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Question Number 55141 by 0201011563081 last updated on 18/Feb/19
prove the following identities a.((sin θ)/(1+cos 2θ))=tan θ b.((1−cos 2θ−sin θ)/(sin 2θ−cos θ))=tan θ c.((cos (x+y)+sin (x−y))/(cos 2ycos 2x))=(1/(cos (x+y)sin (y−x)))
$${prove}\:{the}\:{following}\:{identities} \\ $$$$ \\ $$$${a}.\frac{\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}=\mathrm{tan}\:\theta \\ $$$${b}.\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta−\mathrm{sin}\:\theta}{\mathrm{sin}\:\mathrm{2}\theta−\mathrm{cos}\:\theta}=\mathrm{tan}\:\theta \\ $$$${c}.\frac{\mathrm{cos}\:\left({x}+{y}\right)+\mathrm{sin}\:\left({x}−{y}\right)}{\mathrm{cos}\:\mathrm{2}{y}\mathrm{cos}\:\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}+{y}\right)\mathrm{sin}\:\left({y}−{x}\right)} \\ $$
Commented by math1967 last updated on 18/Feb/19
I think identity should be ((sin 2θ)/(1+cos 2θ)) =((2sinθcosθ)/(2cos^2 θ))=tanθ
$${I}\:{think}\:{identity}\:{should}\:{be} \\ $$$$\:\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}\:=\frac{\mathrm{2}{sin}\theta{cos}\theta}{\mathrm{2}{cos}^{\mathrm{2}} \theta}={tan}\theta \\ $$
Answered by math1967 last updated on 18/Feb/19
b)((1−cos2θ−sinθ)/(sin2θ−cosθ))=((2sin^2 θ−sinθ)/(2sinθcosθ−cosθ)) =((sinθ(2sinθ−1))/(cosθ(2sinθ−1)))=tanθ
$$\left.{b}\right)\frac{\mathrm{1}−{cos}\mathrm{2}\theta−{sin}\theta}{{sin}\mathrm{2}\theta−{cos}\theta}=\frac{\mathrm{2}{sin}^{\mathrm{2}} \theta−{sin}\theta}{\mathrm{2}{sin}\theta{cos}\theta−{cos}\theta} \\ $$$$=\frac{{sin}\theta\left(\mathrm{2}{sin}\theta−\mathrm{1}\right)}{{cos}\theta\left(\mathrm{2}{sin}\theta−\mathrm{1}\right)}={tan}\theta \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Feb/19
)[cos(x+y)+sin(x−y)]×[cos(x+y)−sin(x−y)] cos^2 (x+y)−sin^2 (x−y) (1/2)[2cos^2 (x+y)−2sin^2 (x−y)] (1/2)[1+cos(2x+2y)−1+cos(2x−2y)] (1/2)[2cos2xcos2y] =cos2xcos2y so ((cos(x+y)+sin(x−y))/(cos2xcos2y))=(1/(cos(x+y)−sin(x−y))) ((cos(x+y)+sin(x−y))/(cos2xcos2y))=(1/(cos(x+y)+sin(y−x))) i think this is the problem... typing error...
$$\left.\right)\left[{cos}\left({x}+{y}\right)+{sin}\left({x}−{y}\right)\right]×\left[{cos}\left({x}+{y}\right)−{sin}\left({x}−{y}\right)\right] \\ $$$${cos}^{\mathrm{2}} \left({x}+{y}\right)−{sin}^{\mathrm{2}} \left({x}−{y}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{cos}^{\mathrm{2}} \left({x}+{y}\right)−\mathrm{2}{sin}^{\mathrm{2}} \left({x}−{y}\right)\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+{cos}\left(\mathrm{2}{x}+\mathrm{2}{y}\right)−\mathrm{1}+{cos}\left(\mathrm{2}{x}−\mathrm{2}{y}\right)\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{2}{cos}\mathrm{2}{xcos}\mathrm{2}{y}\right] \\ $$$$={cos}\mathrm{2}{xcos}\mathrm{2}{y} \\ $$$${so}\:\frac{{cos}\left({x}+{y}\right)+{sin}\left({x}−{y}\right)}{{cos}\mathrm{2}{xcos}\mathrm{2}{y}}=\frac{\mathrm{1}}{{cos}\left({x}+{y}\right)−{sin}\left({x}−{y}\right)} \\ $$$$\frac{{cos}\left({x}+{y}\right)+{sin}\left({x}−{y}\right)}{{cos}\mathrm{2}{xcos}\mathrm{2}{y}}=\frac{\mathrm{1}}{{cos}\left({x}+{y}\right)+{sin}\left({y}−{x}\right)} \\ $$$${i}\:{think}\:{this}\:{is}\:{the}\:{problem}… \\ $$$${typing}\:{error}… \\ $$$$ \\ $$

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