Simplify the given trigonometric expression
We need to solve
cos ( sin − 1 ( x cot ( tan − 1 ( cos ( sin − 1 x ) ) ) ) ) = k , 0 < ∣ x ∣ < 1 2 . \cos\left(\sin^{-1}\left(x\cot\left(\tan^{-1}\left(\cos(\sin^{-1}x)\right)\right)\right)\right)=k,
\quad 0<|x|<\frac{1}{\sqrt2}. cos ( sin − 1 ( x cot ( tan − 1 ( cos ( sin − 1 x ) ) ) ) ) = k , 0 < ∣ x ∣ < 2 1 . Let us simplify step by step.
Compute cos ( sin − 1 x ) \cos(\sin^{-1}x) cos ( sin − 1 x )
If θ = sin − 1 x \theta=\sin^{-1}x θ = sin − 1 x sin θ = x \sin\theta=x sin θ = x θ ∈ [ − π 2 , π 2 ] \theta\in\left[-\frac\pi2,\frac\pi2\right] θ ∈ [ − 2 π , 2 π ]
cos ( sin − 1 x ) = cos θ = 1 − x 2 . \cos(\sin^{-1}x)=\cos\theta=\sqrt{1-x^2}. cos ( sin − 1 x ) = cos θ = 1 − x 2 . So,
tan − 1 ( cos ( sin − 1 x ) ) = tan − 1 ( 1 − x 2 ) . \tan^{-1}(\cos(\sin^{-1}x))=\tan^{-1}(\sqrt{1-x^2}). tan − 1 ( cos ( sin − 1 x )) = tan − 1 ( 1 − x 2 ) .
Compute the cotangent part
Let
ϕ = tan − 1 ( 1 − x 2 ) . \phi=\tan^{-1}(\sqrt{1-x^2}). ϕ = tan − 1 ( 1 − x 2 ) . Then
tan ϕ = 1 − x 2 ⇒ cot ϕ = 1 1 − x 2 . \tan\phi=\sqrt{1-x^2}
\quad\Rightarrow\quad
\cot\phi=\frac{1}{\sqrt{1-x^2}}. tan ϕ = 1 − x 2 ⇒ cot ϕ = 1 − x 2 1 . Hence,
x cot ( tan − 1 ( cos ( sin − 1 x ) ) ) = x ⋅ 1 1 − x 2 = x 1 − x 2 . x\cot\left(\tan^{-1}(\cos(\sin^{-1}x))\right)
= x\cdot \frac{1}{\sqrt{1-x^2}}
=\frac{x}{\sqrt{1-x^2}}. x cot ( tan − 1 ( cos ( sin − 1 x )) ) = x ⋅ 1 − x 2 1 = 1 − x 2 x . Therefore the equation becomes
cos ( sin − 1 ( x 1 − x 2 ) ) = k . \cos\left(\sin^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right)=k. cos ( sin − 1 ( 1 − x 2 x ) ) = k .
Use the identity cos ( sin − 1 t ) = 1 − t 2 \cos(\sin^{-1}t)=\sqrt{1-t^2} cos ( sin − 1 t ) = 1 − t 2
For principal values,
cos ( sin − 1 t ) = 1 − t 2 . \cos(\sin^{-1}t)=\sqrt{1-t^2}. cos ( sin − 1 t ) = 1 − t 2 . Here
t = x 1 − x 2 . t=\frac{x}{\sqrt{1-x^2}}. t = 1 − x 2 x . So
k = 1 − x 2 1 − x 2 = 1 − 2 x 2 1 − x 2 . k=\sqrt{1-\frac{x^2}{1-x^2}}
=\sqrt{\frac{1-2x^2}{1-x^2}}. k = 1 − 1 − x 2 x 2 = 1 − x 2 1 − 2 x 2 . Because 0 < ∣ x ∣ < 1 2 0<|x|<\frac1{\sqrt2} 0 < ∣ x ∣ < 2 1 1 − 2 x 2 > 0 1-2x^2>0 1 − 2 x 2 > 0
Thus,
k 2 = 1 − 2 x 2 1 − x 2 . k^2=\frac{1-2x^2}{1-x^2}. k 2 = 1 − x 2 1 − 2 x 2 .
Solve for x 2 x^2 x 2
Let y = x 2 y=x^2 y = x 2
k 2 = 1 − 2 y 1 − y . k^2=\frac{1-2y}{1-y}. k 2 = 1 − y 1 − 2 y . So,
k 2 ( 1 − y ) = 1 − 2 y k^2(1-y)=1-2y k 2 ( 1 − y ) = 1 − 2 y k 2 − k 2 y = 1 − 2 y k^2-k^2y=1-2y k 2 − k 2 y = 1 − 2 y y ( 2 − k 2 ) = 1 − k 2 y(2-k^2)=1-k^2 y ( 2 − k 2 ) = 1 − k 2 y = 1 − k 2 2 − k 2 . y=\frac{1-k^2}{2-k^2}. y = 2 − k 2 1 − k 2 . Hence
x = ± 1 − k 2 2 − k 2 . x=\pm\sqrt{\frac{1-k^2}{2-k^2}}. x = ± 2 − k 2 1 − k 2 . Therefore the two solutions are
α = 1 − k 2 2 − k 2 , β = − 1 − k 2 2 − k 2 . \alpha=\sqrt{\frac{1-k^2}{2-k^2}},
\qquad
\beta=-\sqrt{\frac{1-k^2}{2-k^2}}. α = 2 − k 2 1 − k 2 , β = − 2 − k 2 1 − k 2 . (Their order may be interchanged.)
So,
α 2 = β 2 = 1 − k 2 2 − k 2 . \alpha^2=\beta^2=\frac{1-k^2}{2-k^2}. α 2 = β 2 = 2 − k 2 1 − k 2 .
Compute the two roots of the quadratic x 2 − b x − 5 = 0 x^2-bx-5=0 x 2 − b x − 5 = 0
The given roots are
1 α 2 + 1 β 2 and α β . \frac1{\alpha^2}+\frac1{\beta^2}
\quad\text{and}\quad
\frac\alpha\beta. α 2 1 + β 2 1 and β α . Since α 2 = β 2 \alpha^2=\beta^2 α 2 = β 2
1 α 2 + 1 β 2 = 2 α 2 = 2 ⋅ 2 − k 2 1 − k 2 . \frac1{\alpha^2}+\frac1{\beta^2}=\frac{2}{\alpha^2}
=2\cdot \frac{2-k^2}{1-k^2}. α 2 1 + β 2 1 = α 2 2 = 2 ⋅ 1 − k 2 2 − k 2 . Also, because β = − α \beta=-\alpha β = − α
α β = − 1. \frac\alpha\beta=-1. β α = − 1. Thus the roots are
2 ⋅ 2 − k 2 1 − k 2 and − 1. 2\cdot \frac{2-k^2}{1-k^2}
\quad\text{and}\quad -1. 2 ⋅ 1 − k 2 2 − k 2 and − 1.
Use product of roots
For
x 2 − b x − 5 = 0 , x^2-bx-5=0, x 2 − b x − 5 = 0 , the product of roots is − 5 -5 − 5
So,
( 2 ⋅ 2 − k 2 1 − k 2 ) ( − 1 ) = − 5. \left(2\cdot \frac{2-k^2}{1-k^2}\right)(-1)=-5. ( 2 ⋅ 1 − k 2 2 − k 2 ) ( − 1 ) = − 5. This gives
2 ⋅ 2 − k 2 1 − k 2 = 5. 2\cdot \frac{2-k^2}{1-k^2}=5. 2 ⋅ 1 − k 2 2 − k 2 = 5. So,
2 ( 2 − k 2 ) = 5 ( 1 − k 2 ) 2(2-k^2)=5(1-k^2) 2 ( 2 − k 2 ) = 5 ( 1 − k 2 ) 4 − 2 k 2 = 5 − 5 k 2 4-2k^2=5-5k^2 4 − 2 k 2 = 5 − 5 k 2 3 k 2 = 1 3k^2=1 3 k 2 = 1 k 2 = 1 3 . k^2=\frac13. k 2 = 3 1 .
Now compute b b b
Sum of roots of x 2 − b x − 5 = 0 x^2-bx-5=0 x 2 − b x − 5 = 0 b b b
Hence,
b = 2 ⋅ 2 − k 2 1 − k 2 − 1. b=2\cdot \frac{2-k^2}{1-k^2}-1. b = 2 ⋅ 1 − k 2 2 − k 2 − 1. Substitute k 2 = 1 3 k^2=\frac13 k 2 = 3 1
b = 2 ⋅ 2 − 1 3 1 − 1 3 − 1 = 2 ⋅ 5 3 2 3 − 1 = 2 ⋅ 5 2 − 1 = 5 − 1 = 4. b=2\cdot \frac{2-\frac13}{1-\frac13}-1
=2\cdot \frac{\frac53}{\frac23}-1
=2\cdot \frac52-1
=5-1=4. b = 2 ⋅ 1 − 3 1 2 − 3 1 − 1 = 2 ⋅ 3 2 3 5 − 1 = 2 ⋅ 2 5 − 1 = 5 − 1 = 4. Therefore,
b k 2 = 4 1 / 3 = 12. \frac{b}{k^2}=\frac{4}{1/3}=12. k 2 b = 1/3 4 = 12.
Final Answer
12 \boxed{12} 12