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Question Number 164945 by mnjuly1970 last updated on 23/Jan/22

$$$$$$\mathrm{I}f\{\begin{array}{l}sin\left(3\theta \right)+cos\left(3\theta \right)=x\\ sin\left(\theta \right)+cos\left(\theta \right)=y\end{array}$$$$then,findarelationship$$$$betweenx,y$$$$indepententof,\theta .$$

Answered by mr W last updated on 23/Jan/22

$$y^2=1+2\sin \theta \cos \theta$$$$\Rightarrow \sin \theta \cos \theta =\frac{y^2-1}{2}$$$${(\sin \theta -\cos \theta )}^2=1-2\sin \theta \cos \rho =2-y^2$$$$\sin \theta -\cos \theta =\pm \sqrt{2-y^2}$$$$x=\sin \left(3\theta \right)+\cos \left(3\theta \right)$$$$=3\sin \theta -4{\sin }^3\theta +4{\cos }^3\theta -3\cos \theta$$$$=3(\sin \theta -\cos \theta )-4({\sin }^3\theta -{\cos }^3\theta )$$$$=3(\sin \theta -\cos \theta )-4(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )$$$$=3(\sin \theta -\cos \theta )-4(\sin \theta -\cos \theta )(1+\sin \theta \cos \theta )$$$$=-(\sin \theta -\cos \theta )(1+4\sin \theta \cos \theta )$$$$=\pm \sqrt{2-y^2}(1+2(y^2-1))$$$$=\pm \sqrt{2-y^2}(2y^2-1)$$$$\Rightarrow x=\pm \sqrt{2-y^2}(2y^2-1)$$

Commented by mnjuly1970 last updated on 23/Jan/22

$$thanksalotsirW...grateful$$

Commented by Tawa11 last updated on 23/Jan/22

$$\mathrm{Weldone}\mathrm{sir}$$

Answered by TheSupreme last updated on 24/Jan/22

$$sin(\theta +\frac{\pi }{4})=\frac{y}{\sqrt{2}}$$$$sin(3\theta +\frac{\pi }{4})=\frac{x}{\sqrt{2}}$$$$\theta =arcsin\left(\frac{y}{\sqrt{2}}\right)-\frac{\pi }{4}$$$$3\theta =arcsin\left(\frac{x}{\sqrt{2}}\right)-\frac{\pi }{4}$$$$arcsin\left(\frac{y}{\sqrt{2}}\right)=\frac{1}{3}arcsin\left(\frac{x}{\sqrt{2}}\right)$$$$y=\sqrt{2}sin\left(\frac{1}{3}arcsin\left(\frac{x}{\sqrt{20}}\right)\right)$$

Commented by mnjuly1970 last updated on 24/Jan/22

$$merceysir...$$

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