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a) \(\cos^4\alpha-\sin^4\alpha=\left(\cos^2\alpha+\sin^2\alpha\right)\left(\cos^2\alpha-\sin^2\alpha\right)=\cos^2\alpha-\sin^2\alpha\)
\(2\cos^2\alpha-\left(\sin^2\alpha+\cos^2\alpha\right)=2\cos^2\alpha-1\)
b) \(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\)\(\Leftrightarrow\)\(\left(1-\sin\alpha\right)\left(1+\sin\alpha\right)=\cos^2\alpha\)
\(\Leftrightarrow\)\(1-\left(\sin^2\alpha+\cos^2\alpha\right)=0\)\(\Leftrightarrow\)\(1-1=0\) ( luôn đúng )
c) \(\frac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha.\cos\alpha}=\frac{2\cos\alpha.2\sin\alpha}{\sin\alpha.\cos\alpha}=4\)
um, hình như câu b) chỗ 1-.... đó hơi sai nếu viết từ bước trên xuống á bạn!
mình nghĩ là: sau dấu bằng đầu tiên, sau đó là:
\(=cos^2\alpha=1-sin^2\alpha\)(luôn đúng)
CẢM ƠN bạn nhiều lắm luôn nha!!!!!
\(a,1-sin^2\alpha=cos^2\alpha\)
\(b,\left(1-cos\alpha\right)\left(1+cos\alpha\right)=1-cos^2\alpha=sin^2\alpha\)
\(c,1+sin^2\alpha+cos^2\alpha=1+1=2\)
\(d,sin\alpha-sin\alpha.cos^2\alpha=sin\alpha.\left(1-cos^2\alpha\right)=sin\alpha.sin^2\alpha=sin^3\alpha\)
\(e,sin^2\alpha+cos^2\alpha+2sin^2\alpha.cos^2\alpha\)
\(=1+2sin^2\alpha.cos^2\alpha\)
a) ta có : \(sin\alpha.cos\alpha\left(tan\alpha+cot\alpha\right)=sin\alpha.cos\alpha\left(\dfrac{sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{sin\alpha}\right)\)
\(=sin^2\alpha+cos^2\alpha=1\)
b) ta có : \(\left(sin^2\alpha+cos^2\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2\)
\(=1^2+1-2sin\alpha.cos=2\left(1-2sin\alpha.cos\alpha\right)\)
c) ta có : \(tan^2\alpha-sin^2\alpha.tan^2\alpha=tan^2\alpha\left(1-sin^2\alpha\right)\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}.cos^2\alpha=sin^2\alpha\)
tui rất thích lượng giác:
a) = s2 + 2s.c +c2 +s2- 2s.c + c2 =1+1=2
b) = s.c(s/c + c/s) = s.c(s2 + c2) / s.c = 1
.............................bài nào cx dễ
( k có việc j khó, chỉ sợ lòng k bền....)
đáp án :
a) \(cos^2\alpha\)
b) 1
c) \(sin^2\alpha\)
d) \(sin^2\alpha\)
e) 2
g) 1
h) \(sin^3\alpha\)
i) \(sin^2\alpha\)
\(1+\sin^2\alpha+\cos^2\alpha=1+1=2\)
\(\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha\cdot\cos^2\alpha\\ =\left(\sin^2\alpha\right)^2+2\sin^2\alpha\cdot\cos^2\alpha+\left(\cos^2\alpha\right)^2\\ =\left(\sin^2\alpha+\cos^2\alpha\right)^2\\ =1^2=1\)
\(\tan^2\alpha-\sin^2\alpha\cdot\tan^2\alpha\\ =\tan^2\alpha\left(1-\sin^2\alpha\right)\\ =\left(\frac{\sin\alpha}{\cos\alpha}\right)^2\cdot\cos^2\alpha\\ =\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha\\ =\sin^2\alpha\)
\(\cos^2\alpha+\tan^2\alpha\cdot\cos^2\alpha\\ =\cos^2\alpha+\left(\frac{\sin\alpha}{\cos\alpha}\right)^2\cdot\cos^2\alpha\\ =\cos^2\alpha+\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha\\ =\cos^2\alpha+\sin^2\alpha\\ =1\)
\(\tan^2\alpha\cdot\left(2\cos^2\alpha+\sin^2\alpha-1\right)\\ =\tan^2\alpha\cdot\left(2\cos^2\alpha+\sin^2\alpha-\sin^2\alpha-\cos^2\alpha\right)\\ =\tan^2\alpha\cdot\cos^2\alpha\\ =\frac{\sin^2\alpha}{\cos^2\alpha}\cdot\cos^2\alpha=\sin^2\alpha\)