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AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 10 of 160

Marks: +1, -0

if x +

\frac{1}{x}

= 2 sin α and y +

\frac{1}{y}

= 2 cos β , then

x^{3}y^{3}+\frac{1}{x^{3}y^{3}}

=

2 cos 3 (β - α)
2 cos 3 (β + α)
2 sin 3 (β - α)
2 sin 3 (β + α)

Solution:

Consider the expression,

x+\frac{1}{x}=2\sin\alpha

This implies,

x=\sin\alpha+i\cos\alpha

=\cos\left(\frac{\pi}{2}-\alpha\right)+i\sin\left(\frac{\pi}{2}-\alpha\right)

=e^{t\left(\frac{\pi}{2}-\alpha\right)}

And,

y+\frac{1}{y}=2\cos\beta

y=\cos\beta+i\sin\beta

=e^{t\beta}

Now,

x^{3}y^{3}=e^{i\left(\frac{3\pi}{2}-3a\right)}\cdot e^{i3\beta}

=e^{i\left(\frac{3\pi}{2}+3\beta-3\alpha\right)}

=\cos\left(\frac{3\pi}{2}+3\beta-3\alpha\right)+i\sin\left(\frac{3\pi}{2}\right)

+3\beta-3\alpha )

=\sin(3\beta-3\alpha)-i\cos(3\beta-3\alpha)

So,

\frac{1}{x^{3}y^{3}}=\sin(3\beta-3\alpha)+i\cos(3\beta-3\alpha)

\frac{1}{x^{3}y^{3}}+x^{3}y^{3}=2\sin(3\beta-3\alpha)

=2\sin3(\beta-\alpha)

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