Bài tập 1.6 trang 7 SBT Toán 11 Tập 1 Kết nối tri thức

Chứng minh các đẳng thức sau:

a) \({\cos ^4}x - {\sin ^4}x = 2{\cos ^2}x - 1\);

b) \({\tan ^2}x - {\sin ^2}x = {\tan ^2}x.{\sin ^2}x\);

c) \({(\sin x + \cos x)^2} + {(\sin x - \cos x)^2} = 2\).

QUẢNG CÁO

Hướng dẫn giải chi tiết Bài 1.6

a) Ta có:

\(\begin{array}{l}VT = {\cos ^4}x - {\sin ^4}x\\\,\,\,\,\,\,\,\,\,\, = \left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^2}x + {{\sin }^2}x} \right)\\\,\,\,\,\,\,\,\,\,\, = ({\cos ^2}x - {\sin ^2}x).1 = {\cos ^2}x - {\sin ^2}x\\\,\,\,\,\,\,\,\,\,\, = {\cos ^2}x - (1 - {\cos ^2}x) = {\cos ^2}x - 1 + {\cos ^2}x\\\,\,\,\,\,\,\,\,\,\, = 2{\cos ^2}x - 1 = {\rm{VP}}\end{array}\)

b) Ta có:

\(\begin{array}{l}VT = {\tan ^2}x - {\sin ^2}x = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - {\sin ^2}x\\\,\,\,\,\,\,\,\,\, = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - \frac{{{{\sin }^2}x.{{\cos }^2}x}}{{{{\cos }^2}x}} = \frac{{{{\sin }^2}x - {{\sin }^2}x{{\cos }^2}x}}{{{{\cos }^2}x}}\\\,\,\,\,\,\,\,\,\, = \frac{{{{\sin }^2}x(1 - {{\cos }^2}x)}}{{{{\cos }^2}x}} = \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.(1 - {\cos ^2}x)\\\,\,\,\,\,\,\,\,\, = {\tan ^2}x.{\sin ^2}x = {\rm{VP}}{\rm{.}}\end{array}\)

c) Ta có:

\(\begin{array}{l}VT = {(\sin x + \cos x)^2} + {(\sin x - \cos x)^2}\\\,\,\,\,\,\,\,\,\,\, = {\sin ^2}x + 2\sin x{\mathop{\rm cosx}\nolimits} + co{s^2}x + {\sin ^2}x - 2\sin x\cos x + {\cos ^2}x\\\,\,\,\,\,\,\,\,\,\, = 2{\sin ^2}x + 2{\cos ^2}x = 2({\sin ^2}x + {\cos ^2}x) = 2.1 = 2 = {\rm{VP}}{\rm{.}}\end{array}\)

-- Mod Toán 11 HỌC247

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