Cho x+2y=\(\dfrac{\Pi}{2}\). Rút gọn biểu thức:
\(A=\dfrac{Cos\left(x+y\right)-Cosy}{Cos\left(x+y\right)+Cosy}\)
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5 tháng 7 2021
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
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0
8 tháng 4 2022
ơ bạn :\(\dfrac{cos\left(x+y\right)+cosx}{cos\left(x+y\right)-cosx}=\dfrac{2cos\left(\dfrac{2x+y}{2}\right).cos\left(\dfrac{y}{2}\right)}{-2sin\left(\dfrac{2x+y}{2}\right).sin\left(\dfrac{y}{2}\right)}=-2.cot\left(\dfrac{2x+y}{2}\right).cot\left(\dfrac{y}{2}\right)\) L không thể bẳng 0 được
AH
Akai Haruma
Giáo viên
25 tháng 4 2018
Câu a)
Từ \(\tan a=3\Leftrightarrow \frac{\sin a}{\cos a}=3\Rightarrow \sin a=3\cos a\)
Do đó:
\(\frac{\sin a\cos a+\cos ^2a}{2\sin ^2a-\cos ^2a}=\frac{3\cos a\cos a+\cos ^2a}{2(3\cos a)^2-\cos ^2a}\)
\(=\frac{\cos ^2a(3+1)}{\cos ^2a(18-1)}=\frac{4}{17}\)
Câu b)
Có: \(\cot \left(\frac{\pi}{2}-x\right)=\tan x=\frac{\sin x}{\cos x}\)
\(\cos\left(\frac{\pi}{2}+x\right)=-\sin x\)
\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)=\frac{-\sin ^2x}{\cos x}\)
Và:
\(\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{\sin x\cot x}{\cos^2x}=\frac{\sin x.\frac{\cos x}{\sin x}}{\cos^2x}=\frac{1}{\cos x}\)
Do đó:
\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)+\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{1-\sin ^2x}{\cos x}=\frac{\cos ^2x}{\cos x}=\cos x\)
Ta có đpcm.
23 tháng 3 2018
rút gọn biểu thức:
E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))
\(x+2y=\dfrac{\pi}{2}\)
\(\Leftrightarrow x+y=\dfrac{\pi}{2}-y\) thay vào A được:
\(A=\dfrac{cos\left(\dfrac{\pi}{2}-y\right)-cosy}{cos\left(\dfrac{\pi}{2}-y\right)+cosy}\)\(=\dfrac{siny-cosy}{siny+cosy}\)\(=\dfrac{\dfrac{\sqrt{2}}{2}.siny-\dfrac{\sqrt{2}}{2}.cosy}{\dfrac{\sqrt{2}}{2}.siny+\dfrac{\sqrt{2}}{2}cosy}\)\(=\dfrac{cos\dfrac{\pi}{4}.siny-sin\dfrac{\pi}{4}.cosy}{sin\dfrac{\pi}{4}.siny+cos\dfrac{\pi}{4}.cosy}\)
\(=\dfrac{sin\left(y-\dfrac{\pi}{4}\right)}{cos\left(y-\dfrac{\pi}{4}\right)}\)\(=tan\left(y-\dfrac{\pi}{4}\right)\)
\(x+2y=\dfrac{\pi}{2}\Rightarrow x+y=\dfrac{\pi}{2}-y\)
\(\Rightarrow cos\left(x+y\right)=cos\left(\dfrac{\pi}{2}-y\right)\)
\(\Rightarrow cos\left(x+y\right)=siny\)
Do đó: \(A=\dfrac{siny-cosy}{siny+cosy}=\dfrac{\sqrt{2}sin\left(y-\dfrac{\pi}{4}\right)}{\sqrt{2}cos\left(y-\dfrac{\pi}{4}\right)}=tan\left(y-\dfrac{\pi}{4}\right)\)