Question

Given that \tan \alpha \cdotp \tan \beta = \dfrac{\sqrt{3} }{3}, find the value of (2-\cos 2\alpha )(2-\cos 2\beta)

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

\cos 2\alpha = \dfrac{1-\tan^2 \alpha }{1+\tan^2 \alpha}

2-\cos 2\alpha = 2-\dfrac{1-\tan^2 \alpha }{1+\tan^2 \alpha}

=\dfrac{2+2\tan^2 \alpha-1+\tan^2 \alpha}{1+\tan^2 \alpha}

=\dfrac{1+3 \tan^2 \alpha}{1+\tan^2 \alpha}

Similarly,

2-\cos 2\beta = \dfrac{1+3 \tan^2 \beta }{1+\tan^2 \beta }

(2-\cos 2\alpha )(2-\cos 2\beta) =\dfrac{1+3 \tan^2 \alpha}{1+\tan^2 \alpha}\cdotp \dfrac{1+3 \tan^2 \beta }{1+\tan^2 \beta }

=\dfrac{1+9\tan^2 \alpha\tan^2 \beta+3\tan^2 \alpha+3\tan^2 \beta}{1+\tan^2 \alpha\tan^2 \beta+\tan^2 \alpha+\tan^2 \beta}

=\dfrac{4+3\tan^2 \alpha+3\tan^2 \beta}{\dfrac{4}{3} +\tan^2 \alpha+\tan^2 \beta } =3

Steven Zheng posted 1 month ago

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