Small-angle series in a triangle

Right triangle OBC and angle ACBO; C; A; B; 1; x; 1; θ. OC=OA=1; OB=x>1; θ=angle ACB; right angle COB; example θ=0.2 only.Right triangle OBC and angle ACBOCAB1x1θ
Question diagram 1
Right triangle OBC and angle ACBO; C; A; B; 1; x; 1; θ. OC=OA=1; OB=x>1; θ=angle ACB; right angle COB; example θ=0.2 only.Right triangle OBC and angle ACBOCAB1x1θ

Given right triangle OBC: OC=1, OA=1, OB=x>1, A lies on OB, and angle ACB is θ.

(i) Show that the first two non-zero terms of the Maclaurin series for tan⁡θ\tan\theta is given by θ+13θ3\theta+\frac13\theta^3. You may use the standard results given in the List of Formulae (MF26). marks

In the right-angled triangle OBC shown above, point A lies on OB such that OA = 1, OB = xx, where x>1x>1 and OC = 1. It is given that angle COB is π2\frac\pi2 radians and that angle ACB is θ\theta radians (see diagram).

(ii) Show that AB=2tan⁡θ1−tan⁡θ\mathrm{AB}=\frac{2\tan\theta}{1-\tan\theta}. marks

(iii) Given that θ\theta is a sufficiently small angle, show that AB≈aθ+bθ2+cθ3\mathrm{AB}\approx a\theta+b\theta^2+c\theta^3 for exact real constants a,b,ca,b,c to be determined. marks

[2022/ASRJC/Prelim/I/4]

Total marks: 7

Solution

(i)

Series of a quotient

Show that the first two non-zero terms of the Maclaurin series for tan⁡θ\tan\theta are θ+13θ3\theta+\frac13\theta^3.

Explore this idea

Hint

  1. tan⁡θ=sin⁡θ (cos⁡θ)−1\tan\theta=\sin\theta\,(\cos\theta)^{-1}.

  2. Expand each and multiply.

Answer · (i)

tan⁡θ=θ+θ3/3+O(θ5)\tan\theta=\theta+\theta^3/3+O(\theta^5).

Explore the idea

tan θ and its series

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 3
Test value xx = 0.60
Reset the controls
Function
0.68414
P₃(x)
0.672
Error
-0.01214

Inside the stated open interval. A finite truncation can still have a noticeable error.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • Stated open interval. Endpoint convergence requires a separate check
Key idea

Only odd powers appear, because tan⁡θ\tan\theta is odd.

Ideas to try
Raise the degree and slide θ\theta towards π2\frac\pi2.
(ii)

Trigonometry in a triangle

In the right-angled triangle OBCOBC with OA=OC=1OA=OC=1, OB=x>1OB=x>1, ∠COB=π2\angle COB=\frac\pi2, ∠ACB=θ\angle ACB=\theta, show that AB=2tan⁡θ1−tan⁡θAB=\dfrac{2\tan\theta}{1-\tan\theta}.

Hint

  1. ∠OCA=π4\angle OCA=\frac\pi4 because OA=OCOA=OC and the angle at OO is π2\frac\pi2.

  2. So ∠OCB=π4+θ\angle OCB=\frac\pi4+\theta and x=tan⁡(π4+θ)x=\tan\left(\frac\pi4+\theta\right).

Answer · (ii)

AB=2tan⁡θ/(1−tan⁡θ)AB=2\tan\theta/(1-\tan\theta).

(iii)

Small-angle approximation

For small θ\theta, show AB≈aθ+bθ2+cθ3AB\approx a\theta+b\theta^2+c\theta^3 and find aa, bb, cc.

Explore this idea

Hint

  1. Substitute tan⁡θ≈θ+θ33\tan\theta\approx\theta+\frac{\theta^3}3.

  2. Expand 11−t=1+t+t2+⋯\frac1{1-t}=1+t+t^2+\cdots.

Answer · (iii)

AB≈2θ+2θ2+8θ3/3AB\approx2\theta+2\theta^2+8\theta^3/3; a=2,b=2,c=8/3a=2,b=2,c=8/3.

Small angles in radians; geometry 0<θ<π/40<\theta<\pi/4.

Explore the idea

AB and its small-angle series

Compare the function with a finite polynomial. Change the highest included power and test x, then use the error plot to see how the approximation changes at that x.

Highest included powerxⁿ, n = 3
Test value xx = 0.30
Try an example
Function
0.89577
P₃(x)
0.852
Error
-0.04377

Inside the stated open interval. A finite truncation can still have a noticeable error.

Teal marks the function, amber marks the polynomial, and the red segment joins their values at the selected x.

Does keeping more powers help at this x?

Highest included power n runs horizontally; signed error Pₙ(x) − f(x) runs vertically. Zero means agreement at this x. A missing power leaves the polynomial unchanged; individual steps need not improve the error.

  • Function
  • Series truncated at the chosen power
  • Stated open interval. Endpoint convergence requires a separate check
Key idea

Near 0 three terms are accurate; near θ=π4\theta=\frac\pi4 (where x→∞x\to\infty) the series fails.

Ideas to try
Raise the degree from 1 to 3: how much closer is the polynomial at θ=0.5\theta=0.5?