cho \(sinx\) = \(\dfrac{1}{5}\) và \(\dfrac{\pi}{2}\) < x < \(\pi\) tính
a) sin2x, cos2x, tan2x, cot2x
b) \(sin\left(x-\dfrac{\pi}{6}\right)\)
c) \(cos\left(x-\dfrac{\pi}{3}\right)\)
d) \(tan\left(x-\dfrac{\pi}{4}\right)\)
cho sinx = \(-\dfrac{3}{5}\) và \(\pi\) < x < \(\dfrac{3\pi}{2}\) tính
a) \(cos\left(x+\dfrac{\pi}{6}\right)\)
b) \(tan\left(x+\dfrac{\pi}{4}\right)\)
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) $tan(x + \frac{\pi}{4}) = \frac{-3/5 + 1}{1 + (-3/5)(1)} = \frac{-2/5}{2/5} = -1$
cho sinx = \(-\dfrac{3}{5}\) và \(\pi\) < x < \(\dfrac{3\pi}{2}\) tính
a) sin2x, cos2x, tan2x, cot2x
b) \(sin\left(x+\dfrac{\pi}{3}\right)\)
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: \(sin\left(x+\dfrac{pi}{3}\right)=sinx\cdot cos\left(\dfrac{pi}{3}\right)+sin\left(\dfrac{pi}{3}\right)\cdot cosx\)
\(=\dfrac{-3}{5}\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{-4}{5}=\dfrac{-3-4\sqrt{3}}{10}\)
Cho cos2x=-\(\dfrac{4}{5}\), voi \(\dfrac{\pi}{4}< x< \dfrac{\pi}{2}\). Tinh sinx, cosx, sin(x+\(\dfrac{\pi}{3}\)), cos(2x-\(\dfrac{\pi}{4}\)).
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 $\sin x>0$
$\Rightarrow \sin x=\frac{3}{\sqrt{10}}$
$\sin (x+\frac{\pi}{3})=\sin x\cos \frac{\pi}{3}+\cos x\sin \frac{\pi}{3}$
$=\sqrt{\frac{9}{10}}.\frac{1}{2}+\sqrt{\frac{1}{10}}.\frac{\sqrt{3}}{2}=\frac{\sqrt{30}+3\sqrt{10}}{20}$
1. Cho sinx = \(\dfrac{2}{3}\) , x ∈ (0,\(\dfrac{\Pi}{2}\))
Tính cosx, tanx , sin (x+\(\dfrac{\Pi}{4}\))
2. Cho cos = \(\dfrac{1}{4}\) . Tính sinx, cos2x
3. Cho tanx = 2 . Tính cosx, sinx
x ∈ (0,\(\dfrac{\Pi}{2}\))
4. Rút gọn a) A = cos2x - 2cos2x + sinx +1
b) B = \(\dfrac{cos3x+cos2x+cosx}{cos2x}\)
1.
\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)
\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)
\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)
\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)
2.
Đề bài thiếu, cos?x
Và x thuộc khoảng nào?
3.
\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)
\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)
\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)
4.
\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)
\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)
a, Cho 0<x<\(\dfrac{\Pi}{4}\) .Chứng minh : sinx<cosx
b, Cho \(\dfrac{\Pi}{4}< x< \dfrac{\Pi}{2}\) .Chứng minh : sinx> cosx
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
Tìm các giá trị lượng giác, biết:
a) \(cos\alpha=\dfrac{2}{\sqrt{5}}\); \(-\dfrac{\pi}{2}< \alpha< 0\)
b) \(sinx=\dfrac{3}{5};\dfrac{\pi}{2}< x< \pi\)
c) \(tanx=\dfrac{4}{5};-\pi< x< -\dfrac{\pi}{2}\)
d) \(cotx=-\dfrac{3}{4};\dfrac{3\pi}{2}< x< \pi\)
e) \(tanx=\dfrac{4}{5};\pi< x< \dfrac{3\pi}{2}\)
f) \(cosx=\dfrac{4}{5};270^o< x< 360^o\)
g) \(sinx=-\dfrac{3}{5};180^o< x< 270^o\)
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
Tìm tập xác định của hàm số sau
a) y=cot(\(3x+\dfrac{\pi}{6}\)) + \(\dfrac{tan2x}{sinx+1}\)
b) y=\(\sqrt{5+2cot^2x-sinx}\) + cot\(\left(\dfrac{\pi}{2}+x\right)\)
a.
\(\left\{{}\begin{matrix}sin\left(3x+\dfrac{\pi}{6}\right)\ne0\\cos2x\ne0\\sinx\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
b.
Do \(5+2cot^2x-sinx=4+2cot^2x+\left(1-sinx\right)>0\) nên hàm xác định khi:
\(\left\{{}\begin{matrix}sinx\ne0\\sin\left(x+\dfrac{\pi}{2}\right)\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\)
\(\Leftrightarrow x\ne\dfrac{k\pi}{2}\)
cho sinx α =\(\dfrac{-4}{5}\) và π<α<\(\dfrac{\Pi}{2}\) thì sinx α bằng bao nhiêu?
\(sin^2\alpha=1-sin^2\alpha=1-\left(\dfrac{-4}{5}\right)^2=\dfrac{9}{25}\)
vì π<α<\(\dfrac{3\Pi}{2}\)⇒cos α =\(\dfrac{-3}{5}\)
cos2a =1- sin2a =1-\(\left(\dfrac{-4}{5}\right)^2\)=\(\dfrac{3}{5}\)
Vì π<a<\(\dfrac{3\pi}{2}\)
=>cos a =\(\dfrac{-3}{5}\)
sinx + cosx =\(-\dfrac{1}{2}\). Tìm sinx, cosx khi \(\dfrac{3\pi}{2}\)<x<\(2\pi\)
3/2pi<x<2pi
=>sin x<0; cosx>0
sin x+cosx=-1/2
=>(sinx+cosx)^2=1/4
=>1+2*sinx*cosx=1/4
=>2*sin x*cosx=-3/4
=>sinx*cosx=-3/8
mà sin x+cosx=-1/2
nên \(sinx=\dfrac{-1-\sqrt{7}}{4};cosx=\dfrac{-1+\sqrt{7}}{4}\)