Câu hỏi của Bich Hong

Question

Bài 1: Cho a là góc nhọn . Rút gọn biểu thức

A= sin6a + cos6a+ 3sin2a-cos2 a

Mysterious Person · Accepted Answer

ta có : $A=sin^6a+cos^6a+3sin^2a-cos^2a$

$=\left(sin^2a\right)^3+\left(cos^3a\right)^2+3sin^2a-cos^2a$

$=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a-cos^2a$

$=1-3sin^2a.cos^2a+3sin^2a-cos^2a$

$=3sin^2a-3sin^2a.cos^2a+1-cos^2a$

$=3sin^2a\left(1-cos^2a\right)+\left(1-cos^2a\right)$ $=\left(3sin^2a+1\right)\left(1-cos^2a\right)$

$=\left(3sin^2a+1\right)\left(sin^2a\right)=3sin^4a+sin^2a$Câu trả lời: ta có : $A=sin^6a+cos^6a+3sin^2a-cos^2a$

$=\left(sin^2a\right)^3+\left(cos^3a\right)^2+3sin^2a-cos^2a$

$=\left(sin^2a+cos^2a\right)^3-3sin^2a.cos^2a\left(sin^2a+cos^2a\right)+3sin^2a-cos^2a$

$=1-3sin^2a.cos^2a+3sin^2a-cos^2a$

$=3sin^2a-3sin^2a.cos^2a+1-cos^2a$

$=3sin^2a\left(1-cos^2a\right)+\left(1-cos^2a\right)$ $=\left(3sin^2a+1\right)\left(1-cos^2a\right)$

$=\left(3sin^2a+1\right)\left(sin^2a\right)=3sin^4a+sin^2a$

Bài 2: Tỉ số lượng giác của góc nhọn