Giải bài 1.6 trang 7 sách bài tập toán 11 – Kết nối tri thức với cuộc sống

Đề bài

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

Lời giải chi tiết

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

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