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 If  x+\frac{1}{x}=2 \cos \theta, then x^{3}+\frac{1}{x^{3}}  ?                

Option: 1

\cos 3\theta


Option: 2

2 \cos 3\theta


Option: 3

\frac{1}{2} \cos 3\theta


Option: 4

\frac{1}{3} \cos 3\theta


Answers (1)

best_answer

Trigonometric Ratio for Compound Angles (Some more Result} -

Trigonometric Ratio for Compound Angles (Some more Result}

 

\begin{array}{l}{\text { 1. } \sin (A+B) \sin (A-B)=\sin ^{2} A-\sin ^{2} B=\cos ^{2} B-\cos ^{2} A} \\ {\text { 2. } \cos (A+B) \cos (A-B)=\cos ^{2} A-\sin ^{2} B=\cos ^{2} B-\sin ^{2} A} \\ {\text { 3. } \sin (A+B+C)=\sin A \cos B \cos C+\cos A \sin B \cos C+\cos A \cos B \sin C-\sin A \sin B \cos C} \\ {\text { 4. } \cos (A+B+C)=\cos A \cos B \cos C-\cos A \sin B \sin A \cos B \sin C-\sin A \sin B \cos C} \end{array}{\text { 5. } \tan (A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A \tan B \tan C}{1-\tan A \tan B-\tan B \tan C-\tan C \tan A}}

 

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(x+\frac{1}{x})^3= x^3 +\frac{1}{x^3}+3x \cdot \frac{1}{x}(x+\frac{1}{x})\ \ \ \ \because (x+y)^3=x^3+y^3+3xy(x+y)\\ x^3 +\frac{1}{x^3}=(x+\frac{1}{x})^3-3x \cdot \frac{1}{x}(x+\frac{1}{x})\ \ \ \text{Given } (x+\frac{1}{x})=2 \cos \theta \\ x^3 +\frac{1}{x^3}=(2 \cos \theta)^3 -3 \cdot 2 \cos \theta \\ \text{ \ \ \ \ \ \ \ \ \ \ } =2[ 4 \cos^3 \theta-3 \cos \theta ]\\ \text{ \ \ \ \ \ \ \ \ \ \ } =2 \cos 3\theta

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