Inverse Trignometry

[l9876]

Please help me solve these :

1) tan-¹(1-x²/ 2x) + cos-¹ (1-x²/2x)= pi /2
2) 2tan-¹ 1/5 + sec-¹ 5 sqrt 2/7 + 2tan-¹ 1/8 = pi/4
3) tan -¹(3sin2?/5+3cos2?) + tan-¹ (1/4tan?)=?
4) 2tan-¹( Sq.root( a-b )/ a+b tan ?/2) = cos-¹ ( acos? + b / a + bcos? )
Thank you

 

[Deleted member 4993]

Please help me solve these :

1) tan-¹(1-x²/ 2x) + cos-¹ (1-x²/2x)= pi /2
2) 2tan-¹ 1/5 + sec-¹ 5 sqrt 2/7 + 2tan-¹ 1/8 = pi/4
3) tan -¹(3sin2?/5+3cos2?) + tan-¹ (1/4tan?)=?
4) 2tan-¹( Sq.root( a-b )/ a+b tan ?/2) = cos-¹ ( acos? + b / a + bcos? )
Thank you

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[galactus]

1) tan-¹((1-x²)/ 2x) + cos-¹ ((1-x²)/2x)= pi /2

Try solving \(\displaystyle \frac{1-x^{2}}{2x}=0\) for x.

 

[Deleted member 4993]

Some more hints for #1.

\(\displaystyle let \ \ \cos^{-1}\left (\frac{1-x^2}{2x}\right ) \ \ = \ \ \theta\)

\(\displaystyle \cos(\theta) \ \ = \ \ \frac{1-x^2}{2x}\)

then we have

\(\displaystyle \tan^{-1}[\cos(\theta)] \ \ + \ \ \theta \ \ = \ \ \frac{\pi}{2}\)

\(\displaystyle \tan^{-1}[\cos(\theta)] \ \ = \ \ \frac{\pi}{2} \ \ - \ \ \theta\)

\(\displaystyle \tan(\tan^{-1}[\cos(\theta)]) \ \ = \ \ \tan(\frac{\pi}{2} \ \ - \ \ \theta)\)

\(\displaystyle \cos(\theta) \ \ = \ \ \cot(\theta)\)

\(\displaystyle \cos(\theta) \cdot [1 \ \ - \ \ cosec(\theta)] \ \ = \ \ 0\)

Now continue....

 

[BigGlenntheHeavy]

\(\displaystyle 2) \ Is \ this \ (below) \ a \ true \ statement?, \ to \ wit:\)

\(\displaystyle 2tan^{-1}(1/5)+sec^{-1}(5\sqrt2/7)+2tan^{-1}(1/8) \ = \ \pi/4\)

\(\displaystyle 2tan^{-1}(1/5)+cos^{-1}(7/5\sqrt2)+2tan^{-1}(1/8) \ = \ \pi/4\)

\(\displaystyle Now \ (my \ way), \ tan^{-1}(1/5)+\frac{cos^{-1}(7/5\sqrt2)}{2}+tan^{-1}(1/8) \ = \ \pi/8\)

\(\displaystyle Hence, \ we \ let \ \alpha \ = \ tan^{-1}(1/5) \ \implies \ tan(\alpha) \ = \ (1/5)\)

\(\displaystyle And \ let \ \beta \ = \ tan^{-1}(1/8) \ \implies \ tan(\beta) \ = \ (1/8)\)

\(\displaystyle and \ \gamma \ = \ cos^{-1}(7/5\sqrt2) \ \implies \ cos(\gamma) \ = \ (7/5\sqrt2)\)

\(\displaystyle which \ implies \ tan(\gamma) \ = \ (1/7), \ tan(\gamma/2) \ = \ 5\sqrt2-7\)

\(\displaystyle Ergo, \ \alpha +\beta \ = \ \pi/8-\gamma/2\)

\(\displaystyle tan(\alpha+\beta) \ = \ tan(\pi/8-\gamma/2)\)

\(\displaystyle The \ rest \ is \ grunt \ work \ (mostly \ algebra) \ and \ we \ get:\)

\(\displaystyle 1/3 \ = \ 1/3, \ hence \ the \ original \ statement \ is \ true, \ QED\)

\(\displaystyle Is \ there \ an \ easier \ way \ to \ prove \ the \ above \ equation? \ If \ so \ I \ would \ be \ interested.\)

 

[BigGlenntheHeavy]

\(\displaystyle Another, \ easier \ way, \ I \ think, \ to \ wit:\)

\(\displaystyle Prove \ that \ 2tan^{-1}(1/5)+sec^{-1}(5\sqrt2/7)+2tan^{-1}(1/8) \ = \ \pi/4 \ rings \ true.\)

\(\displaystyle This \ can \ be \ rewritten \ as: \ 2tan^{-1}(1/5)+2tan^{-1}(1/8)+tan^{-1}(1/7) \ = \ \pi/4.\)

\(\displaystyle Now, \ let \ 2tan^{-1}(1/5) \ = \ \alpha, \ \implies \ tan^{-1}(1/5) \ = \ \alpha/2, \ \implies \ tan(\alpha/2)=1/5, \ \implies \ tan(\alpha)=5/12.\)

\(\displaystyle Also, \ let \ 2tan^{-1}(1/8) \ = \ \beta, \ \implies \ tan^{-1}(1/8)=\beta/2, \ \implies \ tan(\beta/2)=1/8, \ \implies \ tan(\beta)=16/63.\)

\(\displaystyle and \ finally, \ let \ tan^{-1}(1/7) \ = \ \gamma, \ \implies \ tan(\gamma) \ = \ 1/7.\)

\(\displaystyle Hence, \ \alpha+\beta+\gamma \ = \ \pi/4, \ implies \ \alpha+\beta \ = \ \pi/4-\gamma.\)

\(\displaystyle Taking \ the \ tangent \ of \ both \ sides, \ we \ get: \ tan(\alpha+\beta) \ = \ tan(\pi/4-\gamma).\)

\(\displaystyle With \ a \ little \ algebra \ and \ appropriate \ identities, \ this \ reduces \ to \ 3/4 \ = \ 3/4, \ QED.\)

\(\displaystyle Hence, \ the \ equation \ (above) \ is \ true.\)

\(\displaystyle or \ tan[(\alpha+\beta)+\gamma] \ = \ tan(\pi/4), \ \implies \ 1 \ = \ 1.\)

\(\displaystyle or \ for \ the \ purists, \ let \ \pi/4 \ = \ \theta.\)

\(\displaystyle then \ \alpha+\beta+\gamma \ = \ \theta.\)

\(\displaystyle Ergo, \ tan[(\alpha+\beta)+\gamma] \ = \ tan(\theta).\)

\(\displaystyle tan(\theta) \ = \ 1, \ \implies \ \theta \ = \ \pi/4, \ which \ is \ what \ we \ wanted, \ QED.\)

\(\displaystyle Note: \ To \ make \ this \ problem \ more \ interesting, \ it \ should \ have \ been \ written \ the \ following:\)

\(\displaystyle 2tan^{-1}(1/5)+sec^{-1}(5\sqrt2/7)+2tan^{-1}(1/8) \ = \ \theta, \ solve \ for \ \theta, \ (without \ a \ calculator) \ 0 \ \le \ \theta \ \le \ \pi/2.\)