Đặt \(\sin^2\alpha=x\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)

\(A=x^3+\left(1-x\right)^3+3x-\left(1-x\right)=x^3+1-3x+3x^2-x^3+3x-1+x=3x^2+x\)

Vậy \(A=3\sin^4\alpha+\sin^2\alpha\). NHỚ NHA!

sữa đề chút nha :

+) ta có : \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(sin\alpha+cos\alpha\right)\left(cos\alpha-sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)

+) ta có :

\(B=sin^6\alpha+cos^6\alpha+3sin^2\alpha.cos^2\alpha\)

\(=\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha\left(sin^2\alpha+cos^2\alpha\right)+3sin^2\alpha.cos^2\alpha\)

\(=1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha.cos^2\alpha=1\)

\(A=sin^6\alpha+cos^6\alpha+3sin^2\alpha-cos^2\alpha\)

\(=\left(sin^2\alpha\right)^3+\left(cos^2\alpha\right)^3+3sin^2\alpha-cos^2\alpha\)

\(=\left(sin^2\alpha+cos^2\alpha\right)\left(sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha\right)+3sin^2\alpha-cos^2\alpha\)

\(=sin^4\alpha+cos^4\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)

\(=\left(sin^2\alpha+cos^2\alpha\right)^2-2sin^2\alpha.cos^2\alpha-sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha\)

\(1-3sin^2\alpha.cos^2\alpha+3sin^2\alpha-cos^2\alpha=3sin^2\alpha\left(1-cos^2\alpha\right)+\left(1-cos^2\alpha\right)\)

\(=\left(3sin^2\alpha+1\right).sin^2\alpha=0\)

lời giải ở bài 28 bạn nha

\(a,\sqrt{2}sin\left(\alpha+\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha cos\dfrac{\pi}{4}+cos\alpha sin\dfrac{\pi}{4}\right)-cos\alpha\\ =\sqrt{2}\left(sin\alpha\cdot\dfrac{\sqrt{2}}{2}+cos\alpha\cdot\dfrac{\sqrt{2}}{2}\right)-cos\alpha\\ =\sqrt{2}\cdot sin\alpha\cdot\dfrac{\sqrt{2}}{2}+\sqrt{2}\cdot cos\alpha\cdot\dfrac{\sqrt{2}}{2}-cos\alpha\\ =sin\alpha+cos\alpha-cos\alpha\\ =sin\alpha\)

\(b,\left(cos\alpha+sin\alpha\right)^2-sin2\alpha\\ =cos^2\alpha+sin^2\alpha=2cos\alpha sin\alpha-2sin\alpha cos\alpha\\ =sin^2\alpha+cos^2\alpha\\ =1\)

\(\dfrac{\left(sina+cosa\right)^2-\left(sina-cosa\right)^2}{sina.cosa}=4\\ VT=\dfrac{sin^2a+2sinacosa+cos^2a-sin^2a+2sinacosa-cos^2a}{sinacosa}\\ =\dfrac{4sinacosa}{sinacosa}=4=VP\)

a: \(S=cos^2a\left(1+tan^2a\right)=cos^2a\cdot\dfrac{1}{cos^2a}=1\)

b: \(VP=\dfrac{1+sin2a-1+sin2a}{\dfrac{1}{2}\cdot sin2a}=\dfrac{2\cdot sin2a}{\dfrac{1}{2}\cdot sin2a}=4=VT\)

a) S= \(cos^2a\left(tg^2a+1\right)=cos^2a.\dfrac{1}{cos^2a}=1\)

\(sin^4a+cos^4a+2sin^2a.cos^2a=\left(sin^2a+cos^2a\right)^2=1^2=1\)

b) \(sin^6a+cos^6a+3sin^2a.cos^2a=\left(sin^2a+cos^2a\right)\left(sin^4a-sin^2a.cos^2a+cos^4a\right)+3sin^2a.cos^2a=sin^4a+2sin^2a.cos^2a+cos^4a=\left(sin^2a+cos^2a\right)^2=1\)

Lời giải:

\(\sin ^2a\cos ^2a+\sin ^6a+2\sin ^2a\cos ^2a+\cos ^6a\)

\(=(\sin ^2a)^3+(\cos ^2a)^3+3\sin ^2a\cos ^2a\)

\(=(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)+3\sin ^2a\cos ^2a\)

\(=\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a+3\sin ^2a\cos ^2a\)

\(=\sin ^4a+2\sin ^2a\cos ^2a+\cos ^4a\)

\(=(\sin ^2a+\cos ^2a)^2=1^2=1\)