MHT CET · Maths · Trigonometric Equations
If \(2 \cos \theta=x+\frac{1}{x}\), then \(2 \cos 3 \theta=\)
- A \(x^3-\frac{1}{x^3}\)
- B \(\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^3\)
- C \(x+\frac{1}{x}\)
- D \(x^3+\frac{1}{x^3}\)
Answer & Solution
Correct Answer
(D) \(x^3+\frac{1}{x^3}\)
Step-by-step Solution
Detailed explanation
\(\text {We have } \cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right) \)
\( 2 \cos 3 \theta \)
\( =2[4 \cos ^3 \theta-3 \cos \theta]=2\{4[(\frac{1}{2})(x+\frac{1}{x})]^3\)\( -[3(\frac{1}{2})(x+\frac{1}{x})]\} \)
\( =2\left[\left(\frac{1}{2}\right)\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\right]-3\left(x+\frac{1}{x}\right)=\) \(x^3+\frac{1}{x^3}\)
\( 2 \cos 3 \theta \)
\( =2[4 \cos ^3 \theta-3 \cos \theta]=2\{4[(\frac{1}{2})(x+\frac{1}{x})]^3\)\( -[3(\frac{1}{2})(x+\frac{1}{x})]\} \)
\( =2\left[\left(\frac{1}{2}\right)\left(x^3+\frac{1}{x^3}\right)+3\left(x+\frac{1}{x}\right)\right]-3\left(x+\frac{1}{x}\right)=\) \(x^3+\frac{1}{x^3}\)
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