4

\(\sin\alpha=4\cos\alpha\Leftrightarrow\cos\alpha=\dfrac{1}{4}\sin\alpha\)

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

\(\Leftrightarrow\sin^2\alpha+\left(\dfrac{1}{4}\sin\alpha\right)^2=1\\ \Leftrightarrow\sin^2\alpha+\dfrac{1}{16}\sin^2\alpha=1\\ \Leftrightarrow\dfrac{17}{16}\sin^2\alpha=1\\ \Leftrightarrow\sin^2\alpha=\dfrac{16}{17}\)

Vì \(\alpha\) là một góc nhọn => \(\alpha>0\) => \(\sin\alpha=\sqrt{\dfrac{16}{17}}=\dfrac{4\sqrt{17}}{17}\)

\(\cos\alpha=\dfrac{1}{4}\sin\alpha=\dfrac{1}{4}.\dfrac{4\sqrt{17}}{17}=\dfrac{\sqrt{17}}{17}\)

Vậy \(P=3\sin\alpha.\cos\alpha=3.\dfrac{4\sqrt{17}}{17}.\dfrac{\sqrt{17}}{17}=\dfrac{12}{17}\)

a) Có: `1+tan^2a=1/(cos^2a)`

`<=> 1+(3/5)^2=1/(cos^2a)`

`=> cosa=\sqrt10/4`

`=> sina = \sqrt(1-cos^2a) = \sqrt6/4`

b) Có: `sin^2a + cos^2a=1`

`<=> sin^2a + (1/4)^2=1`

`=> sina=\sqrt15/4`

`=> tana = (sina)/(cosa) = \sqrt15`

a) Giả sử tam giác ABC vuông tại B có \(tanA=\dfrac{3}{5}\)

\(\Rightarrow\dfrac{BC}{AB}=\dfrac{3}{5}\Rightarrow BC=\dfrac{3}{5}AB\Rightarrow AC=\sqrt{AB^2+\dfrac{9}{25}AB^2}=\dfrac{\sqrt{34}}{5}AB\)

\(\Rightarrow\dfrac{AB}{AC}=\dfrac{5}{\sqrt{34}}\Rightarrow cosA=\dfrac{5}{\sqrt{34}}\)

\(AC=\dfrac{\sqrt{34}}{5}AB\Rightarrow AC=\dfrac{\sqrt{34}}{5}.\dfrac{5}{3}BC=\dfrac{\sqrt{34}}{3}BC\Rightarrow\dfrac{BC}{AC}=\dfrac{3}{\sqrt{34}}\)

\(\Rightarrow sinA=\dfrac{3}{\sqrt{34}}\)

b) cũng tương tự như câu a thôi,bạn tự tính nha

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

\(\Rightarrow tana=\pm\sqrt{\frac{1}{cos^2a}-1}=\pm\frac{\sqrt{21}}{2}\)

Lời giải:

Do góc $a$ nhọn nên các tỉ số lượng giác mang giá trị dương.

Áp dụng công thức $\sin ^2a+\cos ^2a=1$

$\Rightarrow \cos^2 a=1-\sin ^2a=1-0,28^2=0,9216$

$\Rightarrow \cos a=\frac{24}{25}=0,96$

$\tan a=\frac{\sin a}{\cos a}=\frac{0,28}{0,96}=\frac{7}{24}$

$\cot a=\frac{1}{\tan a}=\frac{24}{7}$

\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\dfrac{16}{25}=\dfrac{9}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{3}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\\ \cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{3}{4}\)

\(\sin A=0,8\Rightarrow A=arcsin0,8_{ }\)

\(\Rightarrow\cos A=cos\left(arcsin0,8\right)=\dfrac{3}{5}\)

     tanA=tan(arcsin0,8)=4/3

     cotA=1:4/3=3/4