Question

Bài 1: Cho sin α=\(\frac{1}{5}\) với 0∠α∠\(\frac{\pi}{2}\). Tính cos (α-\(\frac{\pi}{6}\))

Bài 2: Cho cos x=\(\frac{-2}{3}\) với\(\frac{\pi}{2}\)∠x∠π. Tính tan (\(\frac{\pi}{4}\)+x)

Bài 3: Cho tan α=\(\frac{-4}{7}\) với \(\frac{3\pi}{2}\)∠α∠2π. Tính cos (2α -\(\frac{\pi}{2}\))

Bài 4: Cho sin α =\(\frac{1}{2}\) với 0∠α∠π. Tính tan (2α -\(\frac{\pi}{2}\)) +sin α

Answer 1

Bài 1:

$\cos ^2a=1-\sin ^2a=1-\frac{1}{5^2}=\frac{24}{25}$

Vì $0< a< \frac{\pi}{2}$ nên $\cos a>0$

$\Rightarrow \cos a=\frac{\sqrt{24}}{5}$

\(\cos (a-\frac{\pi}{6})=\cos a.\cos \frac{\pi}{6}+\sin a\sin \frac{\pi}{6}=\cos a.\frac{\sqrt{3}}{2}+\sin a. \frac{1}{2}\)

\(=\frac{\sqrt{24}}{5}.\frac{\sqrt{3}}{2}+\frac{1}{5}.\frac{1}{2}=\frac{1+6\sqrt{2}}{10}\)

Bài 2:

$\cos x=\frac{-2}{3}\Rightarrow \sin ^2x=1-\cos ^2x=\frac{5}{9}$

Vì $x\in (\frac{\pi}{2}; \pi)$ nên $\sin x>0\Rightarrow \sin x=\frac{\sqrt{5}}{3}$

\(\Rightarrow \tan x=\frac{\sin x}{\cos x}=\frac{-\sqrt{5}}{2}\)

Do đó:

\(\tan (\frac{\pi}{4}+x)=\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4}.\tan x}=\frac{1+\tan x}{1-\tan x}=\frac{1-\frac{\sqrt{5}}{2}}{1+\frac{\sqrt{5}}{2}}=-9+4\sqrt{5}\)

Answer 2

Bài 3:

\(\tan a=\frac{-4}{7}=\frac{\sin a}{\cos a}\)

\(\Rightarrow \frac{\sin ^2a}{\cos ^2a}=\frac{16}{49}\Rightarrow \frac{1}{\cos ^2a}=\frac{65}{49}\) \(\Rightarrow \cos ^2a=\frac{49}{65}\)

Kết hợp điều kiện của $a$ suy ra $\cos a>0\Rightarrow \cos a=\frac{7}{\sqrt{65}}$

$\Rightarrow \sin a=\frac{-4}{7}\cos a=\frac{-4}{\sqrt{65}}$

Do đó:

\(\cos (2a-\frac{\pi}{2})=\cos 2a.\cos \frac{\pi}{2}+\sin 2a.\sin \frac{\pi}{2}\)

\(=(\cos ^2a-\sin ^2a).0+2\sin a\cos a.1=2\sin a\cos a=2.\frac{-4}{\sqrt{65}}.\frac{7}{\sqrt{65}}=\frac{56}{65}\)

Answer 3

Bài 4:

$\sin a=\frac{1}{2}$ và $0< a< \pi$ nên $a=\frac{\pi}{6}$ hoặc $a=\frac{5}{6}\pi$

Nếu $a=\frac{\pi}{6}$ thì $\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{-\sqrt{3}}{3}+\frac{1}{2}=\frac{3-2\sqrt{3}}{6}$

Nếu $a=\frac{5\pi}{6}$ thì:

\(\tan (2a-\frac{\pi}{2})+\sin a=\tan (2.\frac{5\pi}{6}-\frac{\pi}{2})+\frac{1}{2}=\frac{\sqrt{3}}{3}+\frac{1}{2}=\frac{3+2\sqrt{3}}{6}\)