Dùng công thức trực quan đi.

a) Ta có: \(\sin^2a^o=\cos^2\left(90^o-a^o\right)\)

Biểu thức trên

\(=\left(\sin^21^o+\sin^o89\right)+\left(\sin^22^o+\sin^288^o\right)+...+\left(\sin^244^o+\sin^246^o\right)+\sin^245^o\)

\(=\left(\sin^21^o+\cos^21^o\right)+\left(\sin^22^o+\cos^22^o\right)+...+\left(\sin^244^o+\cos^246^o\right)+\sin^245^o\)

\(=1+1+..+1+\sin^245^o=44+\frac{1}{2}=\frac{89}{2}\)

b) 

Ta có: \(\sin^2x+\cos^2x=1\)

\(0^o< x< 90^o\)

=> \(0< \sin x;\cos x< 1\)

Ta có:  \(\frac{\sin^2x+\cos^2x}{\text{​​}\text{​​}\sin x.\cos x}=\frac{1}{\frac{12}{25}}=\frac{25}{12}\Leftrightarrow\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{25}{12}\)

\(\Leftrightarrow\tan x+\frac{1}{\tan x}=\frac{25}{12}\Leftrightarrow\tan^2x-\frac{25}{12}\tan x+1=0\)

Đặt t =tan x => có phương trình bậc 2 ẩn t => Giải đen ta => ra đc t => ra đc tan t

\(\Leftrightarrow\orbr{\begin{cases}\tan x=\frac{3}{4}\\\tan x=\frac{4}{3}\end{cases}}\)

ĐKXĐ: ...

\(\Leftrightarrow tan^2x+cot^2x=2\left(cos^4x+sin^4x+2sin^2x.cos^2x\right)\)

\(\Leftrightarrow tan^2x+cot^2x=2\left(sin^2x+cos^2x\right)^2\)

\(\Leftrightarrow tan^2x+cot^2x=2\)

\(\Leftrightarrow\left(tanx-cotx\right)^2=0\)

\(\Leftrightarrow tanx=cotx=tan\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow x=\frac{\pi}{2}-x+k\pi\)

\(\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)

Xét \(\Delta ABC\) vuông tại A có:
\(sin\left(B\right)=\dfrac{AC}{BC}\)
\(cotg\left(C\right)=\dfrac{AC}{AB}\)
BC là cạnh huyền của \(\Delta ABC\) \(\left(AB,AC< BC\right)\)
\(\Rightarrow\dfrac{AC}{BC}< \dfrac{AC}{AB}\Rightarrow sin\left(B\right)< cotg\left(C\right)\)

a/

\(0\le sin^2x\le1\Rightarrow-2\le f\left(x\right)\le1\)

\(f\left(x\right)_{min}=-2\) khi \(sin^2x=1\)

\(f\left(x\right)_{max}=1\) khi \(sin^2x=1\)

b/

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

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

Do \(-1\le cosx\le1\Rightarrow\left\{{}\begin{matrix}cosx-1\le0\\2-cosx>0\end{matrix}\right.\)

\(\Rightarrow\left(cosx-1\right)\left(2-cosx\right)\le0\Rightarrow g\left(x\right)\le1\)

\(g\left(x\right)_{max}=1\) khi \(cosx=1\)

\(g\left(x\right)=-cos^2x+3cosx+4-5=\left(cosx+1\right)\left(4-cosx\right)-5\)

\(\left(cosx+1\right)\left(4-cosx\right)\ge0\Rightarrow g\left(x\right)\ge-5\)

\(g\left(x\right)_{min}=-5\) khi \(cosx=-1\)

Ta có: \(sin^2x\le1\Rightarrow y\ge2sinx+\sqrt{3-1}=2sinx+\sqrt{2}\)

Mặt khác \(sinx\ge-1\Rightarrow y\ge-2+\sqrt{2}\)

\(\Rightarrow y_{min}=-2+\sqrt{2}\) khi \(sinx=-1\)

\(y^2=3sin^2x+3+4sinx\sqrt{3-sin^2x}\)

\(y^2=3sin^2x+3+2\sqrt{2}\left(\sqrt{2}sinx.\sqrt{3-sin^2x}\right)\)

\(y^2\le3sin^2x+3+\sqrt{2}\left(2sin^2x+3-sin^2x\right)=\left(3+\sqrt{2}\right)sin^2x+3+3\sqrt{2}\)

Do \(sin^2x\le1\Rightarrow y^2\le3+\sqrt{2}+3+3\sqrt{2}=6+4\sqrt{2}\)

\(\Rightarrow y\le\sqrt{6+4\sqrt{2}}=2+\sqrt{2}\)

\(y_{max}=2+\sqrt{2}\) khi \(sinx=1\)