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Question: Given \(0<\alpha ,\beta <\pi \) and \(\cos \alpha +\cos \beta -\cos (\alpha -\beta )=\frac{3}{2}\)...

Given 0<α,β<π0<\alpha ,\beta <\pi and cos⁡α+cos⁡β−cos⁡(α−β)=32\cos \alpha +\cos \beta -\cos (\alpha -\beta )=\frac{3}{2}

Answer

No Solution

Explanation

Solution

The given equation is: cos⁡α+cos⁡β−cos⁡(α−β)=32\cos \alpha + \cos \beta - \cos (\alpha -\beta )=\frac{3}{2}

First, expand cos⁡(α−β)\cos(\alpha - \beta): cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

Substitute this into the equation: cos⁡α+cos⁡β−(cos⁡αcos⁡β+sin⁡αsin⁡β)=32\cos \alpha + \cos \beta - (\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \frac{3}{2} cos⁡α+cos⁡β−cos⁡αcos⁡β−sin⁡αsin⁡β=32\cos \alpha + \cos \beta - \cos \alpha \cos \beta - \sin \alpha \sin \beta = \frac{3}{2}

To simplify, multiply the entire equation by 2: 2cos⁡α+2cos⁡β−2cos⁡αcos⁡β−2sin⁡αsin⁡β=32\cos \alpha + 2\cos \beta - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta = 3

Rearrange the terms and move the constant to the left side: 2cos⁡α+2cos⁡β−2cos⁡αcos⁡β−2sin⁡αsin⁡β−3=02\cos \alpha + 2\cos \beta - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta - 3 = 0

Consider the expression: (cos⁡α−1)2+(cos⁡β−1)2+(sin⁡α−sin⁡β)2=0(\cos \alpha - 1)^2 + (\cos \beta - 1)^2 + (\sin \alpha - \sin \beta)^2 = 0

This identity implies that each term must be zero, because squares of real numbers are non-negative. So, cos⁡α−1=0  ⟹  cos⁡α=1\cos \alpha - 1 = 0 \implies \cos \alpha = 1 cos⁡β−1=0  ⟹  cos⁡β=1\cos \beta - 1 = 0 \implies \cos \beta = 1 sin⁡α−sin⁡β=0  ⟹  sin⁡α=sin⁡β\sin \alpha - \sin \beta = 0 \implies \sin \alpha = \sin \beta

Let's check if this satisfies the given conditions 0<α,β<π0 < \alpha, \beta < \pi. If cos⁡α=1\cos \alpha = 1, then α=2nπ\alpha = 2n\pi for integer nn. Since 0<α<π0 < \alpha < \pi, there is no value of α\alpha such that cos⁡α=1\cos \alpha = 1. Similarly, if cos⁡β=1\cos \beta = 1, then β=2mπ\beta = 2m\pi for integer mm. Since 0<β<π0 < \beta < \pi, there is no value of β\beta such that cos⁡β=1\cos \beta = 1.

This means that the expression (cos⁡α−1)2+(cos⁡β−1)2+(sin⁡α−sin⁡β)2(\cos \alpha - 1)^2 + (\cos \beta - 1)^2 + (\sin \alpha - \sin \beta)^2 cannot be zero under the given conditions.

Let's verify the expansion of (cos⁡α−1)2+(cos⁡β−1)2+(sin⁡α−sin⁡β)2(\cos \alpha - 1)^2 + (\cos \beta - 1)^2 + (\sin \alpha - \sin \beta)^2: cos⁡2α−2cos⁡α+1+cos⁡2β−2cos⁡β+1+sin⁡2α−2sin⁡αsin⁡β+sin⁡2β\cos^2 \alpha - 2\cos \alpha + 1 + \cos^2 \beta - 2\cos \beta + 1 + \sin^2 \alpha - 2\sin \alpha \sin \beta + \sin^2 \beta =(cos⁡2α+sin⁡2α)+(cos⁡2β+sin⁡2β)−2cos⁡α−2cos⁡β−2sin⁡αsin⁡β+2= (\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) - 2\cos \alpha - 2\cos \beta - 2\sin \alpha \sin \beta + 2 =1+1−2cos⁡α−2cos⁡β−2sin⁡αsin⁡β+2= 1 + 1 - 2\cos \alpha - 2\cos \beta - 2\sin \alpha \sin \beta + 2 =4−2cos⁡α−2cos⁡β−2sin⁡αsin⁡β= 4 - 2\cos \alpha - 2\cos \beta - 2\sin \alpha \sin \beta

Now, compare this with the equation derived from the problem: 2cos⁡α+2cos⁡β−2cos⁡αcos⁡β−2sin⁡αsin⁡β−3=02\cos \alpha + 2\cos \beta - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta - 3 = 0 So, −2cos⁡α−2cos⁡β−2sin⁡αsin⁡β=3−2cos⁡αcos⁡β-2\cos \alpha - 2\cos \beta - 2\sin \alpha \sin \beta = 3 - 2\cos \alpha \cos \beta

Substitute this into the expanded identity: 4+(3−2cos⁡αcos⁡β)=04 + (3 - 2\cos \alpha \cos \beta) = 0 7−2cos⁡αcos⁡β=07 - 2\cos \alpha \cos \beta = 0 2cos⁡αcos⁡β=72\cos \alpha \cos \beta = 7 cos⁡αcos⁡β=72\cos \alpha \cos \beta = \frac{7}{2}

This is impossible, as the maximum value for cos⁡αcos⁡β\cos \alpha \cos \beta is 1×1=11 \times 1 = 1. Since 72>1\frac{7}{2} > 1, there are no real values of α\alpha and β\beta that satisfy this condition.

Therefore, the given equation has no solution for 0<α,β<π0 < \alpha, \beta < \pi.