Cambridge Year 12 3U Question - Inverse Trig Functions (1 Viewer)

HonestMan · Nov 15, 2015

I have been unable to find solutions to this question. It is one of the extension questions in exercise 1B on inverse trigonometric functions.

Question 22: Given that a² + b² = 1, prove that the expression tan⁻¹(ax/(1-bx)) - tan^-1((x-b)/a) is independent of x.

kawaiipotato · Nov 15, 2015

Let y =tan−1(ax/(1-bx)) - tan-1((x-b)/a) and then take the tangent of both sides
tany = tan( tan−1(ax/(1-bx)) - tan-1((x-b)/a))
Expanding the RHS using the formula:
tan (a-b) = (tana - tanb)/(1+tana tanb) to show tany is independent of x and so y is independent of x

InteGrand · Nov 15, 2015

HonestMan said:
I have been unable to find solutions to this question. It is one of the extension questions in exercise 1B on inverse trigonometric functions.

Question 22: Given that a² + b² = 1, prove that the expression tan⁻¹(ax/(1-bx)) - tan^-1((x-b)/a) is independent of x.

$Using the identity $\tan \left(A-B \right)=\frac{\tan A - \tan B}{1+\tan A \tan B}$ and calling the expression $X$, we have$

$\begin{align*}\tan X &= \frac{\frac{ax}{1-bx}-\frac{x-b}{a}}{1+\frac{ax}{1-bx}\cdot \frac{x-b}{a}} \\ &= \frac{a^2x - (1-bx)(x-b)}{a(1-bx) + (ax)(x-b)}\quad (\text{multiplying numerator and denominator by }a(1-bx)) \\ &= \frac{a^2 x - x + b + bx^2 - b^2 x}{a-abx + ax^2 - abx} \\ &= \frac{bx^2 + \left( a^2 -b^2 - 1\right) x + b}{ax^2 - 2abx + a} \\ &= \frac{bx^2 + \left(1-b^2 - b^2 -1 \right)x+b}{a\left(x^2 - 2bx +1 \right)} \quad (\text{as }a^2 = 1-b^2 \text{ in the numerator}) \\ &= \frac{bx^2 + \left(-2b^2\right)x+b}{a\left(x^2 - 2bx +1 \right)} \\ &= \frac{b\left(x^2 -2bx+1 \right)}{a\left(x^2 - 2bx +1 \right)} \\ & = \frac{b}{a}. \end{align*}$

$Hence $\tan X$ is a constant $\frac{b}{a}$, so $\tan X$ is a constant, i.e. independent of $x$. Note that $X$ will actually depend on $x$, as it can differ in multiples of $\pi$ depending on which period of the tan function $X$ lies in. Since the difference of the two arctans can in general lie in $(-\pi,\pi)$, $X$ will be able to be in this range and differ by $\pi$ within this range for given $a$ and $b$, depending on $x$ \Big{(}due to the general solution formula $X = \tan^{-1}\left( \frac{b}{a}\right)+n\pi$\Big{)}.$

$For example, take $a=b=\frac{1}{\sqrt{2}}$. Then if we take $x=3$, we get $X=-\frac{3 \pi}{4}$. If we now take $x=0.3$, we get $X=\frac{\pi}{4}$.$

Cambridge Year 12 3U Question - Inverse Trig Functions (1 Viewer)

HonestMan

New Member

kawaiipotato

Well-Known Member

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)