If $\cos \left(x-\frac{\pi}{3}\right), \cos x, \cos \left(x+\frac{\pi}{3}\right)$ are in a harmonic…
If $\cos \left(x-\frac{\pi}{3}\right), \cos x, \cos \left(x+\frac{\pi}{3}\right)$ are in a harmonic progression, then $\cos x=$
- $\frac{3}{2}$
- 1
- $\frac{\sqrt{3}}{2}$
- $\sqrt{\frac{3}{2}}$
Solution
Let
\(a=\cos \left(x-\frac{\pi}{3}\right), \quad b=\cos x, \quad c=\cos \left(x+\frac{\pi}{3}\right)\)
If \(a, b, c\) are in HP, then their reciprocals are in AP:
\(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)
So,
\(\frac{2}{\cos x}=\frac{1}{\cos (x-\pi / 3)}+\frac{1}{\cos (x+\pi / 3)}\)
\(\frac{1}{\cos A}+\frac{1}{\cos B}=\frac{2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}}{\cos A \cos B}\)
Here,
\(\begin{gathered}
A=x-\frac{\pi}{3}, \quad B=x+\frac{\pi}{3} \\
\frac{A+B}{2}=x, \quad \frac{A-B}{2}=-\frac{\pi}{3} \\
\cos \left(-\frac{\pi}{3}\right)=\frac{1}{2}
\end{gathered}\)
So RHS becomes:
\(\frac{2 \cos x \cdot \frac{1}{2}}{\cos (x-\pi / 3) \cos (x+\pi / 3)}=\frac{\cos x}{\cos (x-\pi / 3) \cos (x+\pi / 3)}\)
\(\begin{gathered}
\frac{2}{\cos x}=\frac{\cos x}{\cos (x-\pi / 3) \cos (x+\pi / 3)} \\
2 \cos (x-\pi / 3) \cos (x+\pi / 3)=\cos ^2 x
\end{gathered}\)
step 4. Wae ide-n'ty
\(\cos (A-B) \cos (A+B)=\cos ^2 A-\sin ^2 B\)
Here \(B=\frac{\pi}{3}\) :
\(\cos (x-\pi / 3) \cos (x+\pi / 3)=\cos ^2 x-\sin ^2 \frac{\pi}{3}=\cos ^2 x-\frac{3}{4}\)
Substitute:
\(\begin{gathered}
2\left(\cos ^2 x-\frac{3}{4}\right)=\cos ^2 x \\
2 \cos ^2 x-\frac{3}{2}=\cos ^2 x \\
\cos ^2 x=\frac{3}{2}
\end{gathered}\)
Final Answer
\(\cos x=\sqrt{\frac{3}{2}}\)
Asked in: MHT CET Full Test 7
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