tính B=cos\(\frac{\text{π}}{7}\)*cos\(\frac{3\text{π}}{7}\)*cos\(\frac{5\text{π}}{7}\)
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26 tháng 7 2023
a: pi/2<a<pi
=>sin a>0
\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)
\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)
\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)
b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
c: \(sin\left(a-\dfrac{pi}{3}\right)\)
\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)
\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)
d: \(cos\left(a-\dfrac{pi}{6}\right)\)
\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 6 2020
\(A=\frac{1}{2}+\frac{1}{2}cos2x+\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}+\frac{1}{2}cos2x+cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}+\frac{1}{2}cos2x-\frac{1}{2}cos2x=\frac{3}{2}\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{4\pi}{3}\right)+\frac{1}{2}-\frac{1}{2}cos\left(2x-\frac{4\pi}{3}\right)\)
\(=\frac{3}{2}-\frac{1}{2}cos2x-cos2x.cos\frac{4\pi}{3}\)
\(=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x=\frac{3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 6 2020
\(sinx+cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}+cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(=\sqrt{2}cos\left(\frac{\pi}{2}-\left(x+\frac{\pi}{4}\right)\right)=\sqrt{2}cos\left(\frac{\pi}{4}-x\right)=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
\(sinx-cosx=\sqrt{2}\left(\frac{\sqrt{2}}{2}sinx-\frac{\sqrt{2}}{2}cosx\right)=\sqrt{2}\left(sinx.cos\frac{\pi}{4}-cosx.sin\frac{\pi}{4}\right)=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\)
\(=-\sqrt{2}sin\left(\frac{\pi}{4}-x\right)=-\sqrt{2}cos\left(\frac{\pi}{2}-\left(\frac{\pi}{4}-x\right)\right)=-\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\)
\(sin^4x-cos^4x=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+sin2x\)
\(=sin^2x-cos^2x+sin2x=sin2x-cos2x\)
\(=\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)
Bạn ghi ko đúng đề
12 tháng 5 2022
\(\cos a=\dfrac{-12}{13}\)
\(\sin b=\dfrac{4}{5}\)
\(\sin\left(a+b\right)=\sin a\cos b+\sin b\cos a\)
\(=\dfrac{5}{13}\cdot\dfrac{3}{5}+\dfrac{4}{5}\cdot\dfrac{-12}{13}=\dfrac{-45}{65}=\dfrac{-9}{13}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 6 2020
\(cot1,25.tan\left(4\pi+1,25\right)-sin\left(x+\frac{\pi}{2}\right).cos\left(6\pi-x\right)=0\)
\(\Leftrightarrow cot1,25.tan1,25-cosx.cos\left(-x\right)=0\)
\(\Leftrightarrow1-cos^2x=0\)
\(\Leftrightarrow sin^2x=0\Rightarrow sinx=0\Rightarrow tanx=0\)
AH
Akai Haruma
Giáo viên
24 tháng 7 2020
Lê Huy Hoàng:
a) ĐK: $x\in\mathbb{R}\setminus \left\{k\pi\right\}$ với $k$ nguyên
PT $\Leftrightarrow \tan ^2x-4\tan x+5=0$
$\Leftrightarrow (\tan x-2)^2+1=0$
$\Leftrightarrow (\tan x-2)^2=-1< 0$ (vô lý)
Do đó pt vô nghiệm.
AH
Akai Haruma
Giáo viên
24 tháng 7 2020
c)
ĐK:.............
PT $\Leftrightarrow 1+\frac{\sin ^2x}{\cos ^2x}-1+\tan x-\sqrt{3}(\tan x+1)=0$
$\Leftrightarrow \tan ^2x+\tan x-\sqrt{3}(\tan x+1)=0$
$\Leftrightarrow \tan ^2x+(1-\sqrt{3})\tan x-\sqrt{3}=0$
$\Rightarrow \tan x=\sqrt{3}$ hoặc $\tan x=-1$
$\Rightarrow x=\pi (k-\frac{1}{4})$ hoặc $x=\pi (k+\frac{1}{3})$ với $k$ nguyên
d)
ĐK:.......
PT $\Leftrightarrow \tan x-\frac{2}{\tan x}+1=0$
$\Leftrightarrow \tan ^2x+\tan x-2=0$
$\Leftrightarrow (\tan x-1)(\tan x+2)=0$
$\Rightarrow \tan x=1$ hoặc $\tan x=-2$
$\Rightarrow x=k\pi +\frac{\pi}{4}$ hoặc $x=k\pi +\tan ^{-2}(-2)$ với $k$ nguyên.
\(B=cos\frac{\pi}{7}.cos\left(\pi-\frac{4\pi}{7}\right).cos\left(\pi-\frac{2\pi}{7}\right)\)
\(B=cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(B.sin\frac{\pi}{7}=sin\frac{\pi}{7}.cos\frac{\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(B.sin\frac{\pi}{7}=\frac{1}{2}sin\frac{2\pi}{7}.cos\frac{2\pi}{7}.cos\frac{4\pi}{7}\)
\(B.sin\frac{\pi}{7}=\frac{1}{4}sin\frac{4\pi}{7}.cos\frac{4\pi}{7}=\frac{1}{8}sin\frac{8\pi}{7}\)
\(B.sin\frac{\pi}{7}=\frac{1}{8}sin\left(\pi+\frac{\pi}{7}\right)=-\frac{1}{8}sin\frac{\pi}{7}\)
\(\Rightarrow B=-\frac{1}{8}\)