Rút gọn các biểu thức sau A= cos7x – cos8x – cos9x + cos10x / sin7x – sin8x – sin9x + sin10x B = sin2x + 2sin3x + sin4x / sin3x +2sin4x + sin5x C= 1+c

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1. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   \cos x + \cos y = 2\cos \dfrac{{x + y}}{2}.\cos \dfrac{{x – y}}{2}\\
   \cos x – \cos y =  – 2\sin \dfrac{{x + y}}{2}.\sin \dfrac{{x – y}}{2}\\
   \sin x + \sin y = 2\sin \dfrac{{x + y}}{2}.\cos \dfrac{{x – y}}{2}\\
   \sin x – \sin y = 2.\cos \dfrac{{x + y}}{2}.\sin \dfrac{{x – y}}{2}\\
   A = \dfrac{{\cos 7x – \cos 8x – \cos 9x + \cos 10x}}{{\sin 7x – \sin 8x – \sin 9x + \sin 10x}}\\
    = \dfrac{{ – \left( {\cos 9x – \cos 7x} \right) + \left( {\cos 10x – \cos 8x} \right)}}{{ – \left( {\sin 9x – \sin 7x} \right) + \left( {\sin 10x – \sin 8x} \right)}}\\
    = \dfrac{{2.sin8x.\sin x – 2\sin 9x.\sin x}}{{ – 2.\cos 8x.\sin x + 2.\cos 9x.\sin x}}\\
    = \dfrac{{2\sin x.\left( {\sin 8x – \sin 9x} \right)}}{{2\sin x.\left( {\cos 9x – \cos 8x} \right)}}\\
    = \dfrac{{\sin 8x – \sin 9x}}{{\cos 9x – \cos 8x}}\\
    = \dfrac{{2.cos\dfrac{{17x}}{2}.\sin \dfrac{{ – x}}{2}}}{{ – 2\sin \dfrac{{17x}}{2}.\sin \dfrac{x}{2}}}\\
    = \dfrac{{\cos \dfrac{{17x}}{2}}}{{\sin \dfrac{{17x}}{2}}}\\
    = \cot \dfrac{{17x}}{2}\\
   B = \dfrac{{\sin 2x + 2\sin 3x + \sin 4x}}{{\sin 3x + 2\sin 4x + \sin 5x}}\\
    = \dfrac{{\left( {\sin 2x + \sin 4x} \right) + 2\sin 3x}}{{\left( {\sin 3x + \sin 5x} \right) + 2\sin 4x}}\\
    = \dfrac{{2.\sin 3x.\cos x + 2\sin 3x}}{{2\sin 4x.\cos x + 2\sin 4x}}\\
    = \dfrac{{2\sin 3x.\left( {\cos x + 1} \right)}}{{2\sin 4x.\left( {\cos x + 1} \right)}}\\
    = \dfrac{{\sin 3x}}{{\sin 4x}}\\
   C = \dfrac{{1 + \cos x + \cos 2x + \cos 3x}}{{\cos x + 2{{\cos }^2}x – 1}}\\
    = \dfrac{{\left( {1 + \cos 2x} \right) + \left( {\cos x + \cos 3x} \right)}}{{\cos x + \left( {2{{\cos }^2}x – 1} \right)}}\\
    = \dfrac{{\left( {1 + 2{{\cos }^2}x – 1} \right) + 2.\cos 2x.\cos x}}{{\cos x + \cos 2x}}\\
    = \dfrac{{2\cos x\left( {\cos x + \cos 2x} \right)}}{{\cos x + \cos 2x}}\\
    = 2\cos x\\
   D = \dfrac{{\sin 4x + \sin 5x + \sin 6x}}{{\cos 4x + \cos 5x + \cos 6x}}\\
    = \dfrac{{\left( {\sin 4x + \sin 6x} \right) + \sin 5x}}{{\left( {\cos 4x + \cos 6x} \right) + \cos 5x}}\\
    = \dfrac{{2.sin5x.\cos x + \sin 5x}}{{2.\cos 5x.\cos x + \cos 5x}}\\
    = \dfrac{{\sin 5x\left( {2\cos x + 1} \right)}}{{\cos 5x\left( {2\cos x + 1} \right)}}\\
    = \dfrac{{\sin 5x}}{{\cos 5x}}\\
    = \tan 5x
   \end{array}\)

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