\(B=\sin^247^o\times\cos45^o+\sin45^o\times\cos^247^o\)
\(B=\sin^247^o\times\cos45^o+\cos45^o\times\cos^247^o\)
\(B=\cos45^o\left(\sin^247^o+\cos^247^o\right)\)
\(B=\cos45^o.1=\cos45^o\)

\(gi\text{ỏi}-qu\text{á}-nh\text{ỉ}\)

Thanks :) <3

245sin

Bạn giải từng bước hộ mik đc ko

mik hay milk

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

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

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

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

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

a: \(=\dfrac{\sqrt{2}}{2}\left(cos^252^0+sin^252^0\right)=\dfrac{\sqrt{2}}{2}\)

b: \(=\dfrac{\sqrt{2}}{2}\left(cos^247^0+sin^247^0\right)=\dfrac{\sqrt{2}}{2}\)

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

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

tks bạn!

a: \(=\dfrac{a^2-b^2}{\dfrac{\sqrt{2}}{2}a+b\cdot0-2a\cdot0}=\dfrac{a^2-b^2}{\dfrac{\sqrt{2}}{2}a}\)

b: \(=3a+b-a=2a+b\)