C = sin (x+10°)cos (2x-80°) + sin (x+100°)cos (2x+10°) F = sin (x-π/3)cos (9π/4-x) – sin (5π/4-x)cos (5π/3+x) 05/09/2021 Bởi Cora C = sin (x+10°)cos (2x-80°) + sin (x+100°)cos (2x+10°) F = sin (x-π/3)cos (9π/4-x) – sin (5π/4-x)cos (5π/3+x)
$\begin{array}{l} C = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) + \sin \left( {x + {{100}^o}} \right)\cos \left( {2x + {{10}^o}} \right)\\ \;\;\; = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) + \sin \left( {{{90}^o} + x + {{10}^o}} \right)\cos \left( {{{90}^o} + 2x – {{80}^o}} \right) = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) + \cos \left( {x + {{10}^o}} \right)\left[ { – \sin \left( {2x – {{80}^o}} \right)} \right]\\ C = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) – \cos \left( {x + {{10}^o}} \right)\sin \left( {2x – {{80}^o}} \right) = \sin \left( {x + {{10}^o} – 2x + {{80}^o}} \right) = \sin \left( {{{90}^o} – x} \right) = \cos x \end{array}$ d)$\begin{array}{l} F = \sin \left( {x – \dfrac{\pi }{3}} \right)\cos \left( {\dfrac{{9\pi }}{4} – x} \right) – \sin \left( {\dfrac{{5\pi }}{4} – x} \right)\cos \left( {\dfrac{{5\pi }}{3} + x} \right) = \sin \left( {x – \dfrac{\pi }{3}} \right)\cos \left( {\dfrac{\pi }{4} – x} \right) – \sin \left( {\pi + \dfrac{\pi }{4} – x} \right)\cos \left( {2\pi – \dfrac{\pi }{3} + x} \right)\\ F = \sin \left( {x – \dfrac{\pi }{3}} \right)\cos \left( {\dfrac{\pi }{4} – x} \right) + \sin \left( {\dfrac{\pi }{4} – x} \right)\cos \left( {x – \dfrac{\pi }{3}} \right) = \sin \left( {x – \dfrac{\pi }{3} + \dfrac{\pi }{4} – x} \right) = \sin \left( {\dfrac{\pi }{4} – \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{4}\cos \dfrac{\pi }{3} – \sin \dfrac{\pi }{3}\cos \dfrac{\pi }{4} = \dfrac{{\sqrt 2 }}{2}.\dfrac{1}{2} – \dfrac{{\sqrt 2 }}{2}.\dfrac{{\sqrt 3 }}{2} = \dfrac{{ – \sqrt 6 + \sqrt 2 }}{4} \end{array}$ Bình luận
$\begin{array}{l} C = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) + \sin \left( {x + {{100}^o}} \right)\cos \left( {2x + {{10}^o}} \right)\\ \;\;\; = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) + \sin \left( {{{90}^o} + x + {{10}^o}} \right)\cos \left( {{{90}^o} + 2x – {{80}^o}} \right) = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) + \cos \left( {x + {{10}^o}} \right)\left[ { – \sin \left( {2x – {{80}^o}} \right)} \right]\\ C = \sin \left( {x + {{10}^o}} \right)\cos \left( {2x – {{80}^o}} \right) – \cos \left( {x + {{10}^o}} \right)\sin \left( {2x – {{80}^o}} \right) = \sin \left( {x + {{10}^o} – 2x + {{80}^o}} \right) = \sin \left( {{{90}^o} – x} \right) = \cos x \end{array}$
d)$\begin{array}{l} F = \sin \left( {x – \dfrac{\pi }{3}} \right)\cos \left( {\dfrac{{9\pi }}{4} – x} \right) – \sin \left( {\dfrac{{5\pi }}{4} – x} \right)\cos \left( {\dfrac{{5\pi }}{3} + x} \right) = \sin \left( {x – \dfrac{\pi }{3}} \right)\cos \left( {\dfrac{\pi }{4} – x} \right) – \sin \left( {\pi + \dfrac{\pi }{4} – x} \right)\cos \left( {2\pi – \dfrac{\pi }{3} + x} \right)\\ F = \sin \left( {x – \dfrac{\pi }{3}} \right)\cos \left( {\dfrac{\pi }{4} – x} \right) + \sin \left( {\dfrac{\pi }{4} – x} \right)\cos \left( {x – \dfrac{\pi }{3}} \right) = \sin \left( {x – \dfrac{\pi }{3} + \dfrac{\pi }{4} – x} \right) = \sin \left( {\dfrac{\pi }{4} – \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{4}\cos \dfrac{\pi }{3} – \sin \dfrac{\pi }{3}\cos \dfrac{\pi }{4} = \dfrac{{\sqrt 2 }}{2}.\dfrac{1}{2} – \dfrac{{\sqrt 2 }}{2}.\dfrac{{\sqrt 3 }}{2} = \dfrac{{ – \sqrt 6 + \sqrt 2 }}{4} \end{array}$