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MATH1011 Chap.7 Trigonometric Functions and Identities

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Chapter 7 of 12 · MATH1011

Trigonometric Functions and Identities

Rebuild the exact values instead of trusting your memory

Two quick sketches replace the whole table of exact values. Halve an equilateral triangle of side 2 and the piece you keep has sides 1, root 3 and 2, which hands you every ratio at 30 and 60 degrees. Cut a unit square along its diagonal and the piece has sides 1, 1 and root 2, which hands you 45 degrees.

Keep the answers as surds: a decimal such as 0.866 will never cancel against a root 3 further down the page, and most questions here are built so that something does cancel.

For an acute angle, one ratio pins the other two. If the sine is 9/41 then the third side is 40, so the cosine is 40/41 and the tangent is 9/40, without a calculator.

Tangent is never an independent third fact about an angle: it is the sine divided by the cosine, which is also why it has nothing to report wherever the cosine is 0.

The circle takes over where the triangle runs out

A right triangle can host an angle up to 90 degrees and no further, yet questions happily ask for the cosine of 300 degrees. So the ratios are rebuilt on a circle of radius 1 centred at the origin.

Turn a ray anticlockwise from the positive x axis through the angle; where it meets the circle, the first coordinate is the cosine, the second is the sine, and their quotient is the tangent. Acute angles give back the triangle values unchanged, while obtuse, reflex and negative angles now have values too.

Two small facts then do the work.

Coordinates carry signs, so the quadrant settles every sign: all three ratios positive in the first quadrant, only the sine in the second, only the tangent in the third, only the cosine in the fourth. The acute angle between the ray and the horizontal axis, the reference angle, settles the size. Any ratio of any angle is therefore an entry from the exact table with a sign attached.

Fix the size first and attach the sign afterwards; doing both in one step is where most of the marks in this chapter are lost.

Radians are not a style choice

Measure an angle by the length of arc it cuts on a circle of radius 1 and you are working in radians, so a full turn measures 2π and a half turn measures π. Multiply by π/180 to leave degrees and by 180/π to come back. That part is bookkeeping.

This part is not: the rule that the derivative of sin x is cos x, and that the derivative of cos x is minus sin x, holds for radian input only. Differentiate the sine of an angle measured in degrees and a factor of π/180, about 0.0175, appears in front, so a gradient found in degree mode comes out roughly 57 times too small. The arc length s = rθ and the sector area A = ½r²θ carry the same requirement.

Read the question for a degree symbol, set the calculator once, and leave it.

Identities shrink the problem, periods count the answers

Three families cover nearly all the manipulation here. The relation sin²θ + cos²θ = 1 converts between squares, and dividing it through by cos²θ or by sin²θ produces the secant and cosecant versions on the spot.

The compound angle formulas split or join a sum of two angles, and they are how an exact value such as sin 75 degrees is reached: write it as sin(45 + 30) degrees and out comes (root 6 plus root 2)/4, about 0.9659. The double angle forms are those same formulas with the two angles set equal, and the three versions of cos 2θ exist so that a square can be traded for a first power whenever an equation holds both.

Periods decide how many answers a question has.

Sine and cosine repeat every 2π, that is every 360 degrees, and never leave the band from -1 to 1. Tangent repeats every π, that is every 180 degrees, which is half as far: a half turn reverses the sign of both coordinates, and a quotient of two reversed signs is unchanged. Any summary printing 360 degrees for all three functions is wrong on its tangent line.

The practical consequence is that solutions of a tangent equation sit π apart, so treating tangent like sine loses exactly half of them on any interval.

In this chapter

What this chapter covers

  • 01

    Sine, cosine and tangent as ratios in a right triangle

  • 02

    The unit circle, and the sign each quadrant forces

  • 03

    Reference angles for angles that are not acute

  • 04

    Pythagorean, compound angle and double angle identities

  • 05

    Degrees and radians, and why calculus needs radians

  • 06

    Graphs and periods: 2π for sine and cosine, π for tangent

Worked example · free

Two given ratios, then a difference and a double

Q [7 marks]. The marks shown are not a University mark allocation. In this AskSia-authored item, angle A lies in the second quadrant with sin A = 8/17 and angle B lies in the fourth quadrant with cos B = 3/5. Find cos A, sin B, sin(A - B), sin 2A and cos 2A exactly, then state the quadrant of 2A.
The ray at 210 degrees meets the circle at cosine -0.866 and sine -0.500.The ray at 210 degrees meets the circle at cosine -0.866 and sine -0.500. Its acute angle to the horizontal axis is 30 degrees, which is why both coordinates match the 30 degree row of the exact table apart from sign.Quadrant Iall three positiveQuadrant IIsine positive onlyQuadrant IIItangent positive onlyQuadrant IVcosine positive onlysigns by quadrant210 degrees30 degreescos -0.866, sin -0.500x axisy axis
The ray at 210 degrees meets the circle at cosine -0.866 and sine -0.500. Its acute angle to the horizontal axis is 30 degrees, which is why both coordinates match the 30 degree row of the exact table apart from sign.
  • 2Apply sin²+cos²=1 to each angle, then let the stated quadrant choose the sign of the square root.
  • 2Expand sin(A - B) = sin A cos B - cos A sin B and substitute the four exact values.
  • 2Use sin 2A = 2 sin A cos A and cos 2A = 1 - 2sin²A.
  • 1Read the quadrant of 2A off the pair of signs you have just produced.
(a) From sin²A + cos²A = 1, cos²A = 1 - 64/289 = 225/289, so the cosine has size 15/17, and the second quadrant makes it negative: cos A = -15/17. In the same way sin²B = 1 - 9/25 = 16/25, and the fourth quadrant gives sin B = -4/5. (b) sin(A - B) = (8/17)(3/5) - (-15/17)(-4/5) = 24/85 - 60/85 = -36/85, about -0.4235. As a check, cos(A - B) = (-15/17)(3/5) + (8/17)(-4/5) = -45/85 - 32/85 = -77/85, and 36² + 77² = 1296 + 5929 = 7225 = 85², so the two answers are consistent. (c) sin 2A = 2(8/17)(-15/17) = -240/289, about -0.8304, and cos 2A = 1 - 2(64/289) = 161/289, about 0.5571. (d) The sine of 2A is negative while the cosine of 2A is positive, which is the fourth quadrant. In decimals A is about 151.93 degrees, so 2A is about 303.86 degrees, and that lands in the fourth quadrant as predicted.
Sia tip — Take the square root and attach the quadrant sign on the same line of working. Writing cos A = 15/17 now and promising yourself a minus sign later is the commonest way this question is thrown away, because that sign has to survive two further substitutions before it reaches the answer.
Glossary

Key terms

Radian
A radian is the angle whose arc on a circle of radius 1 has length 1. The full circumference is 2π, so one complete turn measures 2π radians and 180 degrees is the same rotation as π radians.
Unit Circle
The unit circle is the circle of radius 1 centred at the origin, on which the ratios are defined for an angle of any size. A ray at angle θ meets it at the point with coordinates cos θ and sin θ.
Reference Angle
A reference angle is the acute angle between the ray of a given angle and the horizontal axis. It supplies the size of every ratio, while the quadrant supplies the sign.
Pythagorean Identity
The Pythagorean identity is the statement that sin²θ + cos²θ = 1 holds for every angle θ. Dividing it by cos²θ or by sin²θ produces the secant and cosecant versions.
Compound Angle Formula
A compound angle formula writes a ratio of the sum or difference of two angles in terms of the ratios of the separate angles. Setting the two angles equal converts each one into a double angle formula.
Period
A period is the smallest positive shift of the input that leaves every output of a repeating function unchanged. Sine and cosine have period 2π, while tangent has period π.
FAQ

Trigonometric Functions and Identities FAQ

Why do the calculus formulas insist on radians rather than degrees?

Because the derivative rules for sine and cosine come from a limit that only behaves simply when the angle is measured in radians. Work in degrees and a factor of π/180 appears in front of every one of those derivatives, so a gradient read off a calculator left in degree mode is about 57 times too small.

Does tangent repeat every 360 degrees, the way sine and cosine do?

No. Tangent returns to the same value after only 180 degrees, that is π radians, because a half turn reverses the sign of both coordinates on the circle and a quotient of two reversed signs is unchanged. Solutions of a tangent equation therefore sit π apart rather than 2π apart.

How do I find a ratio when the angle is not acute?

Split the job into size and sign. The acute angle between the ray and the horizontal axis gives the size directly from the exact table, and the quadrant the angle lands in gives the sign. Put the two together at the end instead of guessing both of them at once.

Study strategy

Exam move

Redraw the half equilateral triangle and the half unit square from memory, then list the sixteen angles in one full turn that are multiples of 30 or 45 degrees, write each in both units, and read off sine, cosine and tangent with the sign included. Time the run.

Under three minutes means the exact values will cost you nothing under exam pressure, and the mistakes that remain will almost all be quadrant signs rather than sizes, which is a much cheaper fault to drill away.

Working through Trigonometric Functions and Identities in MATH1011? Sia is AskSia’s AI Mathematics tutor — ask any MATH1011 Trigonometric Functions and Identities question and get a clear, step-by-step explanation grounded in how MATH1011 is taught and assessed. Read this chapter free, then take your hardest questions to Sia.

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