cho sinx = \(-\dfrac{3}{5}\) và \(\pi\) < x < \(\dfrac{3\pi}{2}\) tính
a) sin2x, cos2x, tan2x, cot2x
b) \(sin\left(x+\dfrac{\pi}{3}\right)\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
H
19 tháng 8 2023
a)
$cos\left(x+\frac{\pi }{6}\right)=\frac{4}{5}cos\left(\frac{\pi }{6}\right)-\left(-\frac{3}{5}\right)sin\left(\frac{\pi }{6}\right)=\frac{4}{5}.\frac{\sqrt{3}}{2}+\frac{3}{5}.\frac{1}{2}=\frac{3+4\sqrt{3}}{10}$
b) $tan(x + \frac{\pi}{4}) = \frac{-3/5 + 1}{1 + (-3/5)(1)} = \frac{-2/5}{2/5} = -1$
19 tháng 8 2023
a: pi/2<x<pi
=>cosx<0
=>\(cosx=-\sqrt{1-\left(\dfrac{1}{5}\right)^2}=-\dfrac{2\sqrt{6}}{5}\)
\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{5}\cdot\dfrac{-2\sqrt{6}}{5}=\dfrac{-4\sqrt{6}}{25}\)
\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{24}{25}-1=\dfrac{48}{25}-1=\dfrac{23}{25}\)
\(tan2x=-\dfrac{4\sqrt{6}}{25}:\dfrac{23}{25}=-\dfrac{4\sqrt{6}}{23}\)
\(cot2x=1:\dfrac{-4\sqrt{6}}{23}=\dfrac{-23}{4\sqrt{6}}\)
b: \(sin\left(x-\dfrac{pi}{6}\right)=sinx\cdot cos\left(\dfrac{pi}{6}\right)-cosx\cdot sin\left(\dfrac{pi}{6}\right)\)
\(=sinx\cdot\dfrac{\sqrt{3}}{2}-cosx\cdot\dfrac{1}{2}\)
\(=\dfrac{1}{5}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{-2\sqrt{6}}{5}\cdot\dfrac{1}{2}=\dfrac{\sqrt{3}+2\sqrt{6}}{10}\)
c: \(cos\left(x-\dfrac{pi}{3}\right)=cosx\cdot cos\left(\dfrac{pi}{3}\right)+sinx\cdot sin\left(\dfrac{pi}{3}\right)\)
\(=-\dfrac{2\sqrt{6}}{5}\cdot\dfrac{1}{2}+\dfrac{1}{5}\cdot\dfrac{1}{2}=\dfrac{-2\sqrt{6}+1}{10}\)
d: \(tan\left(x-\dfrac{pi}{4}\right)=\dfrac{tanx-tan\left(\dfrac{pi}{4}\right)}{1+tanx\cdot tan\left(\dfrac{pi}{4}\right)}\)
\(=\dfrac{tanx-1}{1+tanx}\)
\(=\dfrac{\dfrac{1}{-2\sqrt{6}}-1}{1+\dfrac{1}{-2\sqrt{6}}}=\dfrac{-25-4\sqrt{6}}{23}\)
AH
Akai Haruma
Giáo viên
18 tháng 8 2021
Lời giải:
$-\frac{4}{5}=\cos 2x=2\cos ^2x-1$
$\Leftrightarrow \cos ^2x=\frac{1}{10}$
Vì $x\in (\frac{\pi}{4}; \frac{\pi}{2})$ nên $\cos x>0$
$\Rightarrow \cos x=\sqrt{\frac{1}{10}}$
$\sin^2x=1-\cos ^2x=\frac{9}{10}$
Vì $x\in (\frac{\pi}{4}; \frac{\pi}{2})$ nên $\sin x>0$
$\Rightarrow \sin x=\frac{3}{\sqrt{10}}$
$\sin (x+\frac{\pi}{3})=\sin x\cos \frac{\pi}{3}+\cos x\sin \frac{\pi}{3}$
$=\sqrt{\frac{9}{10}}.\frac{1}{2}+\sqrt{\frac{1}{10}}.\frac{\sqrt{3}}{2}=\frac{\sqrt{30}+3\sqrt{10}}{20}$
16 tháng 3 2022
\(\cos^2x=\sqrt{1-\dfrac{9}{25}}=\dfrac{16}{25}\)
mà \(\cos x< 0\)
nên \(\cos x=-\dfrac{4}{5}\)
=>\(\tan x=-\dfrac{3}{4};\cot x=-\dfrac{4}{3}\)
18 tháng 6 2023
a: -pi/2<a<0
=>sin a<0
=>sin a=-1/căn 5
tan a=-1/2
cot a=-2
b: pi/2<x<pi
=>cosx<0
=>cosx=-4/5
=>tan x=-3/4
cot x=-4/3
c: -pi<x<-pi/2
=>cosx<0 và sin x<0
1+tan^2x=1/cos^2x
=>1/cos^2x=1+16/25=41/25
=>cosx=-5/căn 41
sin x=-6/căn 41
cot x=5/4
g: 180 độ<x<270 độ
=>cosx <0
=>cosx=-4/5
tan x=3/4
cot x=4/3
NM
1
KQ
29 tháng 7 2023
\(sin\left(x\right)+\left[sin\left(x+\dfrac{2\pi}{5}\right)-sin\left(x+\dfrac{\pi}{5}\right)\right]+\left[sin\left(x+\dfrac{4\pi}{5}\right)-sin\left(x+\dfrac{3\pi}{5}\right)\right]\)
\(=sin\left(x\right)+2cos\left(x+\dfrac{3\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)+2cos\left(x+\dfrac{7\pi}{10}\right)sin\left(\dfrac{\pi}{10}\right)\)
\(=sin\left(x\right)+2sin\left(\dfrac{\pi}{10}\right)\left[cos\left(x+\dfrac{3\pi}{10}\right)+cos\left(x+\dfrac{7\pi}{10}\right)\right]\)
\(=sin\left(x\right)+4sin\left(\dfrac{\pi}{10}\right)cos\left(\dfrac{\pi}{5}\right)cos\left(x+\dfrac{\pi}{2}\right)\)
\(=sin\left(x\right)+cos\left(x+\dfrac{\pi}{2}\right)\)
\(=sin\left(x\right)+cos\left(x\right)cos\left(\dfrac{\pi}{2}\right)-sin\left(x\right)sin\left(\dfrac{\pi}{2}\right)\)
\(=sin\left(x\right)-sin\left(x\right)\)
\(=0\)
22 tháng 8 2023
a: sin x=-6/5=-1,2
mà -1<=sin x<=1
nên \(x\in\varnothing\)
b: sin3x=căn 3/2
=>3x=pi/3+k2pi hoặc 3x=2/3pi+k2pi
=>x=pi/9+k2pi/3 hoặc x=2/9pi+k2pi/3
c: \(sin\left(x+\dfrac{pi}{3}\right)=sin\left(\dfrac{3}{4}pi\right)\)
=>x+pi/3=3/4pi+k2pi hoặc x+pi/3=1/4pi+k2pi
=>x=5/12pi+k2pi hoặc x=-1/12pi+k2pi
d: =>sin(x+5/6pi)=5/4
mà sin(x+5/6pi) thuộc [-1;1]
nên \(x\in\varnothing\)
12 tháng 9 2021
Ta có \(sinx-cosx=\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\)
a, Do \(0< x< \dfrac{\pi}{4}\Rightarrow-\dfrac{\pi}{4}< x-\dfrac{\pi}{4}< 0\)
⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) < 0
⇒ sinx - cosx < 0
=> sinx < cosx
b, Do \(\dfrac{\pi}{4}< x< \dfrac{\pi}{2}\Rightarrow0< x-\dfrac{\pi}{4}< \dfrac{\pi}{4}\)
⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) > 0
⇒ sinx - cosx > 0
=> sinx > cosx
a: pi<x<3/2pi
=>cosx<0
=>\(cosx=-\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)
\(tanx=\dfrac{-3}{5}:\dfrac{-4}{5}=\dfrac{3}{4}\)
cot x=1:3/4=4/3
\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{-3}{5}\cdot\dfrac{-4}{5}=\dfrac{24}{25}\)
\(cos2x=1-2\cdot sin^2x=1-2\cdot\left(-\dfrac{3}{5}\right)^2=\dfrac{7}{25}\)
\(tan2x=\dfrac{24}{25}:\dfrac{7}{25}=\dfrac{24}{7}\)
cot 2x=1:24/7=7/24
b: \(sin\left(x+\dfrac{pi}{3}\right)=sinx\cdot cos\left(\dfrac{pi}{3}\right)+sin\left(\dfrac{pi}{3}\right)\cdot cosx\)
\(=\dfrac{-3}{5}\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{-4}{5}=\dfrac{-3-4\sqrt{3}}{10}\)