Question

Determine the value of \sin^2 20\degree +\cos^2 50\degree +\sin 20\degree \cos 50\degree

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

Using power reducing, product to sum and sum to product identities

\sin^2\alpha = \dfrac{1-\cos (2\alpha)}{2}

\cos^2\alpha = \dfrac{1+\cos (2\alpha)}{2}


\sin \alpha\cos \beta = \dfrac{1}{2} [\sin(\alpha+\beta ) +\sin(\alpha - \beta) ]

\cos \alpha - \cos \beta = -2\sin\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

\sin^2 20\degree +\cos^2 50\degree +\sin 20\degree \cos 50\degree

= \dfrac{1-\cos 40\degree }{2}+ \dfrac{1+\cos 100\degree }{2} +\dfrac{1}{2}[\sin 70\degree +\sin( -30\degree ) ]

= 1-\dfrac{1}{4}+\dfrac{\cos 100\degree-\cos 40\degree}{2} +\dfrac{1}{2}\sin 70\degree

=\dfrac{3}{4}- \sin 70\degree \sin 30\degree +\dfrac{1}{2}\sin 70\degree

=\dfrac{3}{4}

Steven Zheng posted 1 month ago

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