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a, \(\left(1-sin^2x\right)cot^2x+1-cot^2x\)
\(=cot^2x-sin^2x.cot^2x+1-cot^2x\)
\(=1-sin^2x.\frac{\text{cos}^2x}{sin^2x}=1-\text{cos}^2x=sin^2x\)
b,\(\left(tanx+cotx\right)^2-\left(tanx-cotx\right)2\)
\(=tan^2x2.tanx.cotx+cot^2x-tan^2x+2tanx.cotx-cot^2x\)
\(=4tanxcotx=4\)
c,\(\left(xsina-y\text{cos}a\right)^2+\left(x\text{cos}a+ysina\right)^2\)
\(=x^2sin^2a=2xysina\text{cos}a+y^2\text{cos}^2a+2xysina\text{cos}a+y^2sin^2a\)
\(=x^2\left(sin^2a+\text{cos}^2a\right)+y^2\left(sin^2a+\text{cos}^2a\right)\)
\(=x^2+y^2\)
\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)
\(\left(sinx+cosx\right)^2=\frac{25}{16}\Rightarrow sin^2x+cos^2x+2sinxcosx=\frac{25}{16}\)
\(\Rightarrow2sinxcosx=\frac{25}{16}-1=\frac{9}{16}\Rightarrow A=\frac{9}{32}\)
\(B^2=\left(sinx-cosx\right)^2=1-2sinx.cosx=1-\frac{9}{16}=\frac{7}{16}\Rightarrow B=\pm\frac{\sqrt{7}}{4}\)
\(C=\left(sinx+cosx\right)\left(sinx-cosx\right)=\frac{5}{4}.\left(\pm\frac{\sqrt{7}}{4}\right)=\pm\frac{5\sqrt{7}}{16}\)
\(A=\cos\left(\text{π}-\dfrac{x}{2}\right)-\sin\left(\text{π}-x\right)\)
\(=\sin x+\sin x=2\cdot\sin x\)
\(B=\cos\left(2\text{π}+\dfrac{\text{π}}{2}-x\right)+\sin\left(4\text{π}+\dfrac{\text{π}}{2}-x\right)-\cos\left(6\text{π}+\dfrac{3}{2}\text{π}+x\right)-\sin\left(16\text{π}+\dfrac{3}{2}\text{π}+x\right)\)
\(=\sin x+\cos x-\cos\left(\dfrac{3}{2}\text{π}+x\right)-\sin\left(\dfrac{3}{2}\text{π}+x\right)\)
\(=\sin x+\cos x-\cos\left(\text{π}+\dfrac{\text{π}}{2}+x\right)-\sin\left(\text{π}+\dfrac{\text{π}}{2}+x\right)\)
\(=\cos x+\sin x+\cos\left(\dfrac{1}{2}\text{π}+x\right)+\sin\left(\dfrac{1}{2}\text{π}+x\right)\)
\(=\cos x+\sin x-\sin x+\cos x=2\cos x\)
\(A=sin\left(\dfrac{\pi}{2}-\alpha+2\pi\right)+cos\left(\pi+\alpha+12\pi\right)-3sin\left(\alpha-\pi-4\pi\right)\)
\(=sin\left(\dfrac{\pi}{2}-\alpha\right)+cos\left(\pi+\alpha\right)-3sin\left(\alpha-\pi\right)\)
\(=cos\alpha-cos\alpha+3sin\left(\pi-\alpha\right)\)\(=3sin\alpha\)
\(B=sin\left(x+\dfrac{\pi}{2}+42\pi\right)+cos\left(x+\pi+2016\pi\right)+sin^2\left(x+\pi+32\pi\right)+sin^2\left(x-\dfrac{\pi}{2}-2\pi\right)+cos\left(x-\dfrac{\pi}{2}+2\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+cos\left(x+\pi\right)+sin^2\left(x+\pi\right)+sin^2\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\dfrac{\pi}{2}\right)\)
\(=cosx-cosx+sin^2x+cos^2x+sinx\)
\(=1+sinx\)
\(C=sin\left(x+\dfrac{\pi}{2}+1008\pi\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi+2018\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}+4\pi\right)\)
\(=sin\left(x+\dfrac{\pi}{2}\right)+2sin^2\left(\pi-x\right)+cos\left(x+\pi\right)+cos2x+sin\left(x+\dfrac{\pi}{2}\right)\)
\(=cosx+2sin^2x-cosx+1-2sin^2x+cosx\)
\(=1+cosx\)
\(A=cosa\left(sinb.cosc-cosb.sinc\right)+cosb\left(sinc.cosa-cosc.sina\right)+cosc\left(sinacosb-cosasinb\right)\)
\(A=cosasinbcosc-cosacosbsinc+cosacosbsinc-sinacosbcosc+sinacosbcosc-cosasinbcosc\)
\(A=0\)
\(B=sin^2x+\frac{1}{2}\left(cos\frac{2\pi}{3}+cos2x\right)\)
\(B=\frac{1}{2}-\frac{1}{2}cos2x-\frac{1}{4}+\frac{1}{2}cos2x\)
\(B=\frac{1}{4}\)
\(C=\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}+2x\right)+\frac{1}{2}-\frac{1}{2}cos\left(\frac{4\pi}{3}-2x\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-\frac{1}{2}\left(cos\left(\frac{4\pi}{3}+2x\right)+cos\left(\frac{4\pi}{3}-2x\right)\right)\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x-cos\frac{4\pi}{3}.cos2x\)
\(C=\frac{3}{2}-\frac{1}{2}cos2x+\frac{1}{2}cos2x\)
\(C=\frac{3}{2}\)
\(D=\frac{1}{2}\left[\sqrt{2}sin\left(\frac{\pi}{4}+x\right)\right]^2-sin^2x-sinx.\sqrt{2}cos\left(\frac{\pi}{4}+x\right)\)
\(D=\frac{1}{2}\left(sinx+cosx\right)^2-sin^2x-sinx\left(sinx-cosx\right)\)
\(D=\frac{1}{2}\left(1+2sinx.cosx\right)-sin^2x-sin^2x+sinx.cosx\)
\(D=\frac{1}{2}+sinxcosx+sinxcosx=\frac{1}{2}+sin2x\)
Góc độ cao của thang dựa vào tường là 60º và chân thang cách tường 4,6 m. Chiều dài của thang là
D=sin(pi+x)+sinx+cot(pi-x)+tan(pi/2-x)
=-sinx+sinx-cotx+cotx=0
Chọn A.
Ta có: