Ko biết làm

Bài 1: 

\(\cos\alpha=\dfrac{4}{5}\)

\(\tan\alpha=\dfrac{3}{4}\)

\(\cot\alpha=\dfrac{4}{3}\)

1: Ta sẽ dựng ΔABC vuông tại A có AB=6cm; AC=8cm; BC=10cm

Ta sẽ được sin C=AB/BC=6/10=3/5

2: TA có: \(1+\tan^2a=\frac{1}{cos^2a}\)

=>\(1+\tan^2a=\frac{1}{0,6^2}=1:0,36=1:\frac{9}{25}=\frac{25}{9}\)

=>\(\tan^2a=\frac{25}{9}-1=\frac{16}{9}\)

=>tan a=\(\sqrt{\frac{16}{9}}=\frac43\)

\(\cos\alpha=\sqrt{1-\sin^2\alpha}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\)

\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\)

\(\cot=\frac{1}{\tan}=\frac{1}{\frac{2\sqrt{5}}{5}}=\frac{\sqrt{5}}{2}\)

a) (H.a)

– Dựng góc vuông xOy.

-Trên tia Ox đặt OA=2

– Dựng đường tròn (A;3) cắt tia Oy tại B

Khi đó góc OBA = α

Thật vậy

b) (H.b)

Tương tự:

b) (H.b)

c) (H.c)

d) (H.d).

a: Xét ΔABC vuông tại A có \(\hat{B}+\hat{C}=90^0\)

=>cot C=tan B=2

Ta có: \(\tan C\cdot\cot C=1\)

=>\(\tan C=\frac12\)

Ta có: \(1+\tan^2C=\frac{1}{cos^2C}\)

=>\(\frac{1}{cos^2C}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2C=\frac45\)

=>\(cosC=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-\frac45=\frac15\)

=>sin C=\(\frac{1}{\sqrt5}\)

b: \(\sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\frac{\sqrt3}{2}\right)^2=1-\frac34=\frac14\)

=>\(cosB=\frac12\)

tan B=\(\frac{\sin B}{cosB}=\frac{\sqrt3}{2}:\frac12=\sqrt3\)

\(cotB=\frac{1}{\tan B}=\frac{1}{\sqrt3}\)

a: Xét ΔABC vuông tại A có \(\hat{B}+\hat{C}=90^0\)

=>cot C=tan B=2

Ta có: \(\tan C\cdot\cot C=1\)

=>\(\tan C=\frac12\)

Ta có: \(1+\tan^2C=\frac{1}{cos^2C}\)

=>\(\frac{1}{cos^2C}=1+\left(\frac12\right)^2=\frac54\)

=>\(cos^2C=\frac45\)

=>\(cosC=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\)

Ta có: \(\sin^2C+cos^2C=1\)

=>\(\sin^2C=1-\frac45=\frac15\)

=>sin C=\(\frac{1}{\sqrt5}\)

b: \(\sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\frac{\sqrt3}{2}\right)^2=1-\frac34=\frac14\)

=>\(cosB=\frac12\)

tan B=\(\frac{\sin B}{cosB}=\frac{\sqrt3}{2}:\frac12=\sqrt3\)

\(cotB=\frac{1}{\tan B}=\frac{1}{\sqrt3}\)

bài 1 : ta có : \(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\left(0,6\right)^2=\dfrac{16}{25}\)

\(\Rightarrow cosa=\pm\dfrac{4}{5}\)

\(\Rightarrow tanx=\dfrac{sinx}{cosx}=\pm\dfrac{3}{4}\) \(\Rightarrow cotx=\dfrac{1}{tanx}=\pm\dfrac{4}{3}\)

bài 2)

ý 1 : a) ta có : \(\dfrac{1}{cos^2a}=\dfrac{sin^2a+cos^2a}{cos^2a}=tan^2a+1\left(đpcm\right)\)

b) ta có : \(\dfrac{1}{sin^2a}=\dfrac{sin^2a+cos^2a}{sin^2a}=1+cot^2a\left(đpcm\right)\)

c) \(cos^4a-sin^4a=\left(sin^2a+cos^2a\right)\left(cos^2a-sin^2a\right)\)

\(=cos^2a-sin^2a=2cos^2a-cos^2a-sin^2a=2cos^2a-1\left(đpcm\right)\)

ý 2 :

ta có : \(tana=2\Rightarrow cota=\dfrac{1}{2}\)

ta có : \(tan^2a+1=\dfrac{1}{cos^2a}\Leftrightarrow cos^2a=\dfrac{1}{tan^2a+1}=\dfrac{1}{5}\)

\(\Rightarrow cosa=\pm\dfrac{1}{\sqrt{5}}\Rightarrow sin^2a=1-cos^2a=\dfrac{4}{5}\) \(\Rightarrow sina=\pm\dfrac{2}{\sqrt{5}}\)

vậy ............................................................................

bài 3 bạn tự luyện tập như bài 2 cho quen nha :)