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 80519 by mathocean1 last updated on 03/Feb/20

g(x)=2cos^2 x+sin(2x). g′(x)= ..........?

$$\mathrm{g}\left({x}\right)=\mathrm{2}{cos}^{\mathrm{2}} {x}+{sin}\left(\mathrm{2}{x}\right). \\ $$$${g}'\left({x}\right)=\:..........? \\ $$

Commented by mr W last updated on 03/Feb/20

g′(x)=2×2 cos x (−sin x)+cos (2x) 2 =−4 cos x sin x+2 cos (2x) =−2 sin (2x)+2 cos (2x) =2 [cos (2x)−sin (2x)] =2(√2)[(1/(√2)) cos (2x)−(1/(√2)) sin (2x)] =2(√2)[cos (π/4) cos (2x)−sin (π/4) sin (2x)] =2(√2)cos (2x+(π/4))

$${g}'\left({x}\right)=\mathrm{2}×\mathrm{2}\:\mathrm{cos}\:{x}\:\left(−\mathrm{sin}\:{x}\right)+\mathrm{cos}\:\left(\mathrm{2}{x}\right)\:\mathrm{2} \\ $$$$=−\mathrm{4}\:\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}+\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right) \\ $$$$=−\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)+\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right) \\ $$$$=\mathrm{2}\:\left[\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\left(\mathrm{2}{x}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\left[\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\left[\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\:\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)\right] \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right) \\ $$

Commented by mathocean1 last updated on 03/Feb/20

thank you sirs

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sirs} \\ $$

Answered by Rio Michael last updated on 03/Feb/20

g′(x) = −4sin x + 2cos 2x

$${g}'\left({x}\right)\:=\:−\mathrm{4sin}\:{x}\:+\:\mathrm{2cos}\:\mathrm{2}{x} \\ $$

Answered by Henri Boucatchou last updated on 03/Feb/20

g(x)=cos2x+sin2x+1 g′(x)=2(cos2x−sin2x) =2(√2)cos(2x+(π/4))

$${g}\left({x}\right)={cos}\mathrm{2}{x}+{sin}\mathrm{2}{x}+\mathrm{1} \\ $$$${g}'\left({x}\right)=\mathrm{2}\left({cos}\mathrm{2}{x}−{sin}\mathrm{2}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\mathrm{2}}{cos}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com