).
So, cosβ=±21.
Now, we use the condition sinαcosβ≥0.
Since sinα=1 (which is positive), we must have cosβ≥0.
Therefore, cosβ=21.
For cosβ=21 in the interval [0,π], we get β=4π.
Step 5: Calculate the Required Expression
We need to find the value of cos(α+β)−cos(α−β).
Using the trigonometric identity cos(A+B)−cos(A−B)=−2sinAsinB:
Let A=α and B=β.
Then, cos(α+β)−cos(α−β)=−2sinαsinβ.
Substitute the values of α and β we found:
α=2π⟹sinα=sin2π=1.
β=4π⟹sinβ=sin4π=21.
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] for α and β to determine the correct signs and values of their trigonometric functions. For example, sinα=±1 becomes sinα=1 due to α∈[0,π].
Absolute Value in Geometric Mean: Remember that 4x4=∣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 and β. Finally, a standard trigonometric identity is used to compute the required expression with the found values of α and β.