Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\eqalign{ & A\sin {x \over 5} = \sin {x \over 5}\cos {x \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 2}\sin {{2x} \over 5}\cos {{2x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over 4}\sin {{4x} \over 5}\cos {{4x} \over 5}\cos {{8x} \over 5} = {1 \over 8}\sin {{8x} \over 5}\cos {{8x} \over 5} \cr & = {1 \over {16}}\sin {{16x} \over 5} \cr} \)
Suy ra biểu thức rút gọn \(A =\sin{{16x} \over 5}:16\sin {x \over 5}\)
b)\(\eqalign{ & B = \sin {x \over 7} + 2\sin {{3x} \over 7} + \sin {{5x} \over 7} = 2\sin {{3x} \over 7} + (\sin {x \over 7} + \sin {{5x} \over 7}) \cr & = 2\sin {{3x} \over 7} + 2\sin {1 \over 2}({{5x} \over 7} + {x \over 7})cos{1 \over 2}({{5x} \over 7} - {x \over 7}) \cr & = 2\sin {{3x} \over 7}(1 + \cos {{2x} \over 7}) = 4\sin {{3x} \over 7}{\cos ^2}{x \over 7} \cr}\)
a)
\(\cos\dfrac{22\pi}{3}=\cos\left(8\pi-\dfrac{2\pi}{3}\right)\\ =\cos\left(-\dfrac{2\pi}{3}\right)\\ =\cos\left(\dfrac{2\pi}{3}\right)\\ =-\cos\dfrac{\pi}{3}\\ =-\dfrac{1}{2}\)
b)
\(\sin\dfrac{23\pi}{4}=\sin\left(6\pi-\dfrac{\pi}{4}\right)\\ =\sin\left(-\dfrac{\pi}{4}\right)\\ =-\dfrac{\sqrt{2}}{2}\)
c)
\(\sin\dfrac{25\pi}{3}-\tan\dfrac{10\pi}{3}\\ =\sin\left(8\pi+\dfrac{\pi}{3}\right)-\tan\left(3\pi+\dfrac{\pi}{3}\right)\\ =\sin\dfrac{\pi}{3}-\tan\dfrac{\pi}{3}\\ =\dfrac{\sqrt{3}}{2}-\sqrt{3}\\ =\dfrac{-\sqrt{3}}{2}\)
d)
\(\cos^2\dfrac{\pi}{8}-\sin^2\dfrac{\pi}{8}\\ =\cos\dfrac{\pi}{4}\\ =\dfrac{\sqrt{2}}{2}\)
cau a: \(cos\dfrac{22\Pi}{3}=cos\dfrac{24\Pi-2\Pi}{3}=cos\left(8\Pi-\dfrac{2\Pi}{3}\right)=cos\dfrac{2\Pi}{3}=-\dfrac{1}{2}\)
câu b: \(sin\dfrac{23\Pi}{4}=sin\dfrac{24\Pi-\Pi}{4}=sin\left(6\Pi-\dfrac{\Pi}{4}\right)=-sin\dfrac{\Pi}{4}=-\dfrac{\sqrt{2}}{2}\)
cau c: \(=sin\left(8\Pi-\dfrac{\Pi}{3}\right)-tan\left(3\Pi+\dfrac{\Pi}{3}\right)=-sin\dfrac{\Pi}{3}-tan\dfrac{\Pi}{3}=-\dfrac{\sqrt{3}}{2}-\sqrt{3}=\dfrac{-3\sqrt{3}}{2}\)
cau d: \(cos^2\dfrac{\Pi}{8}-sin^2\dfrac{\Pi}{8}=cos2\left(\dfrac{\Pi}{8}\right)=cos\dfrac{\Pi}{4}=\dfrac{\sqrt{2}}{2}\)
a)\(sin^4\dfrac{\pi}{16}+sin^4\dfrac{3\pi}{16}+sin^4\dfrac{5\pi}{16}+sin^4\dfrac{7\pi}{16}\)
\(=\left(sin^4\dfrac{\pi}{16}+sin^4\dfrac{7\pi}{16}\right)+\left(sin^4\dfrac{3\pi}{16}+sin^4\dfrac{5\pi}{16}\right)\)
\(=\left(sin^4\dfrac{\pi}{16}+cos^4\dfrac{\pi}{16}\right)+\left(sin^4\dfrac{3\pi}{16}+cos^4\dfrac{3\pi}{16}\right)\)
\(=1-2sin^2\dfrac{\pi}{16}cos^2\dfrac{\pi}{16}+1-2sin^2\dfrac{3\pi}{16}cos^2\dfrac{3\pi}{16}\)
\(=2-\dfrac{1}{2}sin^2\dfrac{\pi}{8}-\dfrac{1}{2}sin^2\dfrac{3\pi}{8}\)
\(=2-\dfrac{1}{2}\left(sin^2\dfrac{\pi}{8}+sin^2\dfrac{3\pi}{8}\right)\)
\(=2-\dfrac{1}{2}\left(sin^2\dfrac{\pi}{8}+cos^2\dfrac{\pi}{8}\right)\)
\(=2-\dfrac{1}{2}=\dfrac{3}{2}\).
Có: \(cotx-tanx=\dfrac{cosx}{sinx}-\dfrac{sinx}{cosx}=\dfrac{cos^2x-sin^2x}{sinxcosx}=\dfrac{2cos2x}{sin2x}\)
Vì vậy:
\(cot7,5^o+tan67,5^o-tan7,5^o-cot67,5^o\)
\(=\left(cot7,5^o-tan7,5^o\right)-\left(cot67,5^o-tan67,5^o\right)\)
\(=\dfrac{2cos15^o}{sin15^o}-\dfrac{2cos135^o}{sin135^o}\)
\(=2\left(\dfrac{cos15^osin135^o-sin15^ocos135^o}{sin15^osin135^o}\right)\)
\(=2.\dfrac{sin120^o}{\dfrac{1}{2}\left(cos120^o-cos150^o\right)}\)
\(=\dfrac{4.\dfrac{\sqrt{3}}{2}}{\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}}=\dfrac{4\sqrt{3}}{\sqrt{3}-1}\)
Ta có : A+B+C= 180
=>sin(A+B)/2 = sin(180/2 - C/2) = cosC/2
ttcó: sinC/2 = cos(A+B)/2
=> sA+sB+sC =2cosC/2*cos(A-B)/2 + 2cos(A+B)/2*cosC/2
=2cosC/2
=4cosA/2cosB/2cosC/2
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
Bài 1:
a)
\(\sin ^2x+\sin ^2x\cot^2x=\sin ^2x(1+\cot^2x)=\sin ^2x(1+\frac{\cos ^2x}{\sin ^2x})\)
\(=\sin ^2x.\frac{\sin ^2x+\cos^2x}{\sin ^2x}=\sin ^2x+\cos^2x=1\)
b)
\((1-\tan ^2x)\cot^2x+1-\cot^2x\)
\(=\cot^2x(1-\tan^2x-1)+1=\cot^2x(-\tan ^2x)+1=-(\tan x\cot x)^2+1\)
\(=-1^2+1=0\)
c)
\(\sin ^2x\tan x+\cos^2x\cot x+2\sin x\cos x=\sin ^2x.\frac{\sin x}{\cos x}+\cos ^2x.\frac{\cos x}{\sin x}+2\sin x\cos x\)
\(=\frac{\sin ^3x}{\cos x}+\frac{\cos ^3x}{\sin x}+2\sin x\cos x=\frac{\sin ^4x+\cos ^4x+2\sin ^2x\cos ^2x}{\sin x\cos x}=\frac{(\sin ^2x+\cos ^2x)^2}{\sin x\cos x}=\frac{1}{\sin x\cos x}\)
\(=\frac{1}{\frac{\sin 2x}{2}}=\frac{2}{\sin 2x}\)
Bài 2:
Áp dụng BĐT Cauchy Schwarz ta có:
\(P=\frac{a^2}{\sqrt{a(2c+a+b)}}+\frac{b^2}{\sqrt{b(2a+b+c)}}+\frac{c^2}{\sqrt{c(2b+c+a)}}\)
\(\geq \frac{(a+b+c)^2}{\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}}(*)\)
Tiếp tục áp dụng BĐT Cauchy-Schwarz:
\((\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq (a+b+c)(2c+a+b+2a+b+c+2b+c+a)\)
\(\Leftrightarrow (\sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)})^2\leq 4(a+b+c)^2\)
\(\Rightarrow \sqrt{a(2c+a+b)}+\sqrt{b(2a+b+c)}+\sqrt{c(2b+c+a)}\leq 2(a+b+c)(**)\)
Từ \((*); (**)\Rightarrow P\geq \frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=\frac{3}{2}\)
Vậy \(P_{\min}=\frac{3}{2}\)
Dấu "=" xảy ra khi $a=b=c=1$
Chia tử và mẫu cho cosa ta có:
B=\(\dfrac{4\tan a+5}{2\tan a-3}\). Vì \(\cot a=\dfrac{1}{2}\) nên \(\tan a=2\)
=> B=13
chia tử và mẫu của B cho sina khác 0\(B=\dfrac{4\dfrac{sina}{sina}+5\dfrac{cosa}{sina}}{2\dfrac{sina}{sina}-3\dfrac{cosa}{sina}}=\dfrac{4+5cota}{2-3cota}=\dfrac{4+5\dfrac{1}{2}}{2-3\dfrac{1}{2}}=13\)
vay B = 13
ta có:\(sin\alpha.cosb=\dfrac{1}{2}\left[sin\left(a-b\right)+sin\left(a+b\right)\right]\)
\(=\dfrac{1}{2}\left[\dfrac{2}{5}+\left(-\dfrac{3}{5}\right)\right]\)
\(=\dfrac{1}{2}.\left(-\dfrac{1}{5}\right)\)
\(=-\dfrac{1}{10}\)
Ta có \(sin\left(a-b\right)+sin\left(a+b\right)=2sin\left(\dfrac{a-b+a+b}{2}\right)cos\left(\dfrac{a+b-\left(a-b\right)}{2}\right)\\ \Rightarrow2sin\left(a\right).cos\left(b\right)=\dfrac{2}{5}+\left(-\dfrac{3}{5}\right)=-\dfrac{1}{5}\\ \Rightarrow sin\left(a\right)cos\left(b\right)=-\dfrac{1}{10}\)