\(A=1-cos^2x+2cosx+1=3-\left(cosx-1\right)^2\le3\)

\(A_{max}=3\) khi \(cosx=1\)

\(B=1-sin^2x-2sin^2x-3=-1-\left(sinx+1\right)^2\le-1\)

\(B_{max}=-1\) khi \(sinx=-1\)

\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{2}-1\right)}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{2}}}}=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{2}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{4}-1\right)}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{cos^2\frac{x}{4}}}=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{x}{4}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{x}{8}-1\right)}=\sqrt{cos^2\frac{x}{8}}=cos\frac{x}{8}\)

\(B=\sqrt{2+\sqrt{2+\sqrt{2+2\left(2cos^2\frac{a}{2}-1\right)}}}\)

\(=\sqrt{2+\sqrt{2+\sqrt{4cos^2\frac{a}{2}}}}=\sqrt{2+\sqrt{2+2cos\frac{a}{2}}}\)

\(=\sqrt{2+\sqrt{2+2\left(cos^2\frac{a}{4}-1\right)}}=\sqrt{2+\sqrt{4cos^2\frac{a}{4}}}\)

\(=\sqrt{2+2cos\frac{a}{4}}=\sqrt{2+2\left(2cos^2\frac{a}{8}-1\right)}=2cos\frac{a}{8}\)

a/

\(\frac{1}{sinx}+\frac{cosx}{sinx}=\frac{1+cosx}{sinx}=\frac{1+2cos^2\frac{x}{2}-1}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=cot\frac{x}{2}\)

b/

\(\frac{1-cosx}{sinx}=\frac{1-\left(1-2sin^2\frac{x}{2}\right)}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{2sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=tan\frac{x}{2}\)

c/

\(tan\frac{x}{2}\left(\frac{1}{cosx}+1\right)=\left(\frac{1-cosx}{sinx}\right)\left(\frac{1}{cosx}+1\right)=\frac{\left(1-cosx\right)\left(1+cosx\right)}{sinx.cosx}=\frac{1-cos^2x}{sinx.cosx}\)

\(=\frac{sin^2x}{sinx.cosx}=\frac{sinx}{cosx}=tanx\)

d/

\(\frac{sin2a}{2cosa\left(1+cosa\right)}=\frac{2sina.cosa}{2cosa\left(1+2cos^2\frac{a}{2}-1\right)}=\frac{sina}{2cos^2\frac{a}{2}}=\frac{2sin\frac{a}{2}cos\frac{a}{2}}{2cos^2\frac{a}{2}}=tan\frac{a}{2}\)

e/

\(cotx+tan\frac{x}{2}=\frac{cosx}{sin}+\frac{1-cosx}{sinx}=\frac{cosx+1-cosx}{sinx}=\frac{1}{sinx}\)

Các câu c, e đều sử dụng kết quả từ câu b

f/

\(3-4cos2x+cos4x=3-4cos2x+2cos^22x-1\)

\(=2cos^22x-4cos2x+2=2\left(cos^22x-2cos2x+1\right)\)

\(=2\left(cos2x-1\right)^2=2\left(1-2sin^2x-1\right)^2\)

\(=2.\left(-2sin^2x\right)^2=8sin^4x\)

g/

\(\frac{1-cosx}{sinx}=\frac{sinx\left(1-cosx\right)}{sin^2x}=\frac{sinx\left(1-cosx\right)}{1-cos^2x}=\frac{sinx\left(1-cosx\right)}{\left(1-cosx\right)\left(1+cosx\right)}=\frac{sinx}{1+cosx}\)

h/

\(sinx+cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}+cosx.\frac{\sqrt{2}}{2}\right)\)

\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)

i/

\(sinx-cosx=\sqrt{2}\left(sinx.\frac{\sqrt{2}}{2}-cosx.\frac{\sqrt{2}}{2}\right)\)

\(=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)

j/

\(cosx-sinx=\sqrt{2}\left(cosx.\frac{\sqrt{2}}{2}-sinx\frac{\sqrt{2}}{2}\right)\)

\(=\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

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

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{4}}=\sqrt{\frac{1}{2}+\frac{1}{2}\left(cos^2\frac{a}{8}-1\right)}\)

\(=cos\frac{a}{8}\Rightarrow n=8\)

sinα = \(\frac{1}{\sqrt{3}}\) nên cos²α = 1- sin²α = 1 - \(\frac{1}{3}\)= \(\frac{2}{3}\) ⇒ cosα = \(\pm\sqrt{\frac{2}{3}}\)

mà 0 < α < \(\frac{\pi}{2}\) ⇒ cosα > 0, nên cosα = \(\sqrt{\frac{2}{3}}\)

ta có \(cos\left(\alpha+\frac{\pi}{3}\right)\)= \(cos\alpha.cos\frac{\pi}{3}-sin\alpha.sin\frac{\pi}{3}\)=\(\sqrt{\frac{2}{3}}.\frac{1}{2}-\frac{1}{\sqrt{3}}.\frac{\sqrt{3}}{2}\)

=\(\frac{-3+\sqrt{6}}{6}\)

\(\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}\)

\(=\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}\)

\(=\sqrt{\frac{1}{2}-\frac{1}{2}\sqrt{cos^2\frac{a}{2}}}=\sqrt{\frac{1}{2}-\frac{1}{2}cos\frac{a}{2}}\)

\(=\sqrt{\frac{1}{2}-\frac{1}{2}\left(1-2sin^2\frac{a}{4}\right)}=\sqrt{\frac{1}{2}-\frac{1}{2}+sin^2\frac{a}{4}}\)

\(=\sqrt{sin^2\frac{a}{4}}=sin\frac{a}{4}\)

\(C=2sin3x.cos3x.\frac{cos3x}{sin3x}-\left(cos^23x-sin^23x\right)\)

\(=2cos^23x-cos^23x+sin^23x=cos^23x+sin^23x=1\)

\(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)\)

\(=\sqrt{2}\left(sinx.sin\frac{\pi}{4}-cosx.cos\frac{\pi}{4}\right)=-\sqrt{2}\left(cosx.cos\frac{\pi}{4}-sinx.sin\frac{\pi}{4}\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)

Câu này bạn ghi nhầm đề (lưu ý rằng \(sin\frac{\pi}{4}=cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\))

Câu 2b bạn cũng xem lại đề, chắc chắn ko đúng

\(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=90^0\Rightarrow sin\frac{A}{2}=cos\left(\frac{B}{2}+\frac{C}{2}\right)=cos\frac{B}{2}cos\frac{C}{2}-sin\frac{B}{2}sin\frac{C}{2}\)

Câu 3 bạn cũng ghi sai đề luôn

Trong 1 ngày đẹp trời thì câu 4 cũng sai luôn cho đỡ lạc lõng đồng đội:

\(sin\left(a+b-b\right)=sin\left(a+b\right)cosb-cos\left(a+b\right)sinb=2sin\left(a+b\right)\)

\(\Leftrightarrow sin\left(a+b\right)\left[cosb-2\right]=cos\left(a+b\right).sinb\)

\(\Leftrightarrow\frac{sin\left(a+b\right)}{cos\left(a+b\right)}=\frac{sinb}{cosb-2}\Leftrightarrow tan\left(a+b\right)=\frac{sinb}{cosb-2}\)

4 câu bạn ghi đúng đề bài duy nhất câu 1, kinh thiệt :(