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a: \(A=\sqrt{3}\left(\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx\right)+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\dfrac{\sqrt{3}}{2}sinx-\dfrac{3}{2}cosx+\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\)
\(=\sqrt{3}sinx-cosx\)
c: \(=2\left[\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right]+4sinx+1\)
\(=\sqrt{3}sin2x-cos2x+4sinx+1\)
d: \(D=\sqrt{3}cos2x+sin2x+2\cdot\left(\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x\right)\)
\(=\sqrt{3}\cdot cos2x+sin2x+\sqrt{3}\cdot sin2x-cos2x\)
\(=cos2x\left(\sqrt{3}-1\right)+sin2x\left(1+\sqrt{3}\right)\)
\(M=sin^2x+cos^2x+2sinx.cosx+cos^2x-sin^2x\)
\(=\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(cosx+sinx\right)\)
\(=\left(sinx+cosx\right)\left(sinx+cosx+cosx-sinx\right)\)
\(=2cosx\left(sinx+cosx\right)\)
\(=2\sqrt{2}cosx.cos\left(x-\frac{\pi}{4}\right)\)
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$
a) -1 ≤ -0,7 ≤ 1. Có cung α mà sin α = -0,7
b) > 1. Không có cung α có sin nhận giá trị
c) Không. Vì -√2 < -1
d) Không. Vì > 1
\(A=\frac{1-sinx-1+2sin^2x}{2sinx.cosx-cosx}=\frac{sinx\left(2sinx-1\right)}{cosx\left(2sinx-1\right)}=\frac{sinx}{cosx}=tanx\)
\(B=\frac{2sinx.cosx+sinx}{1+2cos^2x-1+cosx}=\frac{sinx\left(2cosx+1\right)}{cosx\left(2cosx+1\right)}=\frac{sinx}{cosx}=tanx\)
\(C=\frac{sina.cosa\left(tana-cota\right)}{sina.cosa\left(tana+cota\right)}+cos2a=\frac{sin^2a-cos^2a}{sin^2a+cos^2a}+cos2a\)
\(=-cos2a+cos2a=0\)
a) \(B=\dfrac{sin^4x-cos^4x+cos^2x}{2\left(1-cosx\right)\left(1+cosx\right)}\)
\(B=\dfrac{\left(sin^2x\right)^2-\left(cos^2x\right)^2+cos^2x}{2\left(1-cos^2x\right)}\)
\(B=\dfrac{\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+cos^2x}{2\left(sin^2x+cos^2x-cos^2x\right)}\)
\(B=\dfrac{sin^2x-cos^2x+cos^2x}{2sin^2x}=\dfrac{sin^2x}{2sin^2x}=\dfrac{1}{2}\)
b) \(\dfrac{1+sin2x-cos2x}{1+sin2x+cos2x}=tanx\)
\(VT=\dfrac{1+2sinx.cosx-\left(1-2sin^2x\right)}{1+2sinx.cosx+2cos^2x-1}\)
\(VT=\dfrac{1+2sinx.cosx-1+2sin^2x}{2sinx.cosx+2cos^2x}\)
\(VT=\dfrac{2sinx.cosx+2sin^2x}{2sinx.cosx+2cos^2x}\)
\(VT=\dfrac{2sinx\left(cosx+sinx\right)}{2cosx\left(sinx+cosx\right)}=\dfrac{sinx}{cosx}=tanx=VP\) ( đpcm )
p/s : sửa \(cos1x\rightarrow cos2x\)
Quên cách giải ptlg rồi nên lm câu 4 =.=
\(\cos3x=\cos\left(2x+x\right)=\cos2x.\cos x-\sin2x.\sin x\)
\(=\left(2\cos^2x-1\right)\cos x-2\sin^2x.\cos x\)
\(=2\cos^3x-\cos x-2\sin^2x.\cos x\)
\(\Rightarrow A=\frac{1+\cos x+2\cos^2x-1+2\cos^3x-\cos x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\sin^2x.\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\left(1-\cos^2x\right).\cos x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos^2x+2\cos^3x-2\cos x+2\cos^3x}{2\cos^2x-1+\cos x}\)
\(=\frac{2\cos x\left(2\cos^2x+\cos x-1\right)}{2\cos^2x-1+\cos x}=2\cos x\)
\(cosx=\dfrac{4}{5}\Rightarrow cos^2x=\dfrac{16}{25}\)
\(\Rightarrow sin^2x=1-cos^2x=1-\dfrac{16}{25}=\dfrac{9}{25}\)
\(\Rightarrow\sqrt{cos2x}=\sqrt{cos^2x-sin^2x}=\sqrt{\dfrac{16}{25}-\dfrac{9}{25}}=\sqrt{\dfrac{7}{25}}=\dfrac{\sqrt{7}}{5}\)
Em mới lớp 8 nên làm không chắc nhé anh/chị.