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Question Number 191344 by mnjuly1970 last updated on 23/Apr/23
if sin(x) + (√3) cos (x) = (1/2) ⇒(√3) sin(4x ) − cos (4x )= ?
$$ \\ $$$$\:\:\:\:\:\:\mathrm{if}\:\:\:{sin}\left({x}\right)\:+\:\sqrt{\mathrm{3}}\:{cos}\:\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\sqrt{\mathrm{3}}\:\:{sin}\left(\mathrm{4}{x}\:\right)\:−\:\:{cos}\:\left(\mathrm{4}{x}\:\right)=\:? \\ $$$$ \\ $$
Answered by mehdee42 last updated on 23/Apr/23
(1/2)sinx+((√3)/2)cosx=(1/4)⇒cos(x−30)=(1/4) ⇒cos(2x−60)=2((1/(16)))−1=−(7/8) ⇒cos(4x−120)=2(((49)/(16)))−1=((41)/8) ∴ 2(((49)/(64)))−1=((17)/(32)) ⇒cos4xcos120+sin4xsin120=((41)/8) ∴ =((17)/(32)) −(1/2)cos4x+((√3)/2)sin4x=((41)/8) ∴ =((17)/(32)) ⇒(√3)sin4x−cos4x=((41)/4) ∴ ((17)/(16))
$$\frac{\mathrm{1}}{\mathrm{2}}{sinx}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosx}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{cos}\left({x}−\mathrm{30}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}{x}−\mathrm{60}\right)=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{16}}\right)−\mathrm{1}=−\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow{cos}\left(\mathrm{4}{x}−\mathrm{120}\right)=\mathrm{2}\left(\frac{\mathrm{49}}{\mathrm{16}}\right)−\mathrm{1}=\frac{\mathrm{41}}{\mathrm{8}}\:\:\therefore\:\:\mathrm{2}\left(\frac{\mathrm{49}}{\mathrm{64}}\right)−\mathrm{1}=\frac{\mathrm{17}}{\mathrm{32}} \\ $$$$\Rightarrow{cos}\mathrm{4}{xcos}\mathrm{120}+{sin}\mathrm{4}{xsin}\mathrm{120}=\frac{\mathrm{41}}{\mathrm{8}}\:\:\therefore\:\:=\frac{\mathrm{17}}{\mathrm{32}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{4}{x}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\mathrm{4}{x}=\frac{\mathrm{41}}{\mathrm{8}}\:\:\therefore\:\:=\frac{\mathrm{17}}{\mathrm{32}} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}{sin}\mathrm{4}{x}−{cos}\mathrm{4}{x}=\frac{\mathrm{41}}{\mathrm{4}}\:\therefore\:\:\frac{\mathrm{17}}{\mathrm{16}} \\ $$$$ \\ $$
Commented by mr W last updated on 23/Apr/23
very nice sir!
$${very}\:{nice}\:{sir}! \\ $$
Commented by mehdee42 last updated on 23/Apr/23
hello sir W my mistakewas in the calculation , which i have corrected with red color.thank you
$${hello}\:{sir}\:\:{W} \\ $$$${my}\:{mistakewas}\:{in}\:{the}\:{calculation}\:,\:{which}\:{i}\:{have}\:{corrected}\: \\ $$$${with}\:{red}\:{color}.{thank}\:{you} \\ $$
Answered by mr W last updated on 23/Apr/23
((sin x)/2)+((√3)/2) cos x=(1/4) sin x sin (π/6)+cos x cos (π/6)=(1/4) cos (x−(π/6))=(1/4) ⇒x−(π/6)=2kπ±cos^(−1) (1/4) ⇒x=2kπ+(π/6)±cos^(−1) (1/4) (√3) sin 4x−cos 4x =2(sin 4x cos (π/6)−cos 4x sin (π/6)) =2 sin (4x−(π/6)) =2 sin (8kπ+(π/2)±4 cos^(−1) (1/4)) =2 cos (4 cos^(−1) (1/4)) =2 [2 cos^2 (2 cos^(−1) (1/4))−1] =2 [2 {2 cos^2 (cos^(−1) (1/4))−1}^2 −1] =2 [2 {2×(1/4^2 )−1}^2 −1] =2 [2 {(7/8)}^2 −1] =((17)/(16)) ✓
$$\frac{\mathrm{sin}\:{x}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:{x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:{x}\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}+\mathrm{cos}\:{x}\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{cos}\:\left({x}−\frac{\pi}{\mathrm{6}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{x}−\frac{\pi}{\mathrm{6}}=\mathrm{2}{k}\pi\pm\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{6}}\pm\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{4}{x} \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}−\mathrm{cos}\:\mathrm{4}{x}\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{4}{x}−\frac{\pi}{\mathrm{6}}\right) \\ $$$$=\mathrm{2}\:\mathrm{sin}\:\left(\mathrm{8}{k}\pi+\frac{\pi}{\mathrm{2}}\pm\mathrm{4}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{4}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\mathrm{2}\:\left[\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{1}\right] \\ $$$$=\mathrm{2}\:\left[\mathrm{2}\:\left\{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\:\left(\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{1}\right\}^{\mathrm{2}} −\mathrm{1}\right] \\ $$$$=\mathrm{2}\:\left[\mathrm{2}\:\left\{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }−\mathrm{1}\right\}^{\mathrm{2}} −\mathrm{1}\right] \\ $$$$=\mathrm{2}\:\left[\mathrm{2}\:\left\{\frac{\mathrm{7}}{\mathrm{8}}\right\}^{\mathrm{2}} −\mathrm{1}\right] \\ $$$$=\frac{\mathrm{17}}{\mathrm{16}}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 23/Apr/23
very nice solution
$$\:{very}\:{nice}\:{solution}\:\: \\ $$
Answered by cortano12 last updated on 23/Apr/23
⇒sin^2 x+3cos^2 x+(√3) sin 2x = (1/4) ⇒(((1−cos 2x)/2))+3(((1+cos 2x)/2))+(√3) sin 2x=(1/4) ⇒4+2cos 2x+2(√3) sin 2x=(1/2) ⇒2cos 2x+2(√3) sin 2x=−(7/2) ⇒4cos^2 2x+12sin^2 2x+4(√3) sin 4x=((49)/4) ⇒4(((1+cos 4x)/2))+12(((1−cos 4x)/2))+4(√3) sin 4x=((49)/4) ⇒8−4cos 4x+4(√3) sin 4x=((49)/4) ⇒(√3) sin 4x−cos 4x=((49)/(16))−2 ⇒(√3) sin 4x−cos 4x=((17)/(16))
$$\:\Rightarrow\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{3cos}\:^{\mathrm{2}} \mathrm{x}+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{2x}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right)+\mathrm{3}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2x}}{\mathrm{2}}\right)+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{2x}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{4}+\mathrm{2cos}\:\mathrm{2x}+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{2x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2cos}\:\mathrm{2x}+\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{2x}=−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4cos}\:^{\mathrm{2}} \mathrm{2x}+\mathrm{12sin}\:^{\mathrm{2}} \mathrm{2x}+\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{4x}=\frac{\mathrm{49}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{4}\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{4x}}{\mathrm{2}}\right)+\mathrm{12}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{4x}}{\mathrm{2}}\right)+\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{4x}=\frac{\mathrm{49}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{8}−\mathrm{4cos}\:\mathrm{4x}+\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{4x}=\frac{\mathrm{49}}{\mathrm{4}} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{4x}−\mathrm{cos}\:\mathrm{4x}=\frac{\mathrm{49}}{\mathrm{16}}−\mathrm{2} \\ $$$$\Rightarrow\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{4x}−\mathrm{cos}\:\mathrm{4x}=\frac{\mathrm{17}}{\mathrm{16}} \\ $$
Commented by mnjuly1970 last updated on 23/Apr/23
thanks alot sir cortano ⋛
$${thanks}\:{alot}\:{sir}\:{cortano}\:\cancel{\lesseqgtr} \\ $$

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