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a) \(sin^6x+cos^6x+3sin^2x.cos^2x\)
\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cox^2x+cos^4x\right)+3sin^2x.cos^2x\)
\(=sin^4x-sin^2x.cox^2x+cos^4x+3sin^2x.cos^2x\)
\(=sin^4x+2sin^2x.cox^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\text{}\text{}\)
b) \(sin^4x-cos^4x-\left(sinx+cosx\right)\left(sinx-cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)\)
\(=1\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)=0\)
c) \(cos^2x+tan^2x.cos^2x\)
\(=cos^2x+\dfrac{sin^2x}{cos^2x}.cos^2x=sin^2x+cos^2x=1\)
Câu 1:
\(\cos a=\sqrt{1-\left(\dfrac{1}{4}\right)^2}=\dfrac{\sqrt{15}}{4}\)
\(A=\sin^2a+3\cos^2a-1=\dfrac{1}{16}+3\cdot\dfrac{15}{16}-1=\dfrac{15}{8}\)
Lời giải:
a) \(\cot ^2a+1=\left(\frac{\cos a}{\sin a}\right)^2+1=\frac{\cos ^2a+\sin ^2a}{\sin ^2a}=\frac{1}{\sin ^2a}\)
b)
\(\tan ^2a+1=\left(\frac{\sin a}{\cos a}\right)^2+1=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}=\frac{1}{\cos ^2a}\)
c) Đề bài sai.
\(\sin ^4a+\cos ^2a=\sin ^2a.\sin ^2a+\cos ^2a\)
\(=\sin ^2a(1-\cos ^2a)+\cos ^2a\)
\(\sin ^2a+\cos ^2a-\sin ^2a\cos ^2a=1-\sin ^2a\cos ^2a\)
d)
\(\frac{1-4\sin ^2a\cos ^2a}{(\sin a+\cos a)^2}=\frac{1-(2\sin a\cos a)^2}{\sin ^2a+2\sin a\cos a+\cos ^2a}=\frac{(1-2\sin a\cos a)(1+2\sin a\cos a)}{1+2\sin a\cos a}\)
\(=1-2\sin a\cos a\)
e) ĐK tồn tại tan là $\cos x\neq 0$
Vì \(\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\tan a\cos a\)
Ta có:
\(\frac{2\sin a\cos a-1}{\cos ^2a-\sin ^2a}=\frac{1-2\sin a\cos a}{\sin ^2a-\cos ^2a}=\frac{\cos ^2a+\sin ^2a-2\sin a\cos a}{(\sin a-\cos a)(\sin a+\cos a)}\)
\(=\frac{(\sin a-\cos a)^2}{(\sin a-\cos a)(\sin a+\cos a)}=\frac{\sin a-\cos a}{\sin a+\cos a}\)
\(=\frac{\tan a\cos a-\cos a}{\tan a\cos a+\cos a}=\frac{\cos a(\tan a-1)}{\cos a(\tan a+1)}\)\(=\frac{\tan a-1}{\tan a+1}\) (đpcm)
a: \(M=\dfrac{1}{tana+cota}=1:\left(\dfrac{sina}{cosa}+\dfrac{cosa}{sina}\right)\)
\(=1:\dfrac{sin^2a+cos^2a}{cosa\cdot sina}=cosa\cdot sina=\dfrac{2\sqrt{2}}{9}\)
b: \(A=\left(sin^2a+cos^2a\right)^3-3\cdot sin^2a\cdot cos^2a+3\cdot sin^2a\cdot cos^2a\)
=1
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