Inverse Trigonometric Functions Test - 53

Since $\displaystyle -\frac { \pi  }{ 2 } \le \sin ^{ -1 }{ x } \le \frac { \pi  }{ 2 }$

Let $f(x)=\sin ^{ -1 }{ x } -\tan ^{ -1 }{ x }$

We have $\sum _{ i=1 }^{ 2n }{ \sin ^{ -1 }{ { x }_{ i } }  } =n\pi$

Let  $y = (\sin^{-1}x)^{3}+(\cos^{-1}x)^{3} = (\sin^{-1}x+\cos^{-1}x)[(\sin^{-1}x)^2+(\cos^{-1}x)^2-\sin^{-1}x.\cos^{-1}x]$
$\Rightarrow y = \cfrac{\pi}{2}[(\sin^{1-}x+\cos^{-1}x)^2-3\sin^{-1}x.\cos^{-1}x]$
$\Rightarrow y = \cfrac{\pi}{2}\left[\cfrac{\pi^2}{4}-3\sin^{-1}x\left(\cfrac{\pi}{2}-\sin^{-1}x\right)\right ]$
$\Rightarrow y = \cfrac{\pi}{2}\left[\cfrac{\pi^2}{16}+3\left(\cfrac{\pi}{4}-\sin^{-1}x\right)^2\right ]$
Now we know $-\cfrac{\pi}{2} \leq \sin^{-1}x \leq \cfrac{\pi}{2}$
Thus Range of $y$ is $\left[\cfrac{\pi^3}{32}, \cfrac{7\pi^3}{8}\right]$
Thus given equation to have solution $a \in \left[\cfrac{1}{32}, \cfrac{7}{8}\right]$

$\left| y \right| =\sin  x,\quad y=\cos ^{ -1 } \left( \cos  x \right) ,\quad -2\pi \leq x\leq 2\pi$
$\left| y \right| =\sin  x\quad ----\boxed { 1 } \\ y=\cos ^{ -1 } \left( \cos  x \right) ,\quad -2\pi \leq x\leq 2\pi \quad ----\boxed { 2 } \\ \rightarrow y=x\quad \quad \quad \quad when\quad 0\leq x\leq \pi \\ \rightarrow y=2\pi -x\quad \quad \quad \quad \quad \quad \pi \leq x\leq 2\pi \\ \rightarrow y=-2\pi -x\quad \quad \quad \quad \quad -2\pi \leq x\leq -\pi \\ \rightarrow y=-x\quad \quad \quad \quad \quad \quad \quad \quad -\pi\leq x\leq 0$
from (1) and (2)
three different cases are $|x| = \sin{x}$, $|2\pi-x| = \sin{x}$ and $|2\pi+x| = \sin{x}$

For the above simultaneous equations, we get exactly 3 solutions.
For $x=0$ we get $y=0$ for both the equations,
For $x=k(2\pi)$ where $k$ is an integer, we get $y=0$ in both the cases.
Since $x\epsilon[-2\pi,2\pi]$ we get
$x=-2\pi$ and $x=2\pi$.
Hence 3 solutions,
at $x=0$
$x=-2\pi$ and
$x=2\pi$

Hence, no of real solutions is 3

For the domain of the following function,
$1-x^{2}\geq 0$ and $1-\frac{x^{2}}{4}\geq0$
Therefore
$1\geq x^2$ and $4\geq x^2$
$|x|\leq 1$ and $|x|\leq 2$
Hence $x\epsilon[-1,1]$
$\displaystyle \:\cos ^{-1}\left \{ \frac{1}{2}x^{2}+\sqrt{1-x^{2}.}\sqrt{1-\frac{x^{2}}{4}} \right \}= \cos ^{-1}\frac{x}{2}-\cos ^{-1}x$
L.H.S is always positive.
$\Rightarrow \cos ^{-1}\frac{x}{2}-\cos ^{-1}x >0$
which is true for all $0\leq x\leq 1$
$\therefore$ given equation holds for $0\leq x\leq 1$

We have $\displaystyle \sin ^{ -1 }{ x } +\sin ^{ -1 }{ y } +\sin ^{ -1 }{ z } =\frac { 3\pi  }{ 2 }$

$sin^{ -1 }cos\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 }  \right) =cot\left( cot^{ -1 }\left( \cfrac { 2-18x }{ 9x }  \right)  \right) +\cfrac { \pi  }{ 2 } \\ \Rightarrow sin^{ -1 }sin\left( \cfrac { \pi  }{ 2 } -\left( \cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 }  \right)  \right) =\cfrac { 2-18x }{ 9x } +\cfrac { \pi  }{ 2 } \\ \Rightarrow \cfrac { \pi  }{ 2 } -\cfrac { 2x^{ 2 }+10x+4 }{ x^{ 2 }+5x+3 } =\cfrac { 2-18x }{ 9x } +\cfrac { \pi  }{ 2 } \\ \Rightarrow -18x^{ 3 }-90x^{ 2 }-36x=2x^{ 2 }+10x+6-18x^{ 3 }-90x-54x\\ \Rightarrow 2x^{ 2 }-8x+6=0\Rightarrow x^{ 2 }-4+3=0\Rightarrow \left( x-1 \right) \left( x-3 \right) =0$
Therefore product of roots is $1\times 3=3$

$ax^{2}+sin^{-1}((x-1)^{2}+1)+cos^{-1}((x-1)^{2}+1)=0$
$ax^{2}+\frac{\pi}{2}=0$
$x^{2}=\frac{-\frac{\pi}{2}}{a}$
$x^{2}=\frac{-\pi}{2a}$ ...(i)
And
$-1\leq(x-1)^{2}+1\leq1$
$-2\leq(x-1)^{2}\leq0$
$(x-1)^{2}\leq0$
There is only one possibility
$(x-1)^{2}=0$
$x=1$
Substituting in 1, we get
$a=-\frac{\pi}{2}$

$sin^{-1}(x)+sin^{-1}(y)=\pi-sin^{-1}(z)$
$sin^{-1}(x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}})=\pi-sin^{-1}(z)$
$x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}}=z$
$x^{2}(1-y^{2})+y^{2}(1-x^{2})+2xy\sqrt{(1-x^{2})(1-y^{2})}=z^2$
$x^{2}+y^{2}-2x^{2}y^{2}+2xy\sqrt{(1-x^{2})(1-y^{2})}=z^{2}$
$x^{2}+y^{2}-z^{2}=2x^{2}y^{2}-2xy\sqrt{(1-x^{2})(1-y^{2})}$
Squaring and simplifying, we get
$x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+z^2y^2+x^2z^2)$