True or False? Small angle approximations only work for positive angles.

False. They hold for small negative angles just as well. Sine and tangent preserve the sign, and cosine is unaffected by it because the angle is squared.

Where do small angle approximations turn up in a proof?

In differentiating trigonometric functions from first principles . The limits that appear in that proof are exactly these small angle results.

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Radian Measure (Edexcel A Level Maths: Pure): Flashcards

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FrontRadian Measure
Define radian.

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Cards in this collection (19)

Define radian.
One radian is the angle in a sector whose radius and arc length are both $1$ .
Radians are an alternative to degrees, and are usually quoted in terms of $π$ .
$π$ radians $=______°$ , so a full turn of $2 π$ radians is $______°$ .
$π$ radians $= 180 °$ , so a full turn of $2 π$ radians is $360 °$ .
Every other conversion follows from $π = 180 °$ , so $\frac{3}{4} π = 135 °$ .
What is the arc length of a sector of radius $r$ and angle $θ$ ?
$s = r θ$ , with $θ$ in radians; in degrees the formula simply does not hold.
A full turn, $θ = 2 π$ , gives $s = 2 π r$ , which is the whole circumference.
What is the area of a sector of radius $r$ and angle $θ$ ?
$A = \frac{1}{2} r^{2} θ$ , again with $θ$ in radians.
Both sector formulae are just the whole-circle value scaled by $\frac{θ}{2 π}$ .
True or False?
One radian is a little under $60 °$ .
True.
Since $π$ radians is $180 °$ , one radian is $\frac{180}{π} \approx 57.3 °$ .
Knowing roughly how big a radian is makes a useful check on an answer.
How can you tell an angle is in radians when no symbol is given?
If it is written in terms of $π$ , it is almost always in radians.
Otherwise the symbol $rad$ is used, or occasionally a superscript $c$ .
How do you find the area of the segment cut off by a chord in a circle?
Work out the sector area, then subtract the triangle formed by the two radii and the chord.
That triangle has area $\frac{1}{2} r^{2} \sin θ$ , since both of its known sides are radii.
Where do the exact values for $30 °$ , $45 °$ and $60 °$ come from?
From SOH CAH TOA applied to two special triangles: half an equilateral triangle gives $30 °$ and $60 °$ , and a right-angled isosceles triangle gives $45 °$ .
Remembering the two triangles is safer than remembering six separate values.
$\sin 30 ° =______$ , $\cos 60 ° =______$ , $\tan 45 ° =______$
$\sin 30 ° = \frac{1}{2}$ , $\cos 60 ° = \frac{1}{2}$ , $\tan 45 ° = 1$
The first two agree because $30 °$ and $60 °$ are the two acute angles of the same triangle, with opposite and adjacent swapping over.
What are $30 °$ , $45 °$ and $60 °$ in radians?
$\frac{π}{6}$ , $\frac{π}{4}$ and $\frac{π}{3}$ .
Exact values are quoted in radians as often as in degrees, so both forms need to be recognised on sight.
How do you recall $\sin$ , $\cos$ and $\tan$ of $0 °$ , $90 °$ and $180 °$ ?
Read them off the graphs of the three functions rather than from a triangle.
No triangle can contain an angle of $0 °$ , $90 °$ or $180 °$ , so the triangle method cannot reach them.
How do you find the exact value of $\sin 150 \circ$ ?
Use the symmetry of the graph: $150 °$ mirrors $30 °$ , and sine is positive there, so $\sin 150 ° = \frac{1}{2}$ .
Any multiple of $30 °$ , $45 °$ or $60 °$ can be reached the same way.
True or False?
$\tan 90 °$ has an exact value.
False.
$\tan 90 °$ is undefined, because $\cos 90 ° = 0$ and the tangent divides by it.
The graph has an asymptote there rather than a value.
For small $θ$ measured in radians:
$\sin θ \approx______$ , $\tan θ \approx______$ , $\cos θ \approx 1 -______$
$\sin θ \approx θ$ , $\tan θ \approx θ$ , $\cos θ \approx 1 - \frac{1}{2} θ^{2}$
Sine and tangent share the same approximation; cosine is the odd one out and is the one worth checking.
What must be true before you can use a small angle approximation?
The angle must be small and measured in radians.
In degrees they collapse completely: $\sin 5 °$ is about $0.087$ , nothing like $5$ .
How do you approximate $\sin 3 θ$ and $\cos 2 θ$ for small $θ$ ?
Replace the whole angle, so $\sin 3 θ \approx 3 θ$ .
For cosine the substituted angle is then squared: $\cos 2 θ \approx 1 - \frac{1}{2} {(2 θ)}^{2} = 1 - 2 θ^{2}$ .
True or False?
Small angle approximations only work for positive angles.
False.
They hold for small negative angles just as well.
Sine and tangent preserve the sign, and cosine is unaffected by it because the angle is squared.
Why is the approximation for $\cos θ$ not simply $1$ ?
Because $\cos θ$ starts at $1$ but curves away from it, and the $- \frac{1}{2} θ^{2}$ term captures that curvature.
Near zero the cosine graph behaves like a negative quadratic.
Where do small angle approximations turn up in a proof?
In differentiating trigonometric functions from first principles.
The limits that appear in that proof are exactly these small angle results.

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