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5.4.3 Small Angle Approximations

Small angle approximations are useful techniques in trigonometry and calculus for simplifying the values of trigonometric functions when the angle θ \theta is small (typically measured in radians). These approximations are particularly helpful when dealing with limits, series expansions, or problems where the angle is close to zero.

1. Key Small Angle Approximations:

For very small angles θ\theta (in radians), the following approximations hold:

  • Sine:

sinθθ\sin \theta \approx \theta

  • Cosine:

cosθ1θ22\cos \theta \approx 1 - \frac{\theta^2}{2}

  • Tangent:

tanθθ\tan \theta \approx \theta

2. Understanding the Approximations:

  • Sine Approximation: sinθθ\sin \theta \approx \theta As θ\theta approaches 0, the sine of the angle becomes nearly equal to the angle itself when θ \theta is in radians. This is because the graph of sinθ\sin \theta is very close to the line y = θ\theta for small values of θ\theta.

  • Cosine Approximation: cosθ1θ22\cos \theta \approx 1 - \frac{\theta^2}{2} For small angles, cosθ\cos \theta remains close to 1, with the first correction term being θ22-\frac{\theta^2}{2}. The cosine graph is nearly flat near θ=0,\theta = 0, which is why the approximation starts with 1.

  • Tangent Approximation: tanθθ\tan \theta \approx \theta Since tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta}, and both sinθθ\sin \theta \approx \theta and cosθ1 \cos \theta \approx 1 for small θ\theta, the tangent function also approximates to the angle itself.

3. Derivation Using Taylor Series:

These approximations can be derived from the Taylor series expansions of the trigonometric functions around θ=0:\theta = 0:

  • Sine:

sinθ=θθ33!+θ55!\sin \theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots

For small θ \theta, higher-order terms like θ36\frac{\theta^3}{6} become negligible, so:

sinθθ\sin \theta \approx \theta

  • Cosine:

cosθ=1θ22!+θ44!\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots

For small θ\theta, higher-order terms like θ424\frac{\theta^4}{24} become negligible, so:

cosθ1θ22\cos \theta \approx 1 - \frac{\theta^2}{2}

  • Tangent:

tanθ=sinθcosθθ1=θ\tan \theta = \frac{\sin \theta}{\cos \theta} \approx \frac{\theta}{1} = \theta

4. Applications of Small Angle Approximations:

  • Physics: Small angle approximations are frequently used in physics, particularly in pendulum motion, where the sine of the angle of displacement is approximated as the angle itself for small oscillations.
  • Engineering: In engineering, small angle approximations simplify the analysis of structures and mechanisms where small angular displacements occur.
  • Astronomy: The approximation is also used in astronomy for calculating the angular diameter of celestial objects when viewed from Earth.

5. Example Problems Using Small Angle Approximations:

infoNote

Example 1: Approximate sin(0.1)\sin(0.1) radians.

Using the small angle approximation:

sin(0.1)0.1\sin(0.1) \approx 0.1

The exact value sin(0.1)0.09983\sin(0.1) \approx 0.09983, showing that the approximation is quite close.

infoNote

Example 2: Approximate cos(0.05) \cos(0.05) radians.

Using the small angle approximation:

cos(0.05)1(0.05)22=10.00252=10.00125=0.99875\cos(0.05) \approx 1 - \frac{(0.05)^2}{2} = 1 - \frac{0.0025}{2} = 1 - 0.00125 = 0.99875

The exact value cos(0.05)0.99875,\cos(0.05) \approx 0.99875, which matches the approximation closely.

infoNote

Example 3: Simplify sinθθ\frac{\sin \theta}{\theta} for small \theta.

Using the small angle approximation:

sinθθθθ=1\frac{\sin \theta}{\theta} \approx \frac{\theta}{\theta} = 1

This approximation is useful in limits, particularly in calculus when evaluating limθ0sinθθ.\lim_{\theta \to 0} \frac{\sin \theta}{\theta}.

Summary:

infoNote
  • Small angle approximations provide a powerful tool for simplifying trigonometric expressions when dealing with angles close to zero.
  • They are derived from the Taylor series expansions of the trigonometric functions and are especially useful in physics, engineering, and calculus.
  • Understanding these approximations allows for quicker and more intuitive problem-solving in scenarios where small angles are involved.

Small Angle Approximations

When dealing with small angles measured in radians, there are certain useful approximations we can use if the context of the problem allows approximations.

Graph: A graph is shown with two lines:

  • The blue line represents y=xy = x.
  • The red curve represents y=sin(θ)y = \sin(\theta). image

The above graph suggests that for small θ,sin(θ)θ\theta, \sin(\theta) \approx \theta.

Using sin(θ)θ\sin(\theta) \approx \theta, we can derive another approximation for cos(θ)\cos(\theta):

cos(θ)=cos(θ2+θ2)=12sin2(θ2)\cos(\theta) = \cos\left(\frac{\theta}{2} + \frac{\theta}{2}\right) = 1 - 2\sin^2\left(\frac{\theta}{2}\right)

(Using double angle formula)

For small θ,sin(θ)θcos(θ)12(θ2)2=12θ24\theta, \sin(\theta) \approx \theta \Rightarrow \cos(\theta) \approx 1 - 2\left(\frac{\theta}{2}\right)^2 = 1 - \frac{2\theta^2}{4}

=1θ22= 1 - \frac{\theta^2}{2}  For small θ,cos(θ)1θ22\therefore \text{ For small } \theta, \cos(\theta) \approx 1 - \frac{\theta^2}{2}
image

For small θtanθθ\theta \quad \\\tan\theta \approx \theta

infoNote

Example Problem

  1. When θ\theta is small, find the approximate values of: a) sin4θtan2θ3θb)1cos2θtan2θsinθc)3tanθθsin2θ\dfrac{\sin 4\theta - \tan 2\theta}{3\theta} \quad b) \dfrac{1 - \cos 2\theta}{\tan 2\theta \sin \theta} \quad c) \dfrac{3 \tan \theta - \theta}{\sin 2\theta}

Solution:

a) As θ0,sin(4θ)4θ,tan(2θ)2θ\theta \rightarrow 0, \sin(4\theta) \rightarrow 4\theta, \tan(2\theta) \rightarrow 2\theta

sin4θtan2θ3θ4θ2θ3θ=2θ3θ=23\therefore \frac{\sin 4\theta - \tan 2\theta}{3\theta} \approx \frac{4\theta - 2\theta}{3\theta} = \frac{2\theta}{3\theta} = \frac{2}{3}

b) As θ0,cos(2θ)1(2θ)22=14θ22=12θ2\theta \rightarrow 0, \cos(2\theta) \rightarrow 1 - \frac{(2\theta)^2}{2} =1- \frac{4\theta^2}{2} = 1 - 2\theta^2

tan(2θ)20,sinθθ\tan(2\theta) \rightarrow 20, \sin\theta \rightarrow \theta1cos2θtan2θsinθ1(12θ2)(2θ)(θ)=2θ22θ2=1\therefore \quad \frac{1 - \cos 2\theta}{\tan 2\theta \sin \theta} \approx \frac{1 - (1 - 2\theta^2)}{(2\theta)(\theta)} = \frac{2\theta^2}{2\theta^2} = 1

c) For small θ,tanθθ,sin(2θ)2θ\theta, \tan \theta \approx \theta, \sin(2\theta) \approx 2\theta

3tanθθsin2θ3θθ2θ=2θ2θ=1\therefore \quad \frac{3 \tan \theta - \theta}{\sin 2\theta} \approx \frac{3\theta - \theta}{2\theta} = \frac{2\theta}{2\theta} = 1

2. When θ\theta is small, show that:

a)sin3θsin4θ34θb)cosθ1tan2θθ4c)tan4θ+θ23θsin2θ4+θa) \frac{\sin 3\theta}{\sin 4\theta} \approx \frac{3}{4\theta} \quad b) \frac{\cos \theta - 1}{\tan 2\theta} \approx -\frac{\theta}{4} \quad c) \frac{\tan 4\theta + \theta^2}{3\theta - \sin 2\theta} \approx 4 + \theta

a) For small θ\theta,

sin(3θ)3θ,sin(4θ)4θ\sin(3\theta) \approx 3\theta, \sin(4\theta) \approx 4\thetasin3θθsin4θ3θθ(4θ)=3θ4θ2=34θ\Rightarrow \frac{\sin 3\theta}{\theta\sin 4\theta} \approx \frac{3\theta}{\theta(4\theta)} = \frac{3\theta}{4\theta^2}=\frac{3}{4\theta}

b) For small θ\theta,

cosθ1θ22,tan2θ2θ\cos \theta \approx 1 - \frac{\theta^2}{2}, \tan 2\theta \approx 2\thetacosθ1tan2θ1(1θ22)2θ=θ222θ=θ240=θ4\Rightarrow \frac{\cos \theta - 1}{\tan 2\theta} \approx \frac{1 - (1 - \frac{\theta^2}{2})}{2\theta} = \frac{\frac{\theta^2}{2}}{2\theta} = \frac{\theta^2}{40}=\frac{\theta}{4}

c) For small θ\theta,

tan4θ4θ,sin2θ2θ\tan 4\theta \approx 4\theta, \sin 2\theta \approx 2\thetatan4θ+θ23θsin2θ4θ+θ23θ2θ=4θ+θ2θ=4+θ\Rightarrow \frac{\tan 4\theta + \theta^2}{3\theta - \sin 2\theta} \approx \frac{4\theta + \theta^2}{3\theta - 2\theta} = \frac{4\theta + \theta^2}{\theta} = 4 + \theta

infoNote

Q1 (OCR H240/03, Sample Question Paper, Q4) Show that, for a small angle θ\theta, where θ\theta is in radians,

1+cosθ3cos2θ1+52θ21 + \cos \theta - 3\cos^2 \theta \approx -1 + \frac{5}{2}\theta^2

Solution:

When θ\theta is small, cosθ1θ22\cos \theta \approx 1 - \frac{\theta^2}{2}

1+cosθ3cos2θ=1+(1θ22)3(1θ22)2\Rightarrow 1 + \cos \theta - 3\cos^2 \theta = 1 + \left(1 - \frac{\theta^2}{2}\right) - 3\left(1 - \frac{\theta^2}{2}\right)^2=2θ223(1θ22θ22+θ44)= 2 - \frac{\theta^2}{2} - 3\left(1 - \frac {\theta^2}{2}-\frac {\theta^2}{2} + \frac{\theta^4}{4}\right)=2θ223(1θ2+θ44)= 2 - \frac{\theta^2}{2} - 3\left(1 - \theta^2 + \frac{\theta^4}{4}\right)=2θ223+3θ23θ44= 2 - \frac{\theta^2}{2} - 3 + 3\theta^2 - \frac{3\theta^4}{4}=1+5θ223θ44= -1 + \frac{5\theta^2}{2} - \xcancel{\frac{3\theta^4}{4}}=1+5θ22= -1 + \frac{5\theta^2}{2}

As θ\theta is small, θ4\theta^4 is insignificantly small.


infoNote

Q3, (OCR H240/02, Practice Paper Set 3, Q3) Use small angle approximations to estimate the solution of the equation:

cos12θ1+sinθ=0.825\frac{\cos \frac{1}{2}\theta}{1 + \sin \theta} = 0.825

if θ\theta is small enough to neglect terms in θ3\theta^3 or above.

When θ\theta is small,

cos(12θ)1(12θ)22=1(θ24)2=1θ28\cos \left(\frac{1}{2}\theta\right) \approx 1 - \frac{\left(\frac{1}{2}\theta\right)^2}{2} = 1 - \frac{\left(\frac{\theta^2}{4}\right)}{2} = 1 - \frac{\theta^2}{8}sinθθ\sin \theta \approx \theta1θ281+θ=0.825\frac{1 - \frac{\theta^2}{8}}{1 + \theta} = 0.8251θ28=0.825(1+θ)\Rightarrow 1 - \frac{\theta^2}{8} = 0.825(1 + \theta)1θ28=0.825+0.825θ\Rightarrow 1 - \frac{\theta^2}{8} = 0.825 + 0.825\theta0=θ28+0.825θ0.1750 = \frac{\theta^2}{8} + 0.825\theta - 0.175θ0.2057,6.806(:highlight[not valid] as θ must be small)\theta \approx 0.2057, \quad \xcancel{- 6.806} \quad \text {(:highlight[not valid] as } \theta \text{ must be small)}

5. a) When θ\theta is small, show that the expression 4cos3θ2+5sinθ1sin2θ\dfrac{4 \cos 3\theta - 2 + 5 \sin \theta}{1 - \sin 2\theta}

can be written as 9θ+29\theta + 2.

b) Hence write down the value of 4cos3θ2+5sinθ1sin2θ\dfrac{4 \cos 3\theta - 2 + 5 \sin \theta}{1 - \sin 2\theta} when θ\theta is small.

When θ\theta is small,

cos3θ1(3θ)22=19θ22\cos 3\theta \approx 1 - \frac{(3\theta)^2}{2} = 1 - \frac{9\theta^2}{2} sinθθ,sin2θ2θ\sin \theta \approx \theta, \quad \sin 2\theta \approx 2\theta 4(19θ22)2+5θ12θ\frac{4(1 - \frac{9\theta^2}{2}) - 2 + 5\theta}{1 - 2\theta} =418θ22+5θ12θ= \frac{4 - 18\theta^2 - 2 + 5\theta}{1 - 2\theta} 18θ25θ212θ=(9θ+2)(2θ1)(2θ1)\Rightarrow -\frac{18\theta^2 - 5\theta - 2}{1 - 2\theta} = \frac {\cancel-(9\theta+2)\cancel{(2\theta-1)}}{\cancel-\cancel{(2\theta-1)}} 9θ+2\Rightarrow 9\theta +2

Assuming θ\theta is small, we can approximate the expression as 9θ+29\theta + 2.

Hence 4cos3θ2+5sinθ1sin2θ9θ+2\dfrac{4\cos 3\theta - 2 + 5\sin \theta}{1 - \sin 2\theta} \approx 9\theta + 2 when θ\theta is small.


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