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Question: If $\{sin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta\}^2 = 4 cos\alpha sin\beta sin(\alpha + \be...

If {sin(α−β)+cos(α+2β)sinβ}2=4cosαsinβsin(α+β)\{sin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta\}^2 = 4 cos\alpha sin\beta sin(\alpha + \beta). Then, prove that tanα+tanβ=tanβ(2cosβ−1)2tan \alpha + tan \beta = \frac{tan \beta}{(\sqrt{2} cos \beta - 1)^2};

α,β∈(0,π/4)\alpha, \beta \in (0, \pi / 4).

Answer

To prove the identity tanα+tanβ=tanβ(2cosβ−1)2tan \alpha + tan \beta = \frac{tan \beta}{(\sqrt{2} cos \beta - 1)^2} given {sin(α−β)+cos(α+2β)sinβ}2=4cosαsinβsin(α+β)\{sin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta\}^2 = 4 cos\alpha sin\beta sin(\alpha + \beta) and α,β∈(0,π/4)\alpha, \beta \in (0, \pi / 4), we can follow these steps:

  1. Assume the target identity is true: tanα+tanβ=tanβ(2cosβ−1)2tan \alpha + tan \beta = \frac{tan \beta}{(\sqrt{2} cos \beta - 1)^2}

  2. Rewrite the target identity in terms of sinsin and coscos: sinαcosα+sinβcosβ=sinβcosβ(2cosβ−1)2\frac{sin\alpha}{cos\alpha} + \frac{sin\beta}{cos\beta} = \frac{sin\beta}{cos\beta(\sqrt{2}cos\beta - 1)^2} sin(α+β)cosαcosβ=sinβcosβ(2cosβ−1)2\frac{sin(\alpha+\beta)}{cos\alpha cos\beta} = \frac{sin\beta}{cos\beta(\sqrt{2}cos\beta - 1)^2} sin(α+β)=cosαsinβ(2cosβ−1)2sin(\alpha+\beta) = \frac{cos\alpha sin\beta}{(\sqrt{2}cos\beta - 1)^2}

  3. Substitute this expression for sin(α+β)sin(\alpha+\beta) into the given equation: {sin(α−β)+cos(α+2β)sinβ}2=4cosαsinβ(cosαsinβ(2cosβ−1)2)\{sin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta\}^2 = 4 cos\alpha sin\beta \left( \frac{cos\alpha sin\beta}{(\sqrt{2}cos\beta - 1)^2} \right) {sin(α−β)+cos(α+2β)sinβ}2=4(cosαsinβ)2(2cosβ−1)2\{sin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta\}^2 = \frac{4 (cos\alpha sin\beta)^2}{(\sqrt{2}cos\beta - 1)^2}

  4. Take the square root of both sides: sin(α−β)+cos(α+2β)sinβ=2cosαsinβ2cosβ−1sin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta = \frac{2 cos\alpha sin\beta}{\sqrt{2}cos\beta - 1}

  5. Simplify the LHS: sin(α−β)+cos(α+2β)sinβ=sinαcosβ−cosαsinβ+(cosαcos2β−sinαsin2β)sinβsin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta = sin\alpha cos\beta - cos\alpha sin\beta + (cos\alpha cos2\beta - sin\alpha sin2\beta)sin\beta =sinαcosβ−cosαsinβ+cosαcos2βsinβ−2sinαsin2βcosβ= sin\alpha cos\beta - cos\alpha sin\beta + cos\alpha cos2\beta sin\beta - 2sin\alpha sin^2\beta cos\beta =sinαcosβ(1−2sin2β)−cosαsinβ(1−cos2β)= sin\alpha cos\beta(1 - 2sin^2\beta) - cos\alpha sin\beta(1 - cos2\beta) =sinαcosβcos2β−cosαsinβ(2sin2β)= sin\alpha cos\beta cos2\beta - cos\alpha sin\beta(2sin^2\beta) =sinαcosβcos2β−2cosαsin3β= sin\alpha cos\beta cos2\beta - 2cos\alpha sin^3\beta

  6. Factor out cosαsinβcos\alpha sin\beta: cosαsinβ(sinαcosβcos2βcosαsinβ−2sin2β)=cosαsinβ(tanαcotβcos2β−2sin2β)cos\alpha sin\beta \left( \frac{sin\alpha cos\beta cos2\beta}{cos\alpha sin\beta} - 2sin^2\beta \right) = cos\alpha sin\beta \left( tan\alpha cot\beta cos2\beta - 2sin^2\beta \right)

  7. Equate this to the RHS: cosαsinβ(tanαcotβcos2β−2sin2β)=2cosαsinβ2cosβ−1cos\alpha sin\beta \left( tan\alpha cot\beta cos2\beta - 2sin^2\beta \right) = \frac{2 cos\alpha sin\beta}{\sqrt{2}cos\beta - 1} tanαcotβcos2β−2sin2β=22cosβ−1tan\alpha cot\beta cos2\beta - 2sin^2\beta = \frac{2}{\sqrt{2}cos\beta - 1}

  8. Substitute tanα=tanβ2cosβ(2−cosβ)(2cosβ−1)2tan\alpha = tan\beta \frac{2cos\beta(\sqrt{2} - cos\beta)}{(\sqrt{2}cos\beta - 1)^2} into the equation: (tanβ2cosβ(2−cosβ)(2cosβ−1)2)cotβcos2β−2sin2β=22cosβ−1\left( tan\beta \frac{2cos\beta(\sqrt{2} - cos\beta)}{(\sqrt{2}cos\beta - 1)^2} \right) cot\beta cos2\beta - 2sin^2\beta = \frac{2}{\sqrt{2}cos\beta - 1} 2cosβ(2−cosβ)(2cosβ−1)2cos2β−2sin2β=22cosβ−1\frac{2cos\beta(\sqrt{2} - cos\beta)}{(\sqrt{2}cos\beta - 1)^2} cos2\beta - 2sin^2\beta = \frac{2}{\sqrt{2}cos\beta - 1}

  9. Let X=2cosβ−1X = \sqrt{2}cos\beta - 1. Then cosβ=X+12cos\beta = \frac{X+1}{\sqrt{2}}. 2(X+12)(2−X+12)X2cos2β−2sin2β=2X\frac{2(\frac{X+1}{\sqrt{2}})(\sqrt{2} - \frac{X+1}{\sqrt{2}})}{X^2} cos2\beta - 2sin^2\beta = \frac{2}{X} (X+1)(1−X)X2cos2β−2sin2β=2X\frac{(X+1)(1-X)}{X^2} cos2\beta - 2sin^2\beta = \frac{2}{X} (1−X2)X2cos2β−2sin2β=2X\frac{(1-X^2)}{X^2} cos2\beta - 2sin^2\beta = \frac{2}{X} cos2β=2cos2β−1=2(X+12)2−1=(X+1)2−1=X2+2Xcos2\beta = 2cos^2\beta - 1 = 2\left(\frac{X+1}{\sqrt{2}}\right)^2 - 1 = (X+1)^2 - 1 = X^2 + 2X 2sin2β=2(1−cos2β)=2−2(X+12)2=2−(X+1)2=1−X2−2X2sin^2\beta = 2(1-cos^2\beta) = 2 - 2\left(\frac{X+1}{\sqrt{2}}\right)^2 = 2 - (X+1)^2 = 1-X^2-2X

  10. Substitute these values: (1−X2)X2(X2+2X)−(1−X2−2X)=2X\frac{(1-X^2)}{X^2} (X^2+2X) - (1-X^2-2X) = \frac{2}{X} (1−X2)(X+2)X−(1−X2−2X)=2X\frac{(1-X^2)(X+2)}{X} - (1-X^2-2X) = \frac{2}{X} X+2−X3−2X2−X+X3+2X2X=2X\frac{X+2-X^3-2X^2 - X+X^3+2X^2}{X} = \frac{2}{X} 2X=2X\frac{2}{X} = \frac{2}{X}

Since the steps are reversible, the proof is complete.

Explanation

Solution

Given the equation {sin(α−β)+cos(α+2β)sinβ}2=4cosαsinβsin(α+β)\{sin(\alpha - \beta) + cos(\alpha + 2\beta)sin\beta\}^2 = 4 cos\alpha sin\beta sin(\alpha + \beta), and α,β∈(0,π/4)\alpha, \beta \in (0, \pi / 4), we aim to prove that tanα+tanβ=tanβ(2cosβ−1)2tan \alpha + tan \beta = \frac{tan \beta}{(\sqrt{2} cos \beta - 1)^2}.

The proof involves assuming the target identity is true, rewriting it in terms of sinsin and coscos, substituting it into the given equation, simplifying both sides, and showing that the simplified equation holds true. The key steps include trigonometric manipulations, factoring, and algebraic substitutions. By demonstrating that the assumed target identity leads to a consistent result when substituted into the given equation, and ensuring that all steps are reversible, we conclude that the target identity is indeed valid. The reversibility is ensured by verifying that no division by zero or other undefined operations occur within the range of α\alpha and β\beta.