Câu 2 Rút gọn a, A = sin^6 x + 2sin^2 x cos^4 x + 3sin^4 x cos^2 x + cos^4 x b, P = (sin x + sin 5x) / (cos 5x – cos x) c, P = (sin 5x + sin x) / (co

Câu 2 Rút gọn
a, A = sin^6 x + 2sin^2 x cos^4 x + 3sin^4 x cos^2 x + cos^4 x
b, P = (sin x + sin 5x) / (cos 5x – cos x)
c, P = (sin 5x + sin x) / (cos 5x + cos x)

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  1. Giải thích các bước giải:

     Ta có:

    \(\begin{array}{l}
    a,\\
    A = {\sin ^6}x + 2{\sin ^2}x.{\cos ^4}x + 3{\sin ^4}x.{\cos ^2}x + {\cos ^4}x\\
     = \left( {{{\sin }^6}x + 3{{\sin }^4}x.{{\cos }^2}x + 3{{\sin }^2}x.{{\cos }^4}x + {{\cos }^6}x} \right) + {\cos ^4}x – {\cos ^6}x – {\sin ^2}x.{\cos ^4}x\\
     = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^3} + {\cos ^4}x.\left( {1 – {{\sin }^2}x} \right) – {\cos ^6}x\\
     = {1^3} + {\cos ^4}x.{\cos ^2}x – {\cos ^6}x\\
     = 1\\
    b,\\
    P = \dfrac{{\sin x + \sin 5x}}{{\cos 5x – \cos x}}\\
     = \dfrac{{2.\sin \dfrac{{5x + x}}{2}.\cos \dfrac{{5x – x}}{2}}}{{ – 2.\sin \dfrac{{5x + x}}{2}.\sin \dfrac{{5x – x}}{2}}}\\
     = \dfrac{{2.\sin 3x.\cos 2x}}{{ – 2.\sin 3x.\sin 2x}}\\
     =  – \dfrac{{\cos 2x}}{{\sin 2x}} =  – \cot 2x\\
    c,\\
    P = \dfrac{{\sin 5x + \sin x}}{{\cos 5x + \cos x}}\\
     = \dfrac{{2\sin \dfrac{{5x + x}}{2}.\cos \dfrac{{5x – x}}{2}}}{{2.\cos \dfrac{{5x + x}}{2}.\cos \dfrac{{5x – x}}{2}}}\\
     = \dfrac{{2.\sin 3x.\cos 2x}}{{2.\cos 3x.\cos 2x}}\\
     = \dfrac{{\sin 3x}}{{\cos 3x}} = \tan 3x
    \end{array}\)

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