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a)
\(4a^2cos^260^o+2ab.cos^2180^o+\dfrac{4}{3}cos^230^o\)
\(=4a^2.\left(\dfrac{1}{2}\right)^2+2ab.\left(-1\right)^2+\dfrac{4}{3}.\left(\dfrac{\sqrt{3}}{2}\right)^2\)
\(=4a^2.\dfrac{1}{4}+2ab+\dfrac{4}{3}.\dfrac{3}{4}\)
\(=a^2+2ab+1\).
b)
\(\left(asin90^o+btan45^o\right)\left(acos0^o+bcos180^o\right)\)
\(=\left(a+b\right)\left(a-b\right)=a^2-b^2\).
\(A=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+sin^2179^o+c\text{os}^2180^o\)
\(=sin^21^o+c\text{os}^22^o+sin^23^o+c\text{os}^24^o+...+c\text{os}^290^o-sin^289^o-c\text{os}^288^o-...-sin^21^o-c\text{os}^20^o\)
\(=c\text{os}^290^o-c\text{os}^20^o\)
\(=-1\)
Chúc bn học tốt
\(A=cos10+cos170+cos40+cos140+cos70+cos110\)
\(A=cos10+cos\left(180-10\right)+cos40+cos\left(180-40\right)+cos70+cos\left(180-70\right)\)
\(A=cos10-cos10+cos40-cos40+cos70-cos70\)
\(A=0\)
\(B=sin5+sin355+sin10+sin350+...+sin175+sin185+sin360\)
\(B=sin5+sin\left(360-5\right)+sin10+sin\left(360-10\right)+...+sin175+sin\left(360-175\right)+sin360\)
\(B=sin5-sin5+sin10-sin10+...+sin175-sin175+sin360\)
\(B=sin360=0\)
\(C=cos^22+cos^288+cos^24+cos^284+...+cos^244+cos^246\)
\(C=cos^22+cos^2\left(90-2\right)+cos^24+cos^2\left(90-4\right)+...+cos^244+cos^2\left(90-44\right)\)
\(C=cos^22+sin^22+cos^24+sin^24+...+cos^244+sin^244\)
\(C=1+1+...+1\) (có \(\frac{44-2}{2}+1=22\) số 1)
\(\Rightarrow C=22\)
a) \(sin\left(270^o-\alpha\right)=sin\left(-90^o-\alpha\right)=-sin\left(90^o+\alpha\right)\)\(=-cos\alpha\).
b) \(cos\left(270^o-\alpha\right)=cos\left(-90^o-\alpha\right)=cos\left(90^o+\alpha\right)\)\(=-sin\alpha\).
c) \(sin\left(270^o+\alpha\right)=sin\left(-90^o+\alpha\right)=-sin\left(90^o-\alpha\right)\)\(=-cos\alpha\).
d) \(cos\left(270^o+\alpha\right)=cos\left(-90^o+\alpha\right)=cos\left(90^o-\alpha\right)\)\(=sin\alpha\).
1/
\(tanx=\frac{sinx}{cosx}=\frac{sin^2x}{sinx.cosx}=\frac{2sin^2x}{2sinx.cosx}\)
\(=\frac{2\left(\frac{1-cos2x}{2}\right)}{sin2x}=\frac{1-cos2x}{sin2x}\)
2/
\(\frac{sin\left(60-x\right)cos\left(30-x\right)+cos\left(60-x\right)sin\left(30-x\right)}{sin4x}=\frac{sin\left(60-x+30-x\right)}{sin4x}=\frac{sin\left(90-2x\right)}{2sin2x.cos2x}\)
\(=\frac{cos2x}{2sin2x.cos2x}=\frac{1}{2sin2x}\)
3/
\(4cos\left(60+a\right)cos\left(60-a\right)+2sin^2a\)
\(=2\left(cos\left(60+a+60-a\right)+cos\left(60+a-60+a\right)\right)+2sin^2a\)
\(=2cos120+2cos2a+2\left(\frac{1-cos2a}{2}\right)\)
\(=-1+2cos2a+1-cos2a=cos2a\)
a)\(sin^2\left(180^o-\alpha\right)+tan^2\left(180-\alpha\right).tan^2\left(270^o+\alpha\right)\)\(+sin\left(90^o+\alpha\right)cos\left(\alpha-360^o\right)\)
\(=sin^2\alpha+tan^2\alpha.cot^2\alpha+cos\alpha cos\alpha\)
\(=sin^2\alpha+cos^2\alpha+\left(tan\alpha cot\alpha\right)^2=1+1=2\).
\(\dfrac{cos\left(\alpha-180^o\right)}{sin\left(180^o-\alpha\right)}+\dfrac{tan\left(\alpha-180^o\right)cos\left(180^o+\alpha\right)sin\left(270^o+\alpha\right)}{tan\left(270^o+\alpha\right)}\)
\(=\dfrac{cos\left(180^o-\alpha\right)}{sin\left(180^o-\alpha\right)}+\dfrac{-tan\left(180^o-\alpha\right).cos\alpha.sin\left(90^o+\alpha\right)}{-tan\left(90^o+\alpha\right)}\)
\(=tan\left(180^o-\alpha\right)+\dfrac{tan\alpha.cos\alpha.cos\alpha}{cot\alpha}\)
\(=-tan\alpha+tan^2\alpha cos^2\alpha\)
\(=tan\alpha\left(-1+tan\alpha cos^2\alpha\right)\)
\(=tan\alpha\left(sin\alpha cos\alpha-1\right)\).
\(A=tan18^otan288+sin32^osin148^o-sin302^osin122^o\)
\(=tan18^o.tan\left(-72^o\right)+sin32^o.sin32^o+sin58^o.sin58^o\)
\(=-tan18^o.cot18^o+sin^232^o+sin^258^o\)
\(=-1+sin^232^o+cos^232^2=-1+1=0\).
b) \(B=\dfrac{1+sin^4\alpha-cos^4\alpha}{1-sin^6\alpha-cos^6\alpha}\)
\(=\dfrac{1+\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-cos^2\alpha\right)}{1-\left(sin^6\alpha+cos^6\alpha\right)}\)
\(=\dfrac{1+sin^2\alpha-cos^2\alpha}{1-\left(sin^2\alpha+cos^2\alpha\right)\left(sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha\right)}\)
\(=\dfrac{sin^2\alpha+1-cos^2\alpha}{1-\left(1-sin\alpha.cos\alpha\right)}\)
\(=\dfrac{sin^2\alpha+sin^2\alpha}{sin\alpha cos\alpha}\)
\(=\dfrac{2sin^2\alpha}{sin\alpha cos\alpha}=\dfrac{2sin\alpha}{cos\alpha}=2tan\alpha\).