If $x+\frac{1}{x}=2 \sin \alpha$ and $y+\frac{1}{y}=2 \cos \beta$, then $x^3 y^3+\frac{1}{x^3 y^3}=$

Question

Aicharya · Accepted Answer

It is given that $\begin{array}{l} x+\frac{1}{x}=2 \sin \alpha \Rightarrow x=\sin \alpha+i \cos \alpha=\cos \left(\frac{\pi}{2}-\alpha\right) +i \sin \left(\frac{\pi}{2}-\alpha\right) \ \Rightarrow \quad x=e^{i\left(\frac{\pi}{2}-\alpha\right)} \end{array}$ and $y+\frac{1}{y}=2 \cos \beta \Rightarrow y=\cos \beta+i \sin \beta=e^{i \beta}$ Now, $x^3 y^3=e^{i\left(\frac{3 \pi}{2}-3 \alpha\right)} \cdot e^{i 3 \beta}=e^{i\left(\frac{3 \pi}{2}+(3 \beta-3 \alpha)\right)}$ $\begin{aligned} & =\cos \left(\frac{3 \pi}{2}+3 \beta-3 \alpha\right)+i \sin \left(\frac{3 \pi}{2}+3 \beta-3 \alpha\right) \ & =\sin (3 \beta-3 \alpha)-i \cos (3 \beta-3 \alpha) \end{aligned}$ and $\frac{1}{x^3 y^3}=\sin (3 \beta-3 \alpha)+i \cos (3 \beta-3 \alpha)$ So, $x^3 y^3+\frac{1}{x^3 y^3}=2 \sin (3 \beta-3 \alpha)=2 \sin 3(\beta-\alpha)$ Hence, option (c) is correct.

If \(x+\frac{1}{x}=2 \sin \alpha\) and \(y+\frac{1}{y}=2 \cos \beta\), then \(x^3 y^3+\frac{1}{x^3 y^3}=\)

Solution