Question

Inverse trigo help! [12]

Answer 1

\[\large ~If~\cos^{-1} \lambda+\cos^{-1} u+\cos^{-1}v=3\pi,then~\lambda u+uv+v \lambda=?\]

Answer 2

--> arccos lambada=arccos u= arccos v=pi... kyunki pi se badi value le nhi sakta arccosx
ab solve krle..

Answer 3

epic :")

Answer 4

its not necessary that all of them are equal to pi :|
it can also be done in this arrangement:
\[\large \pi+\frac{\pi}{2}+\frac{3\pi}{2}\]

Answer 5

3pi/2 arccosx ke range me nhi hai. [0, pi]

Answer 6

hmm :|

Answer 7

sahi keh rha hun @ParthKohli ?

Answer 8

seems legit :-/

Answer 9

but fishy :P

Answer 10

@DLS Aap batayen ki mujhe trig nahi aati

Answer 11

"NEVER CALL PARTH KOHLI FOR TRIGONOMETRY"

Answer 12

:')

Answer 13

@DLS Do you require further help or are you fine?

Answer 14

I want to know how would we do this if we had another number on RHS lets say 6pi or 7pi

Answer 15

No solution. ^

Answer 16

seems legit :P

Answer 17

another question,what If i had arc tangent instead of arc cos x function?

Answer 18

No solution again.

Answer 19

why?

Answer 20

The range of artan(x) is -pi/2 < y < pi/2, i.e. -pi/2 an pi/2 are the asymptotes of arctan(x). For the sum to be 3pi would be impossible since the function gets really close to pi/2 as it goes to infinity, you could think of the values as always being less than pi/2 but very close to pi/2. Therefore, the sum arctan(u) + arctan(v) + arctan(lambada) could at most have a value of a little less than 3pi/2. Therefore, if it was equated with 3pi, there would be no solution.

Answer 21

Do you understand?

Answer 22

yes i do <3 thanks everyone!

Answer 23

and the last case :| what if I had arctan and pi ?

Answer 24

or arccos and pi