Question Number 63426 by Rio Michael last updated on 04/Jul/19
![show that (a) cos[2cos^(−1) (x) +sin^(−1) (x)]= −(√(1−x^2 )) (b) ((sinα + sinβ)/(cosα−cosβ))=cot(((β−α)/2)) (c) 2cos((π/3)+p)≊ 1−(√3) if p is small enough to neglect p^2 . (d) if θ =(1/2)sin^(−1) ((3/4)), show that sinθ−cosθ = ±(1/2) (e)write tan3A in terms of tanA (f) Factorise cosθ − cos3θ−cos5θ+cos7θ (g)i) verify that f(x)=((sin2θ+sin10θ)/(cos2θ+cos10θ))=((2tan3θ)/(1−tan^2 3θ)) ii) hence find in radians the general solution of f(x)=1 sir Forkum Michael](Q63426.png)
$${show}\:{that}\: \\ $$$$\left({a}\right)\:{cos}\left[\mathrm{2}{cos}^{−\mathrm{1}} \left({x}\right)\:+{sin}^{−\mathrm{1}} \left({x}\right)\right]=\:−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\: \\ $$$$\left({b}\right)\:\frac{{sin}\alpha\:+\:{sin}\beta}{{cos}\alpha−{cos}\beta}={cot}\left(\frac{\beta−\alpha}{\mathrm{2}}\right) \\ $$$$\left({c}\right)\:\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{3}}+{p}\right)\approxeq\:\mathrm{1}−\sqrt{\mathrm{3}}\:{if}\:{p}\:{is}\:{small}\:{enough}\:{to}\:{neglect}\:{p}^{\mathrm{2}} . \\ $$$$\left({d}\right)\:{if}\:\theta\:=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right),\:{show}\:{that}\:{sin}\theta−{cos}\theta\:=\:\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({e}\right){write}\:{tan}\mathrm{3}{A}\:{in}\:{terms}\:{of}\:{tanA} \\ $$$$\left({f}\right)\:{Factorise}\:{cos}\theta\:−\:{cos}\mathrm{3}\theta−{cos}\mathrm{5}\theta+{cos}\mathrm{7}\theta \\ $$$$\left.\left({g}\right){i}\right)\:{verify}\:{that}\:{f}\left({x}\right)=\frac{{sin}\mathrm{2}\theta+{sin}\mathrm{10}\theta}{{cos}\mathrm{2}\theta+{cos}\mathrm{10}\theta}=\frac{\mathrm{2}{tan}\mathrm{3}\theta}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}\theta} \\ $$$$\left.\:\:{ii}\right)\:{hence}\:{find}\:{in}\:{radians}\:{the}\:{general}\:{solution}\:{of}\:\:{f}\left({x}\right)=\mathrm{1} \\ $$$${sir}\:{Forkum}\:{Michael} \\ $$
Commented by Tony Lin last updated on 04/Jul/19
![(a) let θ=cos^(−1) (x)⇒cosθ=x let (π/2)−θ=sin^(−1) (x)⇒sinθ=(√(1−x^2 )) cos[2θ+((π/2)−θ)] =cos((π/2)+θ) =−sinθ =−(√(1−x^2 ))](Q63437.png)
$$\left({a}\right) \\ $$$${let}\:\theta={cos}^{−\mathrm{1}} \left({x}\right)\Rightarrow{cos}\theta={x} \\ $$$${let}\:\frac{\pi}{\mathrm{2}}−\theta={sin}^{−\mathrm{1}} \left({x}\right)\Rightarrow{sin}\theta=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${cos}\left[\mathrm{2}\theta+\left(\frac{\pi}{\mathrm{2}}−\theta\right)\right] \\ $$$$={cos}\left(\frac{\pi}{\mathrm{2}}+\theta\right) \\ $$$$=−{sin}\theta \\ $$$$=−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$
Commented by Tony Lin last updated on 04/Jul/19
$$\left({b}\right) \\ $$$$\frac{{sin}\alpha+{sin}\beta}{{cos}\alpha−{cos}\beta} \\ $$$$=\frac{\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)}{−\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}\right){sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)} \\ $$$$=\frac{−{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)}{{sin}\left(\frac{\alpha−\beta}{\mathrm{2}}\right)} \\ $$$$={cot}\left(\frac{\beta−\alpha}{\mathrm{2}}\right) \\ $$
Commented by Tony Lin last updated on 04/Jul/19
$$\left({d}\right) \\ $$$$\theta=\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\mathrm{2}\theta={sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}={sin}\mathrm{2}\theta \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{2}{sin}\theta{cos}\theta \\ $$$$\left({sin}\theta−{cos}\theta\right)^{\mathrm{2}} \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\theta{cos}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{sin}\theta−{cos}\theta=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Tony Lin last updated on 04/Jul/19
$$\left({e}\right) \\ $$$${tan}\mathrm{3}{A} \\ $$$$={tan}\left({A}+\mathrm{2}{A}\right) \\ $$$$=\frac{{tanA}+{tan}\mathrm{2}{A}}{\mathrm{1}−{tanAtan}\mathrm{2}{A}} \\ $$$$=\frac{{tanA}+\frac{\mathrm{2}{tanA}}{\mathrm{1}−{tan}^{\mathrm{2}} {A}}}{\mathrm{1}−\frac{\mathrm{2}{tan}^{\mathrm{2}} {A}}{\mathrm{1}−{tan}^{\mathrm{2}} {A}}} \\ $$$$=\frac{\mathrm{3}{tanA}−{tan}^{\mathrm{3}} {A}}{\mathrm{1}−\mathrm{3}{tan}^{\mathrm{2}} {A}} \\ $$
Commented by Rio Michael last updated on 05/Jul/19
$${please}\:{explain}\:{the}\:{occurance}\:{of}\:\: \\ $$$$\left({sin}\theta−{cos}\theta\right)^{\mathrm{2}} \\ $$
Answered by som(math1967) last updated on 04/Jul/19
$${g}\left({i}\right)\:\frac{{sin}\mathrm{2}\theta+{sin}\mathrm{10}\theta}{{cos}\mathrm{2}\theta+{cos}\mathrm{10}\theta}=\frac{\mathrm{2}{sin}\mathrm{6}\theta{cos}\mathrm{4}\theta}{\mathrm{2}{cos}\mathrm{6}\theta{cos}\mathrm{4}\theta}={tan}\mathrm{6}\theta \\ $$$${again}\:{tan}\mathrm{6}\theta={tan}\mathrm{2}.\mathrm{3}\theta=\frac{\mathrm{2}{tan}\mathrm{3}\theta}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}\theta} \\ $$$$\left.{ii}\right)\:{Again}\:\frac{\mathrm{2}{tan}\mathrm{3}\theta}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{3}\theta}=\mathrm{1} \\ $$$$\Rightarrow{tan}^{\mathrm{2}} \mathrm{3}\theta+\mathrm{2}{tan}\mathrm{3}\theta=\mathrm{1} \\ $$$$\Rightarrow{tan}^{\mathrm{2}} \mathrm{3}\theta+\mathrm{2}{tan}\mathrm{3}\theta+\mathrm{1}=\mathrm{2} \\ $$$$\Rightarrow\left({tan}\mathrm{3}\theta+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\Rightarrow{tan}\mathrm{3}\theta=\sqrt{\mathrm{2}}\:−\mathrm{1}={tan}\frac{\pi}{\mathrm{8}} \\ $$$$\therefore\mathrm{3}\theta={n}\pi+\frac{\pi}{\mathrm{8}}\:\therefore\theta=\frac{\mathrm{1}}{\mathrm{3}}\left({n}\pi+\frac{\pi}{\mathrm{8}}\right) \\ $$