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

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

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

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

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

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

\(=1+sinx\)

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

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

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

\(=1+cosx\)

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

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

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

\(A=0\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(A=\cos\left(\text{π}-\dfrac{x}{2}\right)-\sin\left(\text{π}-x\right)\)

\(=\sin x+\sin x=2\cdot\sin x\)

\(B=\cos\left(2\text{π}+\dfrac{\text{π}}{2}-x\right)+\sin\left(4\text{π}+\dfrac{\text{π}}{2}-x\right)-\cos\left(6\text{π}+\dfrac{3}{2}\text{π}+x\right)-\sin\left(16\text{π}+\dfrac{3}{2}\text{π}+x\right)\)

\(=\sin x+\cos x-\cos\left(\dfrac{3}{2}\text{π}+x\right)-\sin\left(\dfrac{3}{2}\text{π}+x\right)\)

\(=\sin x+\cos x-\cos\left(\text{π}+\dfrac{\text{π}}{2}+x\right)-\sin\left(\text{π}+\dfrac{\text{π}}{2}+x\right)\)

\(=\cos x+\sin x+\cos\left(\dfrac{1}{2}\text{π}+x\right)+\sin\left(\dfrac{1}{2}\text{π}+x\right)\)

\(=\cos x+\sin x-\sin x+\cos x=2\cos x\)

a) P = cos(\(\frac{\Pi}{2}\) + x) + cos(2π - x) + cos(3π + x)   = -sinx + cosx - cosx = -sinx

Bạn xem lại đề hộ mình với. Đây là đẳng thức chứ k phải biểu thức.

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

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

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

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

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

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

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

a: \(\sin\left(a+b\right)\cdot\sin\left(a-b\right)\)

\(=\frac12\cdot\left\lbrack cos\left(a+b-a+b\right)-cos\left(a+b+a-b\right)\right\rbrack\)

\(=\frac12\left\lbrack cos2b-cos2a\right\rbrack=\frac12\cdot\left\lbrack2\cdot cos^2b-1-\left(2\cdot cos^2a-1\right)\right\rbrack\)

\(=\frac12\cdot\left\lbrack2\cdot cos^2b-2\cdot cos^2a\right\rbrack=cos^2b-cos^2a\)

\(=\left(1-\sin^2b\right)-\left(1-\sin^2a\right)=\sin^2a-\sin^2b\)

b: \(4\cdot\sin\left(x+\frac{\pi}{3}\right)\cdot\sin\left(x-\frac{\pi}{3}\right)\)

\(=4\cdot\frac12\cdot\left\lbrack cos\left(x+\frac{\pi}{3}-x+\frac{\pi}{3}\right)-cos\left(x+\frac{\pi}{3}+x+\frac{\pi}{3}\right)\right\rbrack\)

\(=2\cdot\left\lbrack cos\left(\frac23\pi\right)-cos2x\right\rbrack=2\cdot\left\lbrack-\frac12-cos2x\right\rbrack=-1-2\cdot cos2x\)

\(=-1-2\cdot\left(1-2\cdot\sin^2x\right)=-1-2+4\cdot\sin^2x=4\cdot\sin^2x-3\)

c: \(\sin\left(x+\frac{\pi}{4}\right)-\sin\left(x-\frac{\pi}{4}\right)\)

\(=\sin x\cdot cos\left(\frac{\pi}{4}\right)+cosx\cdot\sin\left(\frac{\pi}{4}\right)-\left\lbrack\sin x\cdot cos\left(\frac{\pi}{4}\right)-cosx\cdot\sin\left(\frac{\pi}{4}\right)\right\rbrack\)

\(=\frac{\sqrt2}{2}\cdot\sin x+\frac{\sqrt2}{2}\cdot cosx-\frac{\sqrt2}{2}\cdot\sin x+\frac{\sqrt2}{2}\cdot cosx=\frac{\sqrt2}{2}\cdot2\cdot cosx=\sqrt2\cdot cosx\)