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Question Number 140368 by meetbhavsar25 last updated on 06/May/21
Answered by benjo_mathlover last updated on 07/May/21
cos^(−1) (((3+5cos x)/(5+3cos x))) = u ⇒((3+5cos x)/(5+3cos x)) = cos u ⇒3+5cos x = 5cos u+3cos u cos x ⇒(5−3cos u)cos x = 5cos u−3 ⇒cos x = ((5cos u−3)/(5−3cos u)) ⇒2cos^2 ((x/2))=((5cos u−3)/(5−3cos u)) +((5−3cos u)/(5−3cos u)) ⇒2cos^2 ((x/2))= ((2cos u+2)/(5−3cos u)) ⇒(1/2)sec^2 ((x/2))= ((5−3cos u)/(2cos u+2)) ⇒sec^2 ((x/2)) = ((5−3cos u)/(cos u+1)) ⇒tan^2 ((x/2))= ((5−3cos u)/(cos u+1)) −((cos u+1)/(cos u+1)) ⇒tan^2 ((x/2))=((4−4cos u)/(1+cos u)) = ((4(1−(1−2sin^2 ((u/2))))/(2cos^2 ((u/2)))) ⇒tan^2 ((x/2))=4 tan^2 ((u/2)) ⇒tan ((x/2))= 2tan ((u/2)) ⇒(u/2) = tan^(−1) ((1/2)tan (x/2)) ⇒u = 2tan^(−1) ((1/2)tan (x/2))
$$\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\right)\:=\:\mathrm{u} \\ $$$$\Rightarrow\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\:=\:\mathrm{cos}\:\mathrm{u} \\ $$$$\Rightarrow\mathrm{3}+\mathrm{5cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}+\mathrm{3cos}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\Rightarrow\left(\mathrm{5}−\mathrm{3cos}\:\mathrm{u}\right)\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}−\mathrm{3} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{5cos}\:\mathrm{u}−\mathrm{3}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}} \\ $$$$\Rightarrow\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{5cos}\:\mathrm{u}−\mathrm{3}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}\:+\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}} \\ $$$$\Rightarrow\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\frac{\mathrm{2cos}\:\mathrm{u}+\mathrm{2}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{2cos}\:\mathrm{u}+\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{cos}\:\mathrm{u}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{cos}\:\mathrm{u}+\mathrm{1}}\:−\frac{\mathrm{cos}\:\mathrm{u}+\mathrm{1}}{\mathrm{cos}\:\mathrm{u}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{4}−\mathrm{4cos}\:\mathrm{u}}{\mathrm{1}+\mathrm{cos}\:\mathrm{u}}\:=\:\frac{\mathrm{4}\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{u}}{\mathrm{2}}\right)\right)\right.}{\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{u}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{4}\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{u}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{2tan}\:\left(\frac{\mathrm{u}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\mathrm{u}}{\mathrm{2}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{u}\:=\:\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$
Commented by meetbhavsar25 last updated on 07/May/21
cos^(−1) (((3+5cos x)/(5+3cos x))) = u ⇒((3+5cos x)/(5+3cos x)) = cos u ⇒3+5cos x = 5cos u+3cos u cos x ⇒(5−3cos u)cos x = 5cos u−3 ⇒cos x = ((5cos u−3)/(5−3cos u)) ⇒2cos^2 ((x/2))=((5cos u−3)/(5−3cos u)) +((5−3cos u)/(5−3cos u)) ⇒2cos^2 ((x/2))= ((2cos u+2)/(5−3cos u)) Thank you...much appreciated ⇒(1/2)sec^2 ((x/2))= ((5−3cos u)/(2cos u+2)) ⇒sec^2 ((x/2)) = ((5−3cos u)/(cos u+1)) ⇒tan^2 ((x/2))= ((5−3cos u)/(cos u+1)) −((cos u+1)/(cos u+1)) ⇒tan^2 ((x/2))=((4−4cos u)/(1+cos u)) = ((4(1−(1−2sin^2 ((u/2))))/(2cos^2 ((u/2)))) ⇒tan^2 ((x/2))=4 tan^2 ((u/2)) ⇒tan ((x/2))= 2tan ((u/2)) ⇒(u/2) = tan^(−1) ((1/2)tan (x/2)) ⇒u = 2tan^(−1) ((1/2)tan (x/2)) Thank you
$$\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\right)\:=\:\mathrm{u} \\ $$$$\Rightarrow\frac{\mathrm{3}+\mathrm{5cos}\:\mathrm{x}}{\mathrm{5}+\mathrm{3cos}\:\mathrm{x}}\:=\:\mathrm{cos}\:\mathrm{u} \\ $$$$\Rightarrow\mathrm{3}+\mathrm{5cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}+\mathrm{3cos}\:\mathrm{u}\:\mathrm{cos}\:\mathrm{x} \\ $$$$\Rightarrow\left(\mathrm{5}−\mathrm{3cos}\:\mathrm{u}\right)\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{5cos}\:\mathrm{u}−\mathrm{3} \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{5cos}\:\mathrm{u}−\mathrm{3}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}} \\ $$$$\Rightarrow\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{5cos}\:\mathrm{u}−\mathrm{3}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}\:+\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}} \\ $$$$\Rightarrow\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\frac{\mathrm{2cos}\:\mathrm{u}+\mathrm{2}}{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}} \\ $$$${Thank}\:{you}…{much}\:{appreciated} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{2cos}\:\mathrm{u}+\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sec}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:=\:\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{cos}\:\mathrm{u}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\frac{\mathrm{5}−\mathrm{3cos}\:\mathrm{u}}{\mathrm{cos}\:\mathrm{u}+\mathrm{1}}\:−\frac{\mathrm{cos}\:\mathrm{u}+\mathrm{1}}{\mathrm{cos}\:\mathrm{u}+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{4}−\mathrm{4cos}\:\mathrm{u}}{\mathrm{1}+\mathrm{cos}\:\mathrm{u}}\:=\:\frac{\mathrm{4}\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \left(\frac{\mathrm{u}}{\mathrm{2}}\right)\right)\right.}{\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{u}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{4}\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{u}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{2tan}\:\left(\frac{\mathrm{u}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\frac{\mathrm{u}}{\mathrm{2}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{u}\:=\:\mathrm{2tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$${Thank}\:{you} \\ $$
Commented by meetbhavsar25 last updated on 07/May/21
I found a simpler method.... cos^(−1) {((3+5cos x)/(5+3cos x))} = y ((cos y)/1) = ((3+5cos x)/(5+3cos x)) ((cos y+1)/(cos y−1)) = ((8+8cos x)/(2cos x−2)) (using componendo−dividendo) ((2cos^2 (y/2))/(−2sin^2 (y/2))) = ((8×2cos^2 (x/2))/(−2×2sin^2 (x/2))) cot^2 (y/2) = 4 × cot^2 (x/2) tan^2 (x/2) = 4 × tan^2 (y/2) tan (x/2) = 2 × tan (y/2) tan (y/2) = (1/2) tan (x/2) (y/2) = tan^(−1) {(1/2)tan (x/2)} y = 2tan^(−1) {(1/2)tan (x/2)}
$${I}\:{found}\:{a}\:{simpler}\:{method}…. \\ $$$$\mathrm{cos}^{−\mathrm{1}} \left\{\frac{\mathrm{3}+\mathrm{5cos}\:{x}}{\mathrm{5}+\mathrm{3cos}\:{x}}\right\}\:=\:{y} \\ $$$$\frac{\mathrm{cos}\:{y}}{\mathrm{1}}\:=\:\frac{\mathrm{3}+\mathrm{5cos}\:{x}}{\mathrm{5}+\mathrm{3cos}\:{x}} \\ $$$$\frac{\mathrm{cos}\:{y}+\mathrm{1}}{\mathrm{cos}\:{y}−\mathrm{1}}\:=\:\frac{\mathrm{8}+\mathrm{8cos}\:{x}}{\mathrm{2cos}\:{x}−\mathrm{2}}\:\left({using}\:{componendo}−{dividendo}\right) \\ $$$$\frac{\mathrm{2cos}\:^{\mathrm{2}} \frac{{y}}{\mathrm{2}}}{−\mathrm{2sin}\:^{\mathrm{2}} \frac{{y}}{\mathrm{2}}}\:=\:\frac{\mathrm{8}×\mathrm{2cos}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{−\mathrm{2}×\mathrm{2sin}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$\mathrm{cot}\:^{\mathrm{2}} \frac{{y}}{\mathrm{2}}\:=\:\mathrm{4}\:×\:\mathrm{cot}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:=\:\mathrm{4}\:×\:\mathrm{tan}\:^{\mathrm{2}} \frac{{y}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:=\:\mathrm{2}\:×\:\mathrm{tan}\:\frac{{y}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\frac{{y}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$$\frac{{y}}{\mathrm{2}}\:=\:\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right\} \\ $$$${y}\:=\:\mathrm{2tan}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right\} \\ $$

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