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\(sina\sqrt{1+\frac{sin^2a}{cos^2a}}=sina\sqrt{\frac{cos^2a+sin^2a}{cos^2a}}=\frac{sina}{\left|cosa\right|}=\pm tana\)
\(\frac{1-cos^2x}{1-sin^2x}+tanx.cotx=\frac{sin^2x}{cos^2x}+\frac{sinx}{cosx}.\frac{cosx}{sinx}=tan^2x+1=\frac{1}{cos^2x}\)
\(\frac{1-4sin^2xcos^2x}{\left(sinx+cosx\right)^2}=\frac{\left(1-2sinx.cosx\right)\left(1+2sinx.cosx\right)}{sin^2x+cos^2x+2sinx.cosx}=\frac{\left(1-sin2x\right)\left(1+2sinx.cosx\right)}{1+2sinx.cosx}=1-2sinx\)
\(sin\left(90-x\right)+cos\left(180-x\right)+sin^2x\left(1+tan^2x\right)-tan^2x\)
\(=cosx-cosx+sin^2x.\frac{1}{cos^2x}-tan^2x=tan^2x-tan^2x=0\)
\(A=cos^212+sin^2\left(90-78\right)+cos^21+sin^2\left(90-89\right)\)
\(=cos^212+sin^212+cos^21+sin^21=1+1=2\)
\(B=sin^23+sin^287+sin^215+sin^275\)
\(=sin^23+cos^23+sin^215+cos^215=1+1=2\)
1/ Vì \(\pi< \alpha< \frac{3}{2}\pi\)
\(\Rightarrow\)\(\alpha\in\) góc phần tư thứ 3\(\Rightarrow\sin\alpha< 0;\cos\alpha< 0;\cot\alpha>0\)
2/ Xét 3 trường hợp:
TH1: \(0^0< \alpha< 90^0\) \(\Rightarrow\alpha\in\) góc phần tư thứ nhất\(\Rightarrow\sin\alpha>0;\cos\alpha>0;\cot\alpha>0\)
TH2: \(-90^0< \alpha< 0^0\Rightarrow\alpha\in\) góc phần tư thứ tư
\(\Rightarrow\sin\alpha< 0;\cos\alpha>0;\cot\alpha< 0\)
TH3: \(-170^0< \alpha< -90^0\)\(\Rightarrow\alpha\in\) góc phần tư thứ ba
\(\Rightarrow\sin\alpha< 0;\cos\alpha< 0;\cot\alpha>0\)
3/ Vì...=> \(\alpha\in\) góc phần tư thứ ba
\(\Rightarrow...\)
\(a,\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4sin^2x.cos^2x}=-1\)
\(VT=\left(\frac{tan^2x-1}{2tanx}\right)^2-\frac{1}{4.sin^2x.cos^2x}=\left(\frac{1}{tan2x}\right)^2-\frac{1}{sin^22x}=\left(\frac{cos2x}{sin2x}\right)^2-\frac{1}{sin^22x}=\frac{cos^22x-1}{sin^22x}=\frac{-sin^22x}{sin^22x}=-1=VP\)
b, \(VT=\frac{cos^2x-sin^2x}{sin^4x+cos^4x-sin^2x}=\frac{cos2x}{\left(sin^2x+cos^2x\right)^2-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{1-sin^2x-2.sin^2x.cos^2x}=\frac{cos2x}{cos^2x-2.sin^2x.cos^2x}\)
=\(\frac{cos2x}{cos^2x.\left(1-2.sin^2x\right)}=\frac{cos2x}{cos^2x.cos2x}=\frac{1}{cos^2x}=1+tan^2x=VP\)
d, \(VT=\left(\frac{cosx}{1+sinx}+tanx\right).\left(\frac{sinx}{1+cosx}+cotx\right)=\left(\frac{cosx}{1+sinx}+\frac{sinx}{cosx}\right).\left(\frac{sinx}{1+cosx}+\frac{cosx}{sinx}\right)\)
\(=\left(\frac{cos^2x+sinx.\left(1+sinx\right)}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx.\left(1+cosx\right)}{sinx.\left(1+cosx\right)}\right)=\left(\frac{cos^2x+sinx+sin^2x}{cosx.\left(1+sinx\right)}\right).\left(\frac{sin^2x+cosx+cos^2x}{sinx.\left(1+cosx\right)}\right)\)
=\(\frac{1}{cosx.sinx}=VP\)
e, \(VT=cos^2x.\left(cos^2x+2sin^2x+sin^2x.tan^2x\right)=cos^2x.\left(1+sin^2x.\left(1+tan^2x\right)\right)=cos^2x.\left(1+tan^2x\right)=cos^2x.\frac{1}{cos^2x}=1=VP\)
c, \(VT=\frac{sin^2x}{cosx.\left(1+tanx\right)}-\frac{cos^2x}{sinx.\left(1+cosx\right)}=\frac{sin^3x.\left(1+cosx\right)-cos^3x.\left(1+tanx\right)}{sinx.cosx.\left(1+tanx\right).\left(1+cosx\right)}\)
=\(\frac{sin^3x+sin^3x.cotx-cos^3x-cos^3.tanx}{\left(sinx+cosx\right)^2}=\frac{sin^3x+sin^2xcosx-cos^3x-cos^2sinx}{\left(sinx+cosx\right)^2}=\frac{sin^2x.\left(sinx+cosx\right)-cos^2x.\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}\)
\(=\frac{\left(sin^2x-cos^2x\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=\frac{\left(sinx-cosx\right).\left(sinx+cosx\right).\left(sinx+cosx\right)}{\left(sinx+cosx\right)^2}=sinx-cosx=VP\)
Đây nha bạn
\(\alpha>0\Rightarrow\cos\left(40^0+\alpha\right)>0\Rightarrow\cos\left(40^0+\alpha\right)=\sqrt{1-\left[\sin^2\left(40^0+\alpha\right)\right]}=\sqrt{1-a^2}\)
\(\cos\left(70^0+\alpha\right)=\cos\left(30^0+40^0+\alpha\right)\)
\(=\cos30^0.\cos\left(40^0+\alpha\right)+\sin30^0.\sin\left(40^0+\alpha\right)\)
\(=\frac{\sqrt{3}}{2}.\sqrt{1-a^2}+\frac{1}{2}.a=\frac{1}{2}\left(\sqrt{3\left(1-a^2\right)}+a\right)\)
Do \(0< 18^0< 90^0\Rightarrow cos18^0=\sqrt{1-sin^218^0}=\frac{\sqrt{10+2\sqrt{5}}}{4}\)
\(sin72^0=sin\left(90^0-18^0\right)=cos18^0=...\)
\(sin162^0=sin\left(180^0-18^0\right)=sin18^0=...\)
\(sin108^0=sin\left(90^0+18^0\right)=cos18^0=...\)
\(cos108^0=cos\left(90^0+18^0\right)=-sin18^0=...\)
\(tan72^0=tan\left(90^0-18^0\right)=cot18^0=\frac{cos18^0}{sin18^0}=...\)