Mt12 - Small angle approximations

Mt12 - Small angle approximations

Understand the standard small angle approximations in Mt12 and use them with their radian conditions intact. You will simplify expressions, explain which powers matter and judge the error in a geometric measurement model.

Why small angles behave simply

Can a turning angle be used as a sideways distance? A ray hits a screen 2 m2\,\mathrm m from its turning point. If it turns through 0.080.08 radians from the perpendicular, the exact sideways displacement is 2tan⁡(0.08) m2\tan(0.08)\,\mathrm m. The simple estimate 2(0.08)=0.16 m2(0.08)=0.16\,\mathrm m is tempting. We will decide how much accuracy it costs.

You already know

In a right-angled triangle, tangent is opposite divided by adjacent. On a unit circle, the coordinates are (cos⁡θ,sin⁡θ)(\cos\theta,\sin\theta); radian measure gives sector area θ/2\theta/2.

At 0.10.1 radians, sine is 0.0998334…0.0998334\ldots and tangent is 0.100335…0.100335\ldots. The angle sits between them. This has a geometric explanation: for 0<θ<π/20<\theta<\pi/2, an inscribed triangle, the unit-circle sector and an enclosing tangent triangle have increasing areas. Therefore

12sin⁡θ<12θ<12tan⁡θ⟹cos⁡θ<sin⁡θθ<1\frac12\sin\theta<\frac12\theta<\frac12\tan\theta \quad\Longrightarrow\quad \cos\theta<\frac{\sin\theta}{\theta}<1

As the angle approaches zero, both bounds on the ratio approach 11. Also tan⁡θ/θ=(sin⁡θ/θ)/cos⁡θ\tan\theta/\theta=(\sin\theta/\theta)/\cos\theta approaches 11. Odd symmetry gives the same limiting ratios for negative angles. The unit-circle identity explains cosine:

1−cos⁡θ=sin⁡2θ1+cos⁡θ≈θ221-\cos\theta =\frac{\sin^2\theta}{1+\cos\theta} \approx\frac{\theta^2}{2}

The exact rearrangement avoids subtracting two nearly equal rounded decimals.

Small angle approximation

A local replacement of a trigonometric function by a simpler expression, valid when its argument is sufficiently close to zero in radians. It is an approximation, rather than an identity at every input.

The three standard replacements follow from those ratios and the cosine rearrangement.

Small angle approximations

For ∣θ∣|\theta| small, with θ\theta in radians,

sin⁡θ≈θcos⁡θ≈1−θ22tan⁡θ≈θ\sin\theta\approx\theta \qquad \cos\theta\approx1-\frac{\theta^2}{2} \qquad \tan\theta\approx\theta

In the formula sheet.

At 5∘=π/365^\circ=\pi/36 radians, sine and tangent are close to 0.08730.0873, not to 55. The radian condition does real work. Another useful surprise is that doubling a small angle makes 1−cos⁡θ1-\cos\theta roughly four times larger: the exact differences at 0.050.05 and 0.10.1 radians are 0.00124974…0.00124974\ldots and 0.00499583…0.00499583\ldots. A squared argument explains that factor of four.

Key idea

Check the units and the whole argument before using a small angle formula. “Small” is relative to the accuracy needed; there is no single cutoff that suits every expression.

Explain the geometry behind the condition, rather than just remembering a calculator setting.

Substitution and retained powers

A multiplier inside a trigonometric function belongs to its argument; carry it into the approximation before simplifying.

Thus sin⁡(3x)≈3x\sin(3x)\approx3x, tan⁡(2x)≈2x\tan(2x)\approx2x and cos⁡(3x)≈1−9x2/2\cos(3x)\approx1-9x^2/2. These need the actual arguments to be small. A denominator also needs to remain nonzero in the expression you started with.

Worked example 1: a quotient of small values

Question. Find the leading approximation to tan⁡(2x)sin⁡(3x)\dfrac{\tan(2x)}{\sin(3x)} for small nonzero xx in radians, and state the argument conditions.

Plan. Replace both functions using their complete arguments, then cancel the nonzero factor.

tan⁡(2x)≈2xtangent’s argumentsin⁡(3x)≈3xsine’s argumenttan⁡(2x)sin⁡(3x)≈2x3x=23x≠0\begin{aligned} \tan(2x)&\approx2x && \text{tangent's argument}\\ \sin(3x)&\approx3x && \text{sine's argument}\\ \frac{\tan(2x)}{\sin(3x)}&\approx\frac{2x}{3x}=\frac23 && x\ne0 \end{aligned}

Why this step? Both numerator and denominator approach zero; replacing only the numerator would leave the comparison unfinished.

Answer. The leading approximation is 2/32/3, requiring ∣2x∣|2x| and ∣3x∣|3x| small in radians and x≠0x\ne0.

Check. At x=0.01x=0.01, the exact quotient is 0.666856…0.666856\ldots. At zero the starting quotient is undefined, even though the approximation is constant.

Change only the sine's multiplier from 33 to 44. The limiting constant changes, and the smallness condition changes with it.

Worked example 2: a different denominator scale

Question. Find the leading approximation to tan⁡(2x)sin⁡(4x)\dfrac{\tan(2x)}{\sin(4x)} for small nonzero xx in radians.

Plan. Approximate the whole arguments, keeping the zero-denominator exclusion.

tan⁡(2x)sin⁡(4x)≈2x4xuse both approximations=12x≠0\begin{aligned} \frac{\tan(2x)}{\sin(4x)} &\approx\frac{2x}{4x} && \text{use both approximations}\\ &=\frac12 && x\ne0 \end{aligned}

Why this step? The multiplier belongs inside the sine; sin⁡(4x)≈x\sin(4x)\approx x would miss a factor of 44.

Answer. The leading approximation is 1/21/2, with ∣4x∣|4x| small in radians. This also makes ∣2x∣|2x| small.

Check. At x=0.01x=0.01, the exact quotient is 0.500200…0.500200\ldots, close to 1/21/2.

Retaining powers consistently matters when a constant cancels. For small nonzero xx,

1−cos⁡xsin⁡x≈x2/2x=x2\frac{1-\cos x}{\sin x}\approx\frac{x^2/2}{x}=\frac x2

Using cos⁡x≈1\cos x\approx1 here would destroy the first nonzero term. Conversely, if terms of degree 33 and above are to be neglected,

sin⁡(3x)1+cos⁡x≈3x2(1−x2/4)−1≈3x2(1+x2/4)≈3x2\frac{\sin(3x)}{1+\cos x} \approx\frac{3x}{2}(1-x^2/4)^{-1} \approx\frac{3x}{2}(1+x^2/4) \approx\frac{3x}{2}

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