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Question Number 48526 by Pk1167156@gmail.com last updated on 25/Nov/18

$$\mathrm{The}\mathrm{maximum}\mathrm{and}\mathrm{minimum}\mathrm{values}$$$$\mathrm{of}a\cos 2\theta +b\sin 2\theta \mathrm{are}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18

$$a=rsin\alpha b=rcos\alpha$$$$a^2+b^2=r^{2}sor=\sqrt{a^2+b^2}$$$$S=acos2\theta +bsin2\theta$$$$=rsin\alpha cos2\theta +rcos\alpha sin2\theta$$$$=rsin(\alpha +2\theta )$$$$maxvalueofsin(\alpha +2\theta )=1$$$$minvalueofsin(\alpha +2\theta )=-1$$$$somaxvalueofacos2\theta +bsin2\theta is$$$$=r\times 1$$$$=\sqrt{a^2+b^2}$$$$minvalueofacos2\theta +bsin2\theta$$$$=r\times -1$$$$=-\sqrt{a^2+b^2}$$$$$$

Answered by ajfour last updated on 25/Nov/18

$$a\cos 2\theta +b\sin 2\theta$$$$=\sqrt{a^2+b^2}\sin ({\tan }^{-1}\frac{a}{b}+2\theta )$$$$somaximumvalue=\sqrt{{\mathit{a}}^2+{\mathit{b}}^2}$$$$andminimumvalue=-\sqrt{{\mathit{a}}^2+{\mathit{b}}^2}.$$

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