Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 81354 by jagoll last updated on 12/Feb/20

$$\mathrm{given}\mathrm{y}=\sin (\frac{\pi }{3}-2\mathrm{x})+2\cos (\frac{\pi }{12}+\mathrm{x})+1$$$$\mathrm{where}\mathrm{x}\in (0,2\pi )\mathrm{has}\mathrm{maximum}\mathrm{and}$$$$\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{p}\mathrm{and}\mathrm{q}.$$$$\mathrm{find}{\mathrm{p}}^2-{\mathrm{q}}^2?$$$$\left(\mathrm{A}\right)-18\left(\mathrm{B}\right)-16$$$$\left(\mathrm{C}\right)\frac{63}{4}\left(\mathrm{D}\right)16$$

Commented by john santu last updated on 12/Feb/20

$$y=\sin (-2x+\frac{\pi }{3})+2\cos (x+\frac{\pi }{12})+1$$$$\sin (-2x+\frac{\pi }{3})=\cos (\frac{\pi }{2}-(-2x+\frac{\pi }{3}))$$$$\cos (2x+\frac{\pi }{6})=\cos 2(x+\frac{\pi }{12})$$$$letx+\frac{\pi }{12}=\theta$$$$y=\cos 2\theta +2\cos \theta +1$$$$y=2\cos {}^2\theta +2\cos \theta$$$$for\cos \theta =1\Rightarrow f_1=4\left(max\right)$$$$for\cos \theta =-1\Rightarrow f_2=0$$$$for\cos \theta =-\frac{1}{2}\Rightarrow f_3=2\times \frac{1}{4}+(-1)=-\frac{1}{2}\left(min\right)$$$$p^2-q^2=(p+q)(p-q)=\frac{7}{2}\times \frac{9}{2}=\frac{63}{4}$$$$$$

Commented by jagoll last updated on 12/Feb/20

$$\mathrm{thank}\mathrm{you}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com