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 8403 by rhm last updated on 10/Oct/16

(Q.1) cos 4A=1−8cos^2 A + 8cos^2 A (Q.2) ((sec8 A −1)/(sec4 A −1)) = ((tan 8A)/(tan 2A)) (Q.3) tanA+tan(60^0 +A)+tan(120^0 +4) = 3tan 3A (Q.4) sinA sin (60^0 −A) sin(60^0 +A) = (1/4)sin3A

$$\left({Q}.\mathrm{1}\right)\:{cos}\:\mathrm{4}{A}=\mathrm{1}−\mathrm{8}{cos}^{\mathrm{2}} {A}\:+\:\mathrm{8}{cos}^{\mathrm{2}} \:{A} \\ $$$$\left({Q}.\mathrm{2}\right)\:\:\frac{{sec}\mathrm{8}\:{A}\:−\mathrm{1}}{{sec}\mathrm{4}\:{A}\:−\mathrm{1}}\:=\:\frac{{tan}\:\mathrm{8}{A}}{{tan}\:\mathrm{2}{A}} \\ $$$$\left({Q}.\mathrm{3}\right)\:\:\:{tanA}+{tan}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)+{tan}\left(\mathrm{120}^{\mathrm{0}} \right. \\ $$$$\left.+\mathrm{4}\right)\:=\:\mathrm{3}{tan}\:\mathrm{3}{A}\: \\ $$$$\left({Q}.\mathrm{4}\right)\:\:\:{sinA}\:{sin}\:\left(\mathrm{60}^{\mathrm{0}} −{A}\right)\:\:{sin}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{3}{A} \\ $$

Answered by ridwan balatif last updated on 10/Oct/16

(Q.4) sinA.(sin(60^o −A)sin(60^o +A)) sinA.((1/2)(cos(−2A)−cos(120))) (1/2)sinA(cos2A+(1/2)) (1/2)sinAcos2A+(1/4)sinA (1/2)((1/2)(sin3A+sin(−A)))+(1/4)sinA (1/4)sin3A−(1/4)sinA+(1/4)sinA (1/4)sin3A ∴sinAsin(60^o −A)sin(60^o +A)=(1/4)sin3A(PROVE)

$$\left(\mathrm{Q}.\mathrm{4}\right)\:\mathrm{sinA}.\left(\mathrm{sin}\left(\mathrm{60}^{\mathrm{o}} −\mathrm{A}\right)\mathrm{sin}\left(\mathrm{60}^{\mathrm{o}} +\mathrm{A}\right)\right)\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sinA}.\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(−\mathrm{2A}\right)−\mathrm{cos}\left(\mathrm{120}\right)\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinA}\left(\mathrm{cos2A}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinAcos2A}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sinA} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin3A}+\mathrm{sin}\left(−\mathrm{A}\right)\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sinA} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin3A}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sinA}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sinA} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin3A}\: \\ $$$$\therefore\mathrm{sinAsin}\left(\mathrm{60}^{\mathrm{o}} −\mathrm{A}\right)\mathrm{sin}\left(\mathrm{60}^{\mathrm{o}} +\mathrm{A}\right)=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin3A}\left(\mathrm{PROVE}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by sandy_suhendra last updated on 10/Oct/16

(Q.1) I think cos 4A=1−8cos^2 A+8cos^4 A cos4A=2cos^2 2A−1 =2(2cos^2 A−1)^2 −1 =2(4cos^4 A−4cos^2 A+1)−1 =8cos^4 A−8cos^2 A+2−1 =1−8cos^2 A+8cos^4 A

$$\left(\mathrm{Q}.\mathrm{1}\right)\:\mathrm{I}\:\mathrm{think}\:\mathrm{cos}\:\mathrm{4A}=\mathrm{1}−\mathrm{8cos}^{\mathrm{2}} \mathrm{A}+\mathrm{8cos}^{\mathrm{4}} \mathrm{A} \\ $$$$\mathrm{cos4A}=\mathrm{2cos}^{\mathrm{2}} \mathrm{2A}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{2cos}^{\mathrm{2}} \mathrm{A}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\mathrm{4cos}^{\mathrm{4}} \mathrm{A}−\mathrm{4cos}^{\mathrm{2}} \mathrm{A}+\mathrm{1}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{8cos}^{\mathrm{4}} \mathrm{A}−\mathrm{8cos}^{\mathrm{2}} \mathrm{A}+\mathrm{2}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\mathrm{8cos}^{\mathrm{2}} \mathrm{A}+\mathrm{8cos}^{\mathrm{4}} \mathrm{A} \\ $$

Commented by rhm last updated on 11/Oct/16

yes yov are right

$${yes}\:{yov}\:{are}\:{right} \\ $$

Answered by sandy_suhendra last updated on 10/Oct/16

(Q.2) ((sec8A−1)/(sec4A−1))×((cos8A cos4A)/(cos8A cos4A)) =((cos4A−cos8A cos4A)/(cos8A−cos8A cos4A)) =((cos4A(1−cos8A))/(cos8A(1−cos4A))) =((cos4A(2sin^2 4A))/(cos8A(2sin^2 2A))) =(((cos4A sin4A) (sin4A))/(cos8A sin^2 2A)) =(((1/2)sin8A (2sin2Acos2A))/(cos8A sin^2 2A)) =((sin8A cos2A)/(cos8A sin2A)) =tan8A cot 2A =((tan8A)/(tan2A))

$$\left(\mathrm{Q}.\mathrm{2}\right)\:\frac{\mathrm{sec8A}−\mathrm{1}}{\mathrm{sec4A}−\mathrm{1}}×\frac{\mathrm{cos8A}\:\mathrm{cos4A}}{\mathrm{cos8A}\:\mathrm{cos4A}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cos4A}−\mathrm{cos8A}\:\mathrm{cos4A}}{\mathrm{cos8A}−\mathrm{cos8A}\:\mathrm{cos4A}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cos4A}\left(\mathrm{1}−\mathrm{cos8A}\right)}{\mathrm{cos8A}\left(\mathrm{1}−\mathrm{cos4A}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{cos4A}\left(\mathrm{2sin}^{\mathrm{2}} \mathrm{4A}\right)}{\mathrm{cos8A}\left(\mathrm{2sin}^{\mathrm{2}} \mathrm{2A}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{cos4A}\:\mathrm{sin4A}\right)\:\left(\mathrm{sin4A}\right)}{\mathrm{cos8A}\:\mathrm{sin}^{\mathrm{2}} \mathrm{2A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin8A}\:\left(\mathrm{2sin2Acos2A}\right)}{\mathrm{cos8A}\:\mathrm{sin}^{\mathrm{2}} \mathrm{2A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin8A}\:\mathrm{cos2A}}{\mathrm{cos8A}\:\mathrm{sin2A}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{tan8A}\:\mathrm{cot}\:\mathrm{2A} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{tan8A}}{\mathrm{tan2A}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com