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Question Number 102510 by Study last updated on 09/Jul/20
lim_(△x→0) ((sin^2 ((1/3)x+(1/3)△x)−sin^2 (1/3)x)/(△x))=?
$${li}\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {{m}}\frac{{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{1}}{\mathrm{3}}\bigtriangleup{x}\right)−{sin}^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{3}}{x}}{\bigtriangleup{x}}=? \\ $$
Answered by bemath last updated on 09/Jul/20
lim_(Δx→0) ((sin^2 (1/3)(x+Δx)−sin^2 (1/3)x)/(Δx)) = (d/dx) [ sin^2 ((1/3)x)] = (1/3)sin (((2x)/3))
$$\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{3}}\left({x}+\Delta{x}\right)−\mathrm{sin}\:^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{3}}{x}}{\Delta{x}}\:= \\ $$$$\frac{{d}}{{dx}}\:\left[\:\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{3}}{x}\right)\right]\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\: \\ $$
Answered by Dwaipayan Shikari last updated on 09/Jul/20
Method 1 (2/3)sin(x/3)cos(x/3)=(1/3)sin((2x)/3) Method 2 lim_(△x→0) (((sin((x/3)+△(x/3))+sin(x/3))(sin((x/3)+△(x/3))−sin(x/3)))/(△x)) lim_(△x→0) (((2sin(((((2x)/3)+△(x/3))/2))cos△(x/6))(2cos(((((2x)/3)+△(x/3))/2))sin△(x/6)))/(△x)) 2sin((2x+△x)/3) .(((△x)/6)/(△x))=(1/3)sin((2x)/3)
$${Method}\:\mathrm{1} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{sin}\frac{{x}}{\mathrm{3}}{cos}\frac{{x}}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{3}}{sin}\frac{\mathrm{2}{x}}{\mathrm{3}} \\ $$$${Method}\:\mathrm{2} \\ $$$$\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({sin}\left(\frac{{x}}{\mathrm{3}}+\bigtriangleup\frac{{x}}{\mathrm{3}}\right)+{sin}\frac{{x}}{\mathrm{3}}\right)\left({sin}\left(\frac{{x}}{\mathrm{3}}+\bigtriangleup\frac{{x}}{\mathrm{3}}\right)−{sin}\frac{{x}}{\mathrm{3}}\right)}{\bigtriangleup{x}} \\ $$$$ \\ $$$$\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{2}{sin}\left(\frac{\frac{\mathrm{2}{x}}{\mathrm{3}}+\bigtriangleup\frac{{x}}{\mathrm{3}}}{\mathrm{2}}\right){cos}\bigtriangleup\frac{{x}}{\mathrm{6}}\right)\left(\mathrm{2}{cos}\left(\frac{\frac{\mathrm{2}{x}}{\mathrm{3}}+\bigtriangleup\frac{{x}}{\mathrm{3}}}{\mathrm{2}}\right){sin}\bigtriangleup\frac{{x}}{\mathrm{6}}\right)}{\bigtriangleup{x}} \\ $$$$\mathrm{2}{sin}\frac{\mathrm{2}{x}+\bigtriangleup{x}}{\mathrm{3}}\:.\frac{\frac{\bigtriangleup{x}}{\mathrm{6}}}{\bigtriangleup{x}}=\frac{\mathrm{1}}{\mathrm{3}}{sin}\frac{\mathrm{2}{x}}{\mathrm{3}} \\ $$$$ \\ $$

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