Bài tập Áp dụng công thức lượng giác vào các bài toán rút gọn, chứng minh đẳng thức lượng giác lớp 11 (có lời giải)

a. \(A = \frac{{{{\cos }^3}a + {{\sin }^3}a}}{{\cos a + \sin a}} + \sin a \cdot \cos a = 1\).

b. \(B = 2\left( {{{\sin }^6}a + {{\cos }^6}a} \right) - 3\left( {{{\sin }^4}a + {{\cos }^4}a} \right) = 1\).

c. C = 8(cos⁸a – sin⁸a) – cos6a – 7cos2a = 8.

d. D = cos²x – 2cosa∙cosx∙cos(a + x) + cos²(a + x) = cos²a.

a. \({S_n} = {\sin ^3}\frac{a}{3} + 3{\sin ^3}\frac{a}{{{3^2}}} + ... + {3^{n - 1}}{\sin ^3}\frac{a}{{{3^n}}}\)\( = \frac{1}{4}\left( {{3^n}\sin \frac{a}{{{3^n}}} - \sin a} \right)\).

b. \(S = \frac{1}{{\sin \alpha }} + \frac{1}{{\sin 2\alpha }} + ... + \frac{1}{{\sin {2^{n - 1}}\alpha }}\)= \(\cot \frac{\alpha }{2} - \cot {2^{n - 1}}\alpha \).

c. \({S_n} = {\tan ^2}\frac{a}{2} \cdot \tan a + 2{\tan ^2}\frac{a}{{{2^2}}} \cdot \tan \frac{a}{2} + ... + {2^{n - 1}}{\tan ^2}\frac{a}{{{2^n}}} \cdot \tan \frac{a}{{{2^{n - 1}}}} = \tan a - {2^n}\tan \frac{a}{{{2^n}}}\) .

d. \({P_n} = \cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}} \cdot ... \cdot c{\rm{os}}\frac{x}{{{2^n}}}\)\( = \frac{{\sin x}}{{{2^n} \cdot \sin \frac{x}{{{2^n}}}}}\) .

a. \(\cot x - \tan x - 2\tan 2x = 4\cot 4x\).

b. sin²(a + b) – sin²a – sin²b = 2sina∙sinb∙cos(a + b).

c. \(\sin \alpha \cdot \sin \left( {\frac{\pi }{3} - \alpha } \right) \cdot \sin \left( {\frac{\pi }{3} + \alpha } \right) = \frac{1}{4} \cdot \cos 3\alpha \).

d. cos(a + b)∙sin(a – b) + cos(b + c)∙sin(b – c) + cos(a + c)∙sin(c – a) = 1.

a. \({\sin ^2}\left( {x + \frac{\pi }{4}} \right) + {\sin ^2}\left( {x + \frac{\pi }{2}} \right) = 2 - {\sin ^2}\left( {x + \frac{{3\pi }}{4}} \right) - {\sin ^2}\left( {x + \pi } \right)\).

b. \({\sin ^4}x\left( {1 + {{\sin }^2}x} \right) + {\cos ^4}x\left( {1 + {{\cos }^2}x} \right) + 5\left( {404 + {{\sin }^2}x \cdot {{\cos }^2}x} \right) = 2020\).

c. \({\mathop{\rm s}\nolimits} {\rm{inx}} \cdot \cos x \cdot \cos 2x \cdot \cos 4x = \frac{1}{8}\cos 8x\).

d. \(\cos \left( {x - \frac{\pi }{3}} \right) \cdot \cos \left( {x + \frac{\pi }{4}} \right) + \cos \left( {x + \frac{\pi }{6}} \right) \cdot \cos \left( {x + \frac{{3\pi }}{4}} \right) = \frac{{\sqrt 2 }}{2} \cdot \left( {\frac{{1 - \sqrt 3 }}{2}} \right)\).

a. \(\sin B \cdot \cos C + \sin C \cdot \cos B = \sin A\).

b. cos²A + cos²B + cos²C = 1 − 2cosA∙cosB∙cosC.

c. \[\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C\].

d. \(\tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1\).