Let us break the rules ..
Let the co domain of sin-1x be (pi2 ,3pi2)
tan-1x be (pi2 ,3pi2)
cos-1x be (pi, 2pi)
cot-1x be (pi,2pi)
1)Then prove that f(x) = 3sin-1x - 2cos-1x is an odd function.
2) Find the minimum value of (sin-1x)3 + (cos-1x)3
71This is the one thing that has haunted me always. Now that you have posted it, I'll surely learn something from this post!
30According to mathematician Shubhodip's new rules [3],
sin-1x + cos-1x = 5 pi2
for question 1,
f(x)= 3\sin^{-1}x - 2 \cos^{-1} x
= 3\sin^{-1}x - 2\left(\frac{5\pi}{2} - \sin^{-1}x\right)
= 5\left( \sin^{-1}x - \pi \right)
It is easy to note sin-1x is pi at x=0, taking (0,pi) as origin, function is clearly odd.
30shubhodip I'll try the next one too.. don't post the solution any time soon.
302. (\sin^{-1} x)^3 + (\cos^{-1})^3 = \left(\sin^{-1} x\right) ^3 + \left(\frac{5\pi }{2} - \sin^{-1} x\right)^3
= \left( \sin^{-1}x\right)^3 + \left( \frac{5\pi }{2}\right)^3 - \left( \sin^{-1}x\right)^3 - \frac{15\pi }{2}\sin^{-1}x\left(\frac{5\pi }{2} - \sin^{-1}x \right)
= \frac{125\pi ^3}{8} - \frac{75\pi^2 }{4}\sin^{-1}x + \frac{15\pi }{2}\sin^{-1}x^2
Since \frac{\pi }{2}\leq \sin^{-1}x \leq \frac{3\pi }{2} ,
= \frac{125\pi ^3}{8} - \frac{75\pi^3 }{8} + \frac{15\pi^3 }{8} = \boxed{\frac{65\pi ^3}{8}}
@shubhodip - can you post your solution now? [7]
49first of all sin-1x and cos-1x are positive numbers as described by the sum!
thus,
(sin-1x)3+(cos-1x)3 ≥ 2√(sin-1x)3(cos-1x)3 (simple AM GM)
min value when sin-1x=cos-1x
and min value = 2(sin-1x)3 for that x!
sin and cos become equal at every nπ+π4
so, in given ranges they become equal only at x=5Ï€/4
so, min value=2.125Ï€3/64=12532Ï€3
koi calculation mistakes ho to shama kare!
49yaar answer mil gaya subhodip se! :D
JEE k baad pehli baar confidence re building! :)
JEE me maine 80 marks ka silly mistake kar aaya...and Seoni said "No mistake is a silly mistake!!" :'( :'( :((
30My approach to the 2nd one was wrong I guess. Will keep this in mind! [1]
49perfecting Ashish's proof to first sum...
after f(x)=5[sin-1x - π]
f(-x)=5[sin-1(-x) - π]
let sinθ=x
π2≤θ≤3π2
sin(m-θ)=-x
-π2≥-θ≥-3π2
m-π2≥m-θ≥m-3π2
for m-θ to lie within principle value, m must be equal to 2π
thus, sin-1(-x)=2Ï€-sin-1x
so, f(-x) = 5[2π - sin-1x - π] = 5[π - sin-1x] = -5[sin-1x - π]
or, f(-x) = -f(x)
thus, f(x) is an odd function! [1]
