E2 - Small-angle approximations

E2 - Small-angle approximations

Replace trigonometric functions by simpler expressions when their arguments are small in radians. You will track transformed arguments, avoid losing terms through cancellation, and decide whether an approximation is accurate enough for a particular purpose.

Approximating near zero

Can you estimate sin⁡0.1\sin0.1, cos⁡0.1\cos0.1 and tan⁡0.1\tan0.1 without pressing a trigonometric key? In radians, all three have simple approximations near zero: sine and tangent are close to 0.10.1, while cosine is close to 0.9950.995.

You already know

Radian measure is arc length divided by radius. On the unit circle, sine is the vertical coordinate and cosine is the horizontal coordinate. These definitions are not supplied in the formula booklet.

For a small positive angle on the unit circle, the sine height, arc length and tangent height are close together. More precisely, sin⁡θ<θ<tan⁡θ\sin\theta<\theta<\tan\theta for 0<θ<π/20<\theta<\pi/2. This geometric comparison is not printed in the formula booklet; it explains why the two linear approximations approach from different sides.

Standard small-angle approximations

For θ\theta close to zero, measured in radians,

sin⁡θ≈θ,cos⁡θ≈1−θ22,tan⁡θ≈θ\sin\theta\approx\theta,\qquad \cos\theta\approx1-\frac{\theta^2}{2},\qquad \tan\theta\approx\theta

In the formula booklet.

The cosine correction uses a square because cosine is unchanged by reversing the angle. Sine and tangent change sign for negative arguments. At zero the approximation values equal the function values, but a separate expression with zero in its denominator can still be undefined. The sign ≈\approx means approximately equal, not identical.

Worked example 1: a positive small angle

Question. Use the standard approximations to estimate sin⁡0.1\sin0.1, cos⁡0.1\cos0.1 and tan⁡0.1\tan0.1, with angles in radians.

Plan. Substitute the whole radian angle into each supplied approximation.

sin⁡0.1≈0.1sin⁡θ≈θcos⁡0.1≈1−(0.1)22=0.995keep the squaretan⁡0.1≈0.1tan⁡θ≈θ\begin{aligned} \sin0.1&\approx0.1 && \sin\theta\approx\theta \\ \cos0.1&\approx1-\frac{(0.1)^2}{2}=0.995 && \text{keep the square} \\ \tan0.1&\approx0.1 && \tan\theta\approx\theta \end{aligned}

Why this step? The approximations use the radian number itself; no trigonometric evaluation is needed.

Answer. Approximately 0.10.1, 0.9950.995 and 0.10.1, respectively.

Check. Radian-mode values are 0.0998334…0.0998334\ldots, 0.9950041…0.9950041\ldots and 0.1003346…0.1003346\ldots, all close to the estimates.

Change only the sign of the angle. The odd functions change sign; the cosine estimate does not.

Worked example 2: a negative small angle

Question. Use the standard approximations to estimate sin⁡(−0.1)\sin(-0.1), cos⁡(−0.1)\cos(-0.1) and tan⁡(−0.1)\tan(-0.1), in radians.

Plan. Substitute the negative argument before squaring it for cosine.

sin⁡(−0.1)≈−0.1signed anglecos⁡(−0.1)≈1−(−0.1)22=0.995square is positivetan⁡(−0.1)≈−0.1signed angle\begin{aligned} \sin(-0.1)&\approx-0.1 && \text{signed angle} \\ \cos(-0.1)&\approx1-\frac{(-0.1)^2}{2}=0.995 && \text{square is positive} \\ \tan(-0.1)&\approx-0.1 && \text{signed angle} \end{aligned}

Why this step? Closeness to zero concerns the magnitude of the angle, not whether it is positive.

Answer. Approximately −0.1-0.1, 0.9950.995 and −0.1-0.1, respectively.

Check. The sine and tangent estimates reverse the previous signs; the cosine estimate is unchanged.

A small number in degrees is still a degree number. For 1∘1^\circ, first use θ=π/180\theta=\pi/180, giving sin⁡1∘≈π/180\sin1^\circ\approx\pi/180, about 0.017450.01745, rather than 11.

Watch out

Wrong: sin⁡1∘≈1\sin1^\circ\approx1. The approximation requires radians. Correct: sin⁡1∘≈π/180\sin1^\circ\approx\pi/180. This conversion is not optional even when the degree angle looks small.

Change only the magnitude of the negative radian angle.

Approximating complete expressions

In cos⁡(2x)\cos(2x), the small angle is 2x2x, so its square is 4x24x^2. The coefficient belongs inside the square.

For example, the supplied standard approximations give sin⁡3x≈3x\sin3x\approx3x and cos⁡2x≈1−2x2\cos2x\approx1-2x^2. These substituted forms are not separately printed in the booklet; derive them by replacing θ\theta with the entire argument. Both arguments must be close to zero in radians. A small xx does not automatically make 100x100x small.

Worked example 3: approximating a sum

Question. For small xx in radians, use the standard approximations to simplify sin⁡3x+cos⁡2x\sin3x+\cos2x.

Plan. Approximate each function at its whole argument, then collect terms.

sin⁡3x+cos⁡2x≈3x+(1−(2x)22)whole arguments=1+3x−2x2simplify\begin{aligned} \sin3x+\cos2x &\approx3x+\left(1-\frac{(2x)^2}{2}\right) && \text{whole arguments} \\ &=1+3x-2x^2 && \text{simplify} \end{aligned}

Why this step? Squaring 2x2x before dividing by two preserves the correct cosine correction.

Answer. 1+3x−2x21+3x-2x^2, provided both ∣3x∣|3x| and ∣2x∣|2x| are small.

Check. At x=0.01x=0.01, the approximation is 1.02981.0298, compared with 1.0297955…1.0297955\ldots from the original expression.

Change only the cosine argument from 2x2x to 4x4x. Its quadratic correction becomes four times as large.

Worked example 4: changing one argument

Question. For small xx in radians, simplify sin⁡3x+cos⁡4x\sin3x+\cos4x using the standard approximations.

Plan. Keep the sine estimate and replace the whole cosine argument.

sin⁡3x+cos⁡4x≈3x+(1−(4x)22)new cosine argument=1+3x−8x2square before dividing\begin{aligned} \sin3x+\cos4x &\approx3x+\left(1-\frac{(4x)^2}{2}\right) && \text{new cosine argument} \\ &=1+3x-8x^2 && \text{square before dividing} \end{aligned}

Why this step? Doubling an argument multiplies its square by four, not two.

Answer. 1+3x−8x21+3x-8x^2, with ∣3x∣|3x| and ∣4x∣|4x| small.

Check. At x=0.01x=0.01, the approximation is 1.02921.0292, compared with 1.0291956…1.0291956\ldots.

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