Question:
If \(\sin^4\alpha+4\cos^4\beta+2=4\sqrt2\,\sin\alpha\cos\beta;\ \alpha,\beta\in[0,\pi]\), then \(\cos(\alpha+\beta)-\cos(\alpha-\beta)\) is equal to :
If \(\sin^4\alpha+4\cos^4\beta+2=4\sqrt2\,\sin\alpha\cos\beta;\ \alpha,\beta\in[0,\pi]\), then \(\cos(\alpha+\beta)-\cos(\alpha-\beta)\) is equal to :
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Apply AM-GM to the four terms \(\sin^4\alpha,\ 4\cos^4\beta,\ 1,\ 1\); equality forces \(\sin\alpha=1\) and \(\cos\beta=\frac{1}{\sqrt2}\).
Updated On: Nov 5, 2026
- \(-\sqrt2\)
- 0
- -1
- \(\sqrt2\)
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The Correct Option is A
Solution and Explanation
By AM-GM on four non-negative terms: \(\sin^4\alpha+4\cos^4\beta+1+1\ge4\left(\sin^4\alpha\cdot4\cos^4\beta\right)^{1/4}=4\sqrt2\,\sin\alpha\cos\beta\) (for \(\sin\alpha\cos\beta\ge0\)).
The given relation is the equality case, so \(\sin^4\alpha=4\cos^4\beta=1\). Thus \(\sin\alpha=1\) (\(\alpha=\frac\pi2\), as \(\alpha\in[0,\pi]\)) and \(\cos^2\beta=\frac12\); the right side must be positive, so \(\cos\beta=+\frac{1}{\sqrt2}\), \(\beta=\frac\pi4\).
\(\cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta=-2(1)\left(\frac{1}{\sqrt2}\right)=-\sqrt2\).
Answer: (1) \(-\sqrt2\)
The given relation is the equality case, so \(\sin^4\alpha=4\cos^4\beta=1\). Thus \(\sin\alpha=1\) (\(\alpha=\frac\pi2\), as \(\alpha\in[0,\pi]\)) and \(\cos^2\beta=\frac12\); the right side must be positive, so \(\cos\beta=+\frac{1}{\sqrt2}\), \(\beta=\frac\pi4\).
\(\cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta=-2(1)\left(\frac{1}{\sqrt2}\right)=-\sqrt2\).
Answer: (1) \(-\sqrt2\)
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