Ta có: 

A. \(\alpha< \beta\)

\(\Rightarrow\left(0,3\right)^{\alpha}>\left(0,3\right)^{\beta}\)

Sai 

B. \(\alpha< \beta\)

\(\Rightarrow\pi^{\alpha}< \pi^{\beta}\)

Sai

C. \(\alpha< \beta\)

\(\Rightarrow\left(\sqrt{2}\right)^{\alpha}< \left(\sqrt{2}\right)^{\beta}\)

Đúng

D. \(\alpha< \beta\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{\alpha}>\left(\dfrac{1}{2}\right)^{\beta}\)

Sai 

⇒ Chọn C

Theo định lý py - ta - go ta có :

\(BC=\sqrt{\left(3a\right)^2+\left(4a\right)^2}=\sqrt{25a^2}=5a\)

Theo hệ thức lượng ta có :

\(AH=\dfrac{AB.AC}{BC}=\dfrac{3a.4a}{5a}=\dfrac{12a^2}{5a}=\dfrac{12}{5}a\)

Theo tỉ số lượng giác :

\(\tan ABC=\dfrac{AC}{AB}=\dfrac{4a}{3a}=\dfrac{4}{3}\)

mình nghĩ nên sửa đề là \(-\sqrt{a^2+b^2}\le a\cos\alpha+b\sin\alpha\le\sqrt{a^2+b^2}\)

với a,b,x,y là số thực ta luôn có \(\left(ax+by\right)^2=\left(a^2+b^2\right)\left(x^2+y^2\right)-\left(ay-bx\right)^2\)

từ đẳng thức này ta suy ra \(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)

dấu "=" xảy ra khi \(\left(ax-by\right)^2=0\)

trở lại bài toán ta luôn có \(\left(a\cos\alpha+b\sin\alpha\right)^2\le\left(a^2+b^2\right)\left(\cos^2\alpha+\sin^2\alpha\right)=a^2+b^2\)

từ đó ta có \(-\sqrt{a^2+b^2}\le a\cos\alpha+b\sin\alpha\le\sqrt{a^2+b^2}\)

\(\left(a.cos\alpha+b.sin\alpha\right)^2\le\left(a^2+b^2\right)\left(sin^2a+cos^2a\right)=a^2+b^2\)

\(\Rightarrow-\sqrt{a^2+b^2}\le a.cos\alpha+b.sin\alpha\le\sqrt{a^2+b^2}\)

\(sin\left(\frac{9\pi}{2}+\alpha\right)=sin\left(4\pi+\frac{\pi}{2}+\alpha\right)=sin\left(\frac{\pi}{2}+\alpha\right)=cos\alpha\)

1) \(tan\alpha=\dfrac{2}{3}\)

Mà: \(tan\alpha\cdot cot\alpha=1\)

\(\Rightarrow cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{\dfrac{2}{3}}=\dfrac{3}{2}\) 

Và: \(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)

\(\Rightarrow cos^2\alpha=\dfrac{1}{1+tan^2\alpha}\)

\(\Rightarrow cos\alpha=\sqrt{\dfrac{1}{1+tan^2\alpha}}=\sqrt{\dfrac{1}{1+\left(\dfrac{2}{3}\right)^2}}=\dfrac{3\sqrt{13}}{13}\)

Lại có:

\(tan\alpha=\dfrac{sin\alpha}{cos\alpha}\)

\(\Rightarrow sin\alpha=tan\alpha\cdot cos\alpha=\dfrac{2}{3}\cdot\dfrac{3\sqrt{13}}{13}=\dfrac{2\sqrt{13}}{13}\)