Among the following functions defined on $\mathbb{R}$ into $\mathbb{R}$, the constant function is
Among the following functions defined on $\mathbb{R}$ into $\mathbb{R}$, the constant function is
$\frac{3}{5+4 \sin 3 x}$
$\frac{1}{2-\cos 3 x}$
$\cos ^2 x+\cos ^2\left(x+\frac{\pi}{3}\right)+\sin x \cdot \sin \left(x+\frac{\pi}{3}\right)$
$\frac{15}{3 \sin x+4 \cos x+10}$
Solution
The given expression is \(\cos ^{2}x+\cos ^{2}\left(x+\frac{\pi }{3}\right)+\sin x\cdot \sin \left(x+\frac{\pi }{3}\right)\).
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Step 1: Apply the power-reducing formula for cosine squared
The identity \(\cos ^{2}\theta =\frac{1+\cos (2\theta )}{2}\) is applied to the first two terms. The expression becomes \(\frac{1+\cos (2x)}{2}+\frac{1+\cos \left(2\left(x+\frac{\pi }{3}\right)\right)}{2}+\sin x\cdot \sin \left(x+\frac{\pi }{3}\right)\). This simplifies to \(\frac{1+\cos (2x)}{2}+\frac{1+\cos \left(2x+\frac{2\pi }{3}\right)}{2}+\sin x\cdot \sin \left(x+\frac{\pi }{3}\right)\).
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Step 2: Combine the constant terms and apply the product-to-sum formula for sine The constant terms are combined: \(\frac{1}{2}+\frac{1}{2}=1\). The product \(\sin A\cdot \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]\) is applied to the last term. Here, \(A=x\) and \(B=x+\frac{\pi }{3}\). So, \(\sin x\cdot \sin \left(x+\frac{\pi }{3}\right)=\frac{1}{2}\left[\cos \left(x-\left(x+\frac{\pi }{3}\right)\right)-\cos \left(x+\left(x+\frac{\pi }{3}\right)\right)\right]\).
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This simplifies to \(\frac{1}{2}\left[\cos \left(-\frac{\pi }{3}\right)-\cos \left(2x+\frac{\pi }{3}\right)\right]\). Since \(\cos (-\theta )=\cos (\theta )\), this becomes \(\frac{1}{2}\left[\cos \left(\frac{\pi }{3}\right)-\cos \left(2x+\frac{\pi }{3}\right)\right]\). As \(\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\), the term becomes \(\frac{1}{2}\left[\frac{1}{2}-\cos \left(2x+\frac{\pi }{3}\right)\right]=\frac{1}{4}-\frac{1}{2}\cos \left(2x+\frac{\pi }{3}\right)\).
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Step 3: Substitute back into the expression and group terms The expression now is \(1+\frac{1}{2}\cos (2x)+\frac{1}{2}\cos \left(2x+\frac{2\pi }{3}\right)+\frac{1}{4}-\frac{1}{2}\cos \left(2x+\frac{\pi }{3}\right)\). This can be rearranged as \(\frac{5}{4}+\frac{1}{2}\left[\cos (2x)+\cos \left(2x+\frac{2\pi }{3}\right)-\cos \left(2x+\frac{\pi }{3}\right)\right]\).
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Step 4: Apply the sum-to-product formula for cosine The identity \(\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\) is applied to \(\cos (2x)+\cos \left(2x+\frac{2\pi }{3}\right)\). Here, \(A=2x\) and \(B=2x+\frac{2\pi }{3}\). So, \(\cos (2x)+\cos \left(2x+\frac{2\pi }{3}\right)=2\cos \left(\frac{2x+2x+\frac{2\pi }{3}}{2}\right)\cos \left(\frac{2x-(2x+\frac{2\pi }{3})}{2}\right)\). This simplifies to \(2\cos \left(2x+\frac{\pi }{3}\right)\cos \left(-\frac{\pi }{3}\right)=2\cos \left(2x+\frac{\pi }{3}\right)\cos \left(\frac{\pi }{3}\right)\). Since \(\cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\), this term becomes \(2\cos \left(2x+\frac{\pi }{3}\right)\left(\frac{1}{2}\right)=\cos \left(2x+\frac{\pi }{3}\right)\).
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Step 5: Final simplification Substitute this back into the expression from Step 3: \(\frac{5}{4}+\frac{1}{2}\left[\cos \left(2x+\frac{\pi }{3}\right)-\cos \left(2x+\frac{\pi }{3}\right)\right]\). The terms inside the bracket cancel out, resulting in \(\frac{5}{4}+\frac{1}{2}(0)=\frac{5}{4}\).
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Final Answer
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The simplified value of the expression \(\cos ^{2}x+\cos ^{2}\left(x+\frac{\pi }{3}\right)+\sin x\cdot \sin \left(x+\frac{\pi }{3}\right)\) is \(\frac{5}{4}\).