245sin

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\(A=\frac{\sqrt{2}}{2}cos^252+\frac{\sqrt{2}}{2}sin^252=\frac{\sqrt{2}}{2}\left(sin^252+cos^252\right)=\frac{\sqrt{2}}{2}\)

\(B=\sqrt{3}.cos^247+\sqrt{3}.sin^247=\sqrt{3}\left(sin^247+cos^247\right)=\sqrt{3}\)

Oử đây rất ít người học lớp 9 , bạn lên Học 24h để các giáo viên giải cho nha !

tks bạn!

\(D=\cos45^0\cdot\cos^223^0+\sin45^0\cdot\cos^267^0\)

\(=\dfrac{\sqrt{2}}{2}\left(\cos^223^0+\cos^267^0\right)\)

\(=\dfrac{\sqrt{2}}{2}\)

\(ADCT:\sin^2\alpha+\cos^2\alpha=1\)

\(A=\left(\sin^242^0+\sin^248^0\right)+\left(\sin^243^0+\sin^247^0\right)+\left(\sin^244^0+\sin^246^0\right)+\sin45^0\)

\(A=\left(\sin^242^0+\cos^242^0\right)+\left(\sin^243^0+\cos^243^0\right)+\left(\sin^244^0+\cos^244^0\right)+\frac{\sqrt{2}}{2}\)

\(A=1+1+1+\frac{\sqrt{2}}{2}=\frac{6+\sqrt{2}}{2}\)

Câu b lm tương tự

\(=\left(1-sin^247^0\right)\cdot sin45^0+sin47^0\cdot\left(1-sin^245^0\right)\)

\(=sin45-sin^247^0\cdot\dfrac{\sqrt{2}}{2}+sin47^0-\dfrac{1}{2}\cdot sin47^0\)

\(\simeq0.69\)

Chú ý 2 điều: \(\cos45^o=\sin45^o=\frac{\sqrt{2}}{2}\) và \(\cos^2a+\sin^2a=1\)

Do đó: 

a) \(A=\cos^252^o.\frac{\sqrt{2}}{2}+\sin^252^o.\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\cos^252^o+\sin^252^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)

b) \(B=\frac{\sqrt{2}}{2}.\cos^247^o+\frac{\sqrt{2}}{2}.\sin^247^o=\frac{\sqrt{2}}{2}\left(\cos^247^o+\sin^247^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)