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if $\displaystyle \frac{\sin^3 \beta}{\sin(\beta - 3 \alpha)} = \frac{\cos^3 \beta}{\cos(\beta-3 \alpha)} = \lambda\;,$ then $2\lambda^2-1 = $

options

$(a)\; \lambda \sin \beta$

$(b)\; \lambda \cos \beta$

$(c)\; 2\lambda \cos \beta$

$(d)\; \lambda \tan \beta$

Attempt with ratio and proportion

$\displaystyle \frac{\sin^4 \beta}{\sin \beta \sin(\beta - 3 \alpha)} = \frac{\cos^4 \beta}{\cos \beta \cos(\beta-3 \alpha)} = \frac{\cos^4 \beta - \sin^4 \beta}{\cos \beta \cos(\beta-3 \alpha)-\sin \beta \sin(\beta - 3 \alpha)}$

so $\displaystyle \frac{\sin^4 \beta}{\sin \beta \sin(\beta - 3 \alpha)} = \frac{\cos^4 \beta}{\cos \beta \cos(\beta-3 \alpha)} = \frac{\cos 2 \beta}{\cos (2\beta-3 \alpha)}$

wan,t be able to go further ,could some help me

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    Just check the options with $\beta = 45^\circ$ and $\alpha =0^\circ $. – 2017-02-05

1 Answers 1

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HINT:

Clearly we need to eliminate $\alpha$

Expand $\sin(\beta-3\alpha),\cos(\beta-3\alpha)$ to form two simultaneous equations in $\sin(3\alpha),\cos(3\alpha)$

Solve for $\sin(3\alpha),\cos(3\alpha)$

Use $\sin^2(3\alpha)+\cos^2(3\alpha)=1$ to eliminate $\alpha$