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

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

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

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

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

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

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

a/ Bạn coi lại đề bài, pt này có 1 nghiệm rất xấu ko giải được:

\(\Leftrightarrow1-sin^2x-2\sqrt{3}sinx.cosx=sin^3x+1\)

\(\Leftrightarrow sin^3x+sin^2x+2\sqrt{3}sinx.cosx=0\)

\(\Leftrightarrow sinx\left(sin^2x+sinx+2\sqrt{3}cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sin^2x+sinx+2\sqrt{3}cosx=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow sin^2x+sinx=-2\sqrt{3}cosx\) (\(cosx\le0\))

\(\Leftrightarrow sin^2x\left(sinx+1\right)^2=12cos^2x\)

\(\Leftrightarrow sin^2x\left(sinx+1\right)^2=12\left(1-sinx\right)\left(1+sinx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}1+sinx=0\left(2\right)\\sin^2x\left(sinx+1\right)=12\left(1-sinx\right)\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow x=-\frac{\pi}{2}+k2\pi\) (thỏa mãn)

\(\left(3\right)\Leftrightarrow sin^3x+sin^2x+12sinx-12=0\)

Pt bậc 3 này có nghiệm thực thuộc \(\left(-1;1\right)\) nhưng rất xấu

b/

\(\Leftrightarrow\frac{3}{5}sin2x+\frac{4}{5}cos2x=-cos2003x\)

Đặt \(\frac{3}{5}=cosa\) với \(a\in\left(0;\pi\right)\)

\(\Rightarrow sin2x.cosa+cos2x.sina=-cos2003x\)

\(\Leftrightarrow sin\left(2x+a\right)=sin\left(2003x-\frac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2003x-\frac{\pi}{2}=2x+a+k2\pi\\2003x-\frac{\pi}{2}=\pi-2x-a+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4002}+\frac{a}{2001}+\frac{k2\pi}{2001}\\x=\frac{3\pi}{4010}-\frac{a}{2005}+\frac{k2\pi}{2005}\end{matrix}\right.\)

c/

\(\Leftrightarrow\sqrt{3}sin\left(x-\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}-x\right)=2sin1972x\)

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin\left(x-\frac{\pi}{3}\right)+\frac{1}{2}cos\left(x-\frac{\pi}{3}\right)=sin1972x\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}+\frac{\pi}{6}\right)=sin1972x\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin1972x\)

\(\Leftrightarrow\left[{}\begin{matrix}1972x=x-\frac{\pi}{6}+k2\pi\\1972x=\frac{7\pi}{6}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{11826}+\frac{k2\pi}{1971}\\x=\frac{7\pi}{11838}+\frac{k2\pi}{1973}\end{matrix}\right.\)

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

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

\(A=0\)

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

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

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

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

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

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

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

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

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

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

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

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

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

a) Ta có:

      \(\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x\cos \frac{\pi }{4} + \cos x\sin \frac{\pi }{4}} \right) = \sqrt 2 \left( {\sin x.\frac{{\sqrt 2 }}{2} + \cos x.\frac{{\sqrt 2 }}{2}} \right) = \sin x + \cos x\)

b) Ta có:

\(\tan \left( {\frac{\pi }{4} - x} \right) = \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}} = \frac{{1 - \tan x}}{{1 + \tan x}}\;\)

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

\(=-\frac{1}{4}cosx+\frac{1}{4}\left(cos3x+cosx\right)=\frac{1}{4}cos3x\)

\(sin5x-2sinx\left(cos4x+cos2x\right)=sinx.cos4x+cosx.sin4x-2sinx.cos4x-2sinx.cos2x\)

\(=sin4x.cosx-cos4x.sinx-2sinx.cos2x=sin3x-2sinx.cos2x\)

\(=sinx.cos2x+cosx.sin2x-2sinx.cos2x\)

\(=sin2x.cosx-cos2x.sinx=sinx\)

\(=sin\left(x+\frac{\pi}{2}+42\pi\right)+cos\left(206\pi+\pi+x\right)+sin^2\left(32\pi+\pi+x\right)+sin^2\left(x+\frac{\pi}{2}-2\pi\right)\)

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

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

\(=sin^2x+cos^2x=1\)

\(a,VT=2.tanx+tan\left(-x\right)\\ =2tanx-tanx=tanx\)

\(b,VT=sin\left(2\pi+\dfrac{\pi}{2}-x\right)+cos\left(12\pi+\pi+x\right)-sin\left(x-4\pi-\pi\right)\\ =sin\left(\dfrac{\pi}{2}-x\right)+cos\left(\pi+x\right)+sin\left(\pi-x\right)\\ =cosx-cosx+sinx\\ =sinx=VP\)

a: \(2\cdot cot\left(\dfrac{pi}{2}-x\right)+tan\left(pi-x\right)\)

\(=2\cdot tanx-tanx\)

=tan x

b: \(sin\left(\dfrac{5}{2}pi-x\right)+cos\left(13pi+x\right)-sin\left(x-5pi\right)\)

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

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