1, A = cos2 (73) + cos2 (47) + cos (73).cos (47)
2, B = sin6 (pi/24) + cos6 (pi/24)
3, C = tan2 (pi/12) + tan2 (5pi/12)
4, D = tan6 (20) - 33tan4 (20) +27tan2 (20) - 3
Rút gọn các biểu thức :
a) \(\dfrac{\tan2\alpha}{\tan4\alpha-\tan2\alpha}\)
b) \(\sqrt{1+\sin\alpha}-\sqrt{1-\sin\alpha}\), với \(0< \alpha< \dfrac{\pi}{2}\)
c) \(\dfrac{3-4\cos2\alpha+\cos4\alpha}{3+4\cos2\alpha+\cos4\alpha}\)
d) \(\dfrac{\sin\alpha+\sin3\alpha+\sin5\alpha}{\cos\alpha+\cos3\alpha+\cos5\alpha}\)
a) \(\dfrac{tan2\alpha}{tan4\alpha-tan2\alpha}=\dfrac{sin2\alpha}{cos2\alpha}:\left(\dfrac{sin4\alpha}{cos4\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\right)\)
\(=\dfrac{sin2\alpha}{cos2\alpha}:\dfrac{sin4\alpha cos2\alpha-sin2\alpha cos4\alpha}{cos4\alpha cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos2\alpha}.\dfrac{cos4\alpha.cos2\alpha}{sin2\alpha}=cos4\alpha\).
b) \(\sqrt{1+sin\alpha}-\sqrt{1-sin\alpha}=\sqrt{sin^2\dfrac{\alpha}{2}+2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)\(-\sqrt{sin^2\dfrac{\alpha}{2}-2sin\dfrac{\alpha}{2}cos\dfrac{\alpha}{2}+cos^2\dfrac{\alpha}{2}}\)
\(=\sqrt{\left(sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right)^2}-\sqrt{\left(sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right)^2}\)
\(=\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
Vì \(0< \alpha< \dfrac{\pi}{2}\) nên \(0< \alpha< \dfrac{\pi}{4}\).
Trong \(\left(0;\dfrac{\pi}{4}\right)\) thì \(sin\dfrac{\alpha}{2}\) tăng dần từ 0 tới \(\dfrac{\sqrt{2}}{2}\) và \(cos\dfrac{\alpha}{2}\) giảm dần từ 1 tới \(\dfrac{\sqrt{2}}{2}\) nên \(\left|sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right|=-\left(sin\dfrac{\alpha}{4}-cos\dfrac{\alpha}{4}\right)=cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\).
Vì vậy:
\(\left|sin\dfrac{\alpha}{2}+cos\dfrac{\alpha}{2}\right|-\left|sin\dfrac{\alpha}{2}-cos\dfrac{\alpha}{2}\right|\)
\(=sin\dfrac{\alpha}{4}+cos\dfrac{\alpha}{4}-\left(cos\dfrac{\alpha}{4}-sin\dfrac{\alpha}{4}\right)=2sin\dfrac{\alpha}{4}\).
c) \(\dfrac{3-4cos2\alpha+cos4\alpha}{3+4cos2\alpha+cos4\alpha}\)\(=\dfrac{4-4cos2\alpha+cos4\alpha-1}{4+4cos2\alpha+cos4\alpha-1}\)
\(=\dfrac{4\left(1-cos2\alpha\right)-2sin^22\alpha}{4\left(1+cos2\alpha\right)-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-2sin^22\alpha}{4sin^2\alpha-2sin^22\alpha}\)
\(=\dfrac{4cos^2\alpha-8sin^2\alpha cos^2\alpha}{4sin^2\alpha-8sin^2\alpha cos^2\alpha}\)
\(=\dfrac{4cos^2\alpha\left(1-2sin^2\alpha\right)}{4sin^2\alpha\left(1-2cos^2\alpha\right)}=cot^2\alpha.\dfrac{cos2\alpha}{-cot2\alpha}\)
\(=-cot^2\alpha\).
Sin6(π + x) + cos6(x - π) - 2sin4(x + 2π) - sin4(x - \(\dfrac{3\pi}{2}\)) + cos2(x - \(\dfrac{\pi}{2}\)) . Rút gọn biểu thức trên.
\(sin^6\left(\pi+x\right)=sin^6x,cos^6\left(x-\pi\right)=cos^6\pi\\ sin^4\left(x+2\pi\right)=sin^4x,sin^4\left(x-\dfrac{3\pi}{2}\right)=cos^4x,cos^2\left(x-\dfrac{\pi}{2}\right)=sin^2x.\)
Khi đó \(A=sin^6x+cos^6x-2sin^4x-cos^4x+sin^2x\\ =\left(sin^2x+cos^2x\right)^2-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)-\left(sin^4x+cos^4x\right)-sin^4x+sin^2x\\ =1-3sin^2x.cos^2x-\left[1-2sin^2x.cos^2x\right]-sin^2x.\left(sin^2x-1\right)\\ =1-3sin^2x.cos^2x-1+2sin^2x.cos^2x+sin^2x.cos^2x\\ =0\)
Bài 4. a) Tính giá trị biểu thức:
A = cos2 20° + cos2 40° + cos2 50° + cos2 70°.
b) Rút gọn biểu thức:
B = sin6 a + cos6 a + 3 sin2 a. cos2 a
\(a,A=\left(\cos^220^0+\cos^270^0\right)+\left(\cos^240^0+\cos^250^0\right)\\ A=\left(\cos^220^0+\sin^220^0\right)+\left(\cos^240^0+\sin^240^0\right)=1+1=2\\ b,B=\left(\cos^2\alpha\right)^3+\left(\sin^2\alpha\right)^3+3\sin^2\alpha\cdot\cos^2\alpha\cdot\left(\sin^2\alpha+\cos^2\alpha\right)\\ B=\left(\sin^2\alpha+\cos^2\alpha\right)^3=1^3=1\)
a) Cho tan x=3 và \(\frac{\pi}{6}\)∠x∠\(\frac{\pi}{3}\) . Tính giá trị của biểu thức B =\(\frac{\cos^2x+\cot^2x}{\tan x-\cot x}\)
b) Cho cos α=\(\frac{-4}{5}\) và \(\frac{\pi}{2}\)∠α∠\(\pi\) . Tính giá trị của biểu thức A=\(\frac{3\sin2\alpha-\tan2\alpha}{\cos\alpha-\cos2\alpha}\)
c) Cho tan x=-2 và\(\frac{3\pi}{2}\)∠x∠\(2\pi\) . Tính giá trị của biểu thức B=\(\frac{\cos^2x+\sin2x}{\tan2x-\cos2x}\)
a/ \(\frac{\pi}{6}< x< \frac{\pi}{3}\Rightarrow cosx>0\)
\(cos^2x=\frac{1}{1+tan^2x}=\frac{1}{10}\)
\(cotx=\frac{1}{tanx}=\frac{1}{3}\)
Thay số và bấm máy
b/ \(\frac{\pi}{2}< a< \pi\Rightarrow\left\{{}\begin{matrix}sina>0\\tana< 0\end{matrix}\right.\)
\(sina=\sqrt{1-cos^2a}=\frac{3}{5}\)
\(tana=\frac{sina}{cosa}=-\frac{3}{4}\)
\(A=\frac{6sina.cosa-\frac{2tana}{1-tan^2a}}{cosa-\left(2cos^2a-1\right)}\)
Thay số và bấm máy
c/ \(\frac{3\pi}{2}< x< 2\pi\Rightarrow\left\{{}\begin{matrix}cosx>0\\sinx< 0\end{matrix}\right.\)
\(cosx=\frac{1}{\sqrt{1+tan^2x}}=\frac{1}{\sqrt{5}}\)
\(sinx=cosx.tanx=-\frac{2}{\sqrt{5}}\)
\(B=\frac{cos^2x+2sinx.cosx}{\frac{2tanx}{1-tan^2x}-\left(2cos^2x-1\right)}\)
Thay số
giải phương trình
a) \(cosx=3\)
b) \(cosx=\dfrac{\sqrt{3}}{2}\)
c) \(cos\left(x+\dfrac{\pi}{12}\right)\) = \(cos\dfrac{3\pi}{12}\)
d) \(cos\)(x - 20 độ) = cos70 độ
a: cosx=3
mà -1<=cosx<=1
nên \(x\in\varnothing\)
b: \(cosx=\dfrac{\sqrt{3}}{2}\)
=>\(cosx=cos\left(\dfrac{pi}{6}\right)\)
=>x=pi/6+k2pi hoặc x=-pi/6+k2pi
c: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{pi}{12}=\dfrac{3}{12}pi+k2pi\\x+\dfrac{pi}{12}=-\dfrac{3}{12}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{12}pi+k2pi=\dfrac{1}{6}pi+k2pi\\x=-\dfrac{4}{12}pi+k2pi=-\dfrac{1}{3}pi+k2pi\end{matrix}\right.\)
d: =>x-20 độ=70 độ+k*360 hoặc x-20=-70+k*360
=>x=90+k*360 hoặc x=-50+k*360
Cho tan2α = 2 và π < α < \(\frac{3\pi}{2}\). Biết giá trị của biểu thức M= \(\frac{cos(\alpha+\frac{\pi}{3})+cos(\alpha-\frac{\pi}{3})}{tan(\frac{\pi}{2}-\alpha)+tan(\frac{\pi+\alpha}{2}}=\frac{a}{\sqrt{b}}\) với a, b là các số nguyên. Khi đó, giá trị của biểu thức T = 2a + b là ?
\(\pi< a< \frac{3\pi}{2}\Rightarrow2\pi< 2a< 3\pi\Rightarrow sin2a>0\)
\(cot2a=\frac{1}{2}\Rightarrow sin2a=\frac{1}{\sqrt{1+cot^22a}}=\frac{2\sqrt{5}}{5}\)
\(cos\left(a+\frac{\pi}{3}\right)+cos\left(a-\frac{\pi}{3}\right)=2cosa.cos\frac{\pi}{3}=cosa\)
\(tan\left(\frac{\pi}{2}-a\right)+tan\left(\frac{\pi}{2}+\frac{a}{2}\right)=\frac{-sin\frac{a}{2}}{cos\left(\frac{\pi}{2}-a\right).cos\left(\frac{\pi}{2}+\frac{a}{2}\right)}=\frac{sin\frac{a}{2}}{sina.sin\frac{a}{2}}=\frac{1}{sina}\)
\(\Rightarrow M=sina.cosa=\frac{1}{2}sin2a=\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)
\(\Rightarrow2a+b=7\)
Bài 1: Phương trình căn 2 cot x + căn 2=0 có tổng các nghiệm khi k =0 và k=1?
A.3pi/4 B.5pi/4 C.4pi/3 D.pi/2Bài 2:cho sin a=1/3,0<a<pi/2 tính sin( a-pi/4) Bài 3:cho cos a=-2/3,pi/2<a<pi tính cos ( a+pi/3)
Giúp vs bạn
cho \(\cos\alpha=\dfrac{-12}{13}\) biết \(\pi< \alpha< \dfrac{3\pi}{2}\)
tính \(\sin\alpha,cos2\alpha,tan\left(\alpha-\dfrac{\pi}{3}\right),sin\left(2\alpha+\dfrac{\pi}{6}\right)\)
Có \(a\) thuộc góc phần tư thứ III -> sin\(a\) < 0
+) sin\(a\)=-\(\sqrt{1-cos^2a}\)=-\(\sqrt{1-\left(\dfrac{-12}{13}\right)^2}\)=\(\dfrac{-5}{13}\)
\(cos2a=cos^2a-sin^2a\)=\(\left(\dfrac{-12}{13}\right)^2-\left(\dfrac{-5}{13}\right)^2=\dfrac{119}{169}\)
cho \(sin\alpha+cos\alpha=\frac{1}{2}\) với \(\frac{3\pi}{4}< \alpha< \pi\). tính \(tan2\alpha\)
\(\frac{3\pi}{4}< a< \pi\Rightarrow\frac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)
\(\left(sina+cosa\right)^2=\frac{1}{4}\Leftrightarrow sin^2a+cos^2a+2sina.cosa=\frac{1}{4}\)
\(\Leftrightarrow1+sin2a=\frac{1}{4}\Rightarrow sin2a=-\frac{3}{4}\)
\(\Rightarrow cos2a=\sqrt{1-sin^22a}=\frac{\sqrt{7}}{4}\)
\(\Rightarrow tan2a=\frac{sin2a}{cos2a}=-\frac{3\sqrt{7}}{7}\)