\(S=sin\left(x+2017\pi\dfrac{ }{ }2\right)+2sin^2\left(x-\pi\right)+cos\left(x...">
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5 tháng 4 2017

a) \(A=sin\left(\dfrac{\pi}{4}+x\right)-cos\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-\left(cos\dfrac{\pi}{4}.cosx+sin\dfrac{\pi}{4}.sinx\right)\)

\(\Leftrightarrow A=sin\dfrac{\pi}{4}.cosx+cos\dfrac{\pi}{4}.sinx-cos\dfrac{\pi}{4}.cosx-sin\dfrac{\pi}{4}.sinx\)

\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2}.cosx+\dfrac{\sqrt{2}}{2}.sinx-\dfrac{\sqrt{2}}{2}.cosx-\dfrac{\sqrt{2}}{2}.sinx\)

\(\Leftrightarrow A=0\)

b) \(B=cos\left(\dfrac{\pi}{6}-x\right)-sin\left(\dfrac{\pi}{3}+x\right)\)

\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-\left(sin\dfrac{\pi}{3}.cosx+cos\dfrac{\pi}{3}.sinx\right)\)

\(\Leftrightarrow B=cos\dfrac{\pi}{6}.cosx+sin\dfrac{\pi}{6}.sinx-sin\dfrac{\pi}{3}.cosx-cos\dfrac{\pi}{3}.sinx\)

\(\Leftrightarrow B=\dfrac{\sqrt{3}}{2}.cosx+\dfrac{1}{2}.sinx-\dfrac{\sqrt{3}}{2}.cosx-\dfrac{1}{2}.sinx\)

\(\Leftrightarrow B=0\)

c) \(C=sin^2x+cos\left(\dfrac{\pi}{3}-x\right).cos\left(\dfrac{\pi}{3}+x\right)\)

\(\Leftrightarrow C=sin^2x+\left(cos\dfrac{\pi}{3}.cosx+sin\dfrac{\pi}{3}.sinx\right).\left(cos\dfrac{\pi}{3}.cosx-sin\dfrac{\pi}{3}.sinx\right)\)

\(\Leftrightarrow C=sin^2x+\left(\dfrac{1}{2}.cosx+\dfrac{\sqrt{3}}{2}.sinx\right).\left(\dfrac{1}{2}.cosx-\dfrac{\sqrt{3}}{2}.sinx\right)\)

\(\Leftrightarrow C=sin^2x+\dfrac{1}{4}.cos^2x-\dfrac{3}{4}.sin^2x\)

\(\Leftrightarrow C=\dfrac{1}{4}.sin^2x+\dfrac{1}{4}.cos^2x\)

\(\Leftrightarrow C=\dfrac{1}{4}\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow C=\dfrac{1}{4}\)

d) \(D=\dfrac{1-cos2x+sin2x}{1+cos2x+sin2x}.cotx\)

\(\Leftrightarrow D=\dfrac{1-\left(1-2sin^2x\right)+2sinx.cosx}{1+2cos^2a-1+2sinx.cosx}.cotx\)

\(\Leftrightarrow D=\dfrac{2sin^2x+2sinx.cosx}{2cos^2x+2sinx.cosx}.cotx\)

\(\Leftrightarrow D=\dfrac{2sinx\left(sinx+cosx\right)}{2cosx\left(cosx+sinx\right)}.cotx\)

\(\Leftrightarrow D=\dfrac{sinx}{cosx}.cotx\)

\(\Leftrightarrow D=tanx.cotx\)

\(\Leftrightarrow D=1\)

NV
2 tháng 4 2019

\(A=2cosx-3cosx-sin\left(3\pi+\frac{\pi}{2}-x\right)+tan\left(\pi+\frac{\pi}{2}-x\right)\)

\(A=-cosx+sin\left(\frac{\pi}{2}-x\right)+tan\left(\frac{\pi}{2}-x\right)\)

\(A=-cosx+cosx+cotx=cotx\)

\(B=2cosx+sin\left(4\pi+\pi-x\right)+sin\left(2\pi-\frac{\pi}{2}+x\right)-sinx\)

\(B=2cosx+sin\left(\pi-x\right)+sin\left(-\frac{\pi}{2}+x\right)-sinx\)

\(B=2cosx+sinx-sin\left(\frac{\pi}{2}-x\right)-sinx\)

\(B=2cosx-cosx=cosx\)

5 tháng 7 2021

\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)

\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)

\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)

\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)

\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)

\(=cosx-cosx+sin^2x+cos^2x+sinx\)

\(=1+sinx\)

\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)

\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)

\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)

\(=1+cosx\)

5 tháng 7 2021

bị bỏ gp chị nhắn tin vs mấy ad ấy, nhanh ko ấy mà chị =))

26 tháng 4 2017

Giải bài 3 trang 154 SGK Đại Số 10 | Giải toán lớp 10

23 tháng 3 2018

rút gọn biểu thức:

E=cos(\(\dfrac{3\pi}{3}-\alpha\))-sin(\(\dfrac{3\pi}{2}-\alpha\))+sin(\(\alpha+4\pi\))

25 tháng 7 2018

bài 1) ta có : \(G=cos\left(\alpha-5\pi\right)+sin\left(\dfrac{-3\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{3\pi}{2}-\alpha\right)\)

\(G=cos\left(\alpha-\pi\right)+sin\left(\dfrac{\pi}{2}+\alpha\right)-tan\left(\dfrac{\pi}{2}+\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\)

\(G=cos\left(\pi-\alpha\right)+sin\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)-tan\left(\pi+\alpha-\dfrac{\pi}{2}\right).cot\left(\dfrac{\pi}{2}-\alpha\right)\) \(G=cos\left(\alpha\right)+cos\left(\alpha\right)+tan\left(\dfrac{\pi}{2}-\alpha\right).cot\left(\dfrac{\pi}{2}-\alpha\right)=2cos\alpha+1\) bài 2) ta có : \(H=cot\left(\alpha\right).cos\left(\alpha+\dfrac{\pi}{2}\right)+cos\left(\alpha\right)-2sin\left(\alpha-\pi\right)\) \(H=cot\left(\alpha\right).cos\left(\dfrac{\pi}{2}-\left(-\alpha\right)\right)+cos\left(\alpha\right)+2sin\left(\pi-\alpha\right)\) \(H=-cot\left(\alpha\right).sin\left(\alpha\right)+cos\left(\alpha\right)+2sin\left(\alpha\right)\) \(H=-cos\alpha+cos\alpha+2sin\alpha=2sin\alpha\)

NV
17 tháng 4 2019

\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)

\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)

\(A=0\)

\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)

\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)

\(B=\frac{1}{4}\)

\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)

\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)

\(C=\frac{3}{2}\)

\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)

\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)

\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)

\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)

30 tháng 4 2019

Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là

NV
22 tháng 6 2020

\(2sin\left(\frac{\pi}{2}+x\right)+sin\left(3\pi-x\right)+sin\left(\frac{3\pi}{2}+x\right)+cos\left(\frac{\pi}{2}+x\right)\)

\(=2cosx+sinx-cosx-sinx\)

\(=cosx\)