Trigonometric functions and identities — CSCA (China)

In this lesson you will learn

Convert between degrees and radians and find arc lengths and sector areas
Find sin, cos and tan of any angle with the unit circle, signs by quadrant and special angles
Simplify expressions with the basic identities and reduction formulas
Apply the sum, difference and double-angle formulas and a·sin x + b·cos x = R sin(x + φ)

The CSCA forbids calculators, yet it asks for cos 75°, or for sin 2α when only cos α is given. None of these values is memorised: they are built from five special angles and a handful of formulas, and this lesson is that toolkit. The graphs of sin, cos and tan and y = A sin(ωx + φ) are in «Graphs of trigonometric functions»; triangles are in «Solving triangles: the law of sines and the law of cosines».

Radians and the unit circle

Definition

Radian (弧度)

The central angle whose arc is as long as the radius. A full turn is 2π rad = 360°, so 180° = π rad and 1 rad ≈ 57.3°. The unit is usually left out: sin 1 means the sine of 1 radian, not of 1°. Angles that differ by whole turns, α + 2kπ (k ∈ Z), share the same terminal side (终边).

α (rad) = α° · π/180 l = |α| · r S = ½ · l · r = ½ · |α| · r²α (rad) = α° · π/180 l = |α| · r S = ½ · l · r = ½ · |α| · r²

where:

αthe angle in radians
lthe arc length
rthe radius
Sthe area of the sector

In radians the arc and sector formulas need no 360°: this is why calculus and Chinese textbooks (弧度制) prefer radians.

Example 1: radians, arcs and sectors

1) Convert 135° to radians.
2) Convert 11π/6 to degrees.
3) A sector has radius 10 cm and arc length 15 cm. Find its central angle and its area.

Show solution

1) 135 · π/180 = 3π/4.
2) (11/6) · 180° = 330°.
3) α = l/r = 15/10 = 1.5 rad; S = ½ · 15 · 10 = 75 cm².

Definition

Trigonometric functions of any angle (三角函数)

Put the vertex of α at the origin and its initial side along the positive x-axis. If the terminal side meets the unit circle (单位圆) at P(x, y), then sin α = y, cos α = x and tan α = y/x (x ≠ 0).

sin α = y/r cos α = x/r tan α = y/x r = √(x² + y²)sin α = y/r cos α = x/r tan α = y/x r = √(x² + y²)

where:

(x, y)any point on the terminal side of α (other than the origin)
rthe distance from the point to the origin

On the unit circle r = 1, so the point is simply (cos α, sin α). The signs of x and y give the signs by quadrant: I all positive, II only sin, III only tan, IV only cos.

Example 2: values from a point, and the quadrant

1) The terminal side of α passes through P(5, −12). Find sin α, cos α and tan α.
2) cos α < 0 and tan α < 0. In which quadrant is the terminal side of α?

Show solution

1) r = √(25 + 144) = 13: sin α = −12/13, cos α = 5/13, tan α = −12/5 (quadrant IV: only cos is positive).
2) cos < 0 in quadrants II and III; tan < 0 in II and IV. Both hold only in quadrant II.

α	0	π/6 (30°)	π/4 (45°)	π/3 (60°)	π/2 (90°)	π (180°)
sin α	0	1/2	√2/2	√3/2	1	0
cos α	1	√3/2	√2/2	1/2	0	−1
tan α	0	√3/3	1	√3	undefined	0

Know this table by heart: every other exact value in CSCA comes from it through the formulas below.

Interactive

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210° = 7π/6 lies in quadrant III: sin = −1/2 and cos = −√3/2 are negative, tan = √3/3 is positive. Drag the angle through the four quadrants and watch the signs.

Basic identities and reduction formulas

sin²α + cos²α = 1 tan α = sin α / cos αsin²α + cos²α = 1 tan α = sin α / cos α

where:

αany angle (for tan, cos α ≠ 0)

The basic identities of one angle (同角三角函数关系). The first is Pythagoras on the unit circle, x² + y² = 1. From one value you get the other two; the quadrant decides the sign.

Example 3: from one value to the others

1) cos α = −12/13 and α is in quadrant III. Find sin α and tan α.
2) tan α = −2. Find (2 sin α − cos α)/(sin α + 3 cos α) and sin α cos α.

Show solution

1) sin²α = 1 − 144/169 = 25/169; in quadrant III sine is negative: sin α = −5/13; tan α = (−5/13) ÷ (−12/13) = 5/12.
2) Divide the numerator and the denominator by cos α: (2 tan α − 1)/(tan α + 3) = (−4 − 1)/(−2 + 3) = −5.
For sin α cos α divide by 1 = sin²α + cos²α: sin α cos α / (sin²α + cos²α) = tan α / (tan²α + 1) = −2/5.

Reduction formulas (诱导公式) turn a trigonometric function of any angle into one of an acute angle. They follow from the symmetry of the unit circle: the point for −α is the mirror image of the point for α in the x-axis, the point for π − α is its mirror image in the y-axis, and the point for π + α is symmetric to it about the origin.

Angle	sin	cos	tan
−α	−sin α	cos α	−tan α
π − α	sin α	−cos α	−tan α
π + α	−sin α	−cos α	tan α
2π − α	−sin α	cos α	−tan α
π/2 − α	cos α	sin α	cot α
π/2 + α	cos α	−sin α	−cot α
3π/2 − α	−cos α	−sin α	cot α
3π/2 + α	−cos α	sin α	−cot α

Also: adding 2kπ changes nothing, and tan(α + kπ) = tan α; cot α = 1/tan α.

1
Make a negative angle positive
sin(−α) = −sin α, cos(−α) = cos α, tan(−α) = −tan α — or simply add 2π until the angle is positive.
2
Remove whole turns
Subtract multiples of 2π (360°) until the angle lies in [0, 2π).
3
Reduce to an acute angle
Write the angle as π − α, π + α or 2π − α with α acute and use the table (sign by quadrant).
4
Read the value
Use the special-angle table.

Example 4: from any angle to an acute one

1) cos 225° 2) sin(−13π/6) 3) tan 300° 4) Simplify sin(π + α) · cos(−α) / [cos(π/2 + α) · cos(π − α)].

Show solution

1) 225° = 180° + 45°, quadrant III: cos 225° = −cos 45° = −√2/2.
2) −13π/6 + 4π = 11π/6 = 2π − π/6: sin(11π/6) = −sin(π/6) = −1/2.
3) 300° = 360° − 60°: tan 300° = −tan 60° = −√3.
4) Numerator: (−sin α)(cos α); denominator: (−sin α)(−cos α) = sin α cos α. The quotient is −1.

Example 5 (CSCA style): reduction plus the basic identity

sin(π/2 + α) = −5/13 and α ∈ (π/2, π). What is tan(π − α)?
A) 12/5 B) −12/5 C) 5/12 D) −5/12

Show solution

sin(π/2 + α) = cos α, so cos α = −5/13. In quadrant II sin α > 0: sin α = 12/13 and tan α = −12/5. Finally tan(π − α) = −tan α = 12/5, answer A.
Why the others tempt: B) is tan α itself, with the sign of the reduction forgotten; C) and D) come from swapping sin and cos (tan = cos/sin).

Sum, difference and double-angle formulas

sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan(α ± β) = (tan α ± tan β) / (1 ∓ tan α tan β)sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan(α ± β) = (tan α ± tan β) / (1 ∓ tan α tan β)

where:

α, βany angles (for tan, all tangents and the denominator must be defined)

The sum and difference formulas (两角和与差公式). In the cosine formula the sign between the products is the opposite of the sign in the bracket. They turn 15°, 75° and 105° into 45° ± 30° or 60° ± 45°.

Where the formulas come from. Take the points P(cos α, sin α) and Q(cos β, sin β) on the unit circle. The angle between OP and OQ is α − β, so the dot product is OP · OQ = 1 · 1 · cos(α − β); in coordinates the same product is cos α cos β + sin α sin β. That is cos(α − β). Replacing β by −β gives cos(α + β); sin x = cos(π/2 − x) gives the sine formulas, and dividing sin by cos gives the tan formula (the dot product is in «Plane vectors»).

Example 6: sums and differences

1) sin 105° 2) tan 75° 3) cos 80° cos 20° + sin 80° sin 20° 4) α and β are acute, sin α = 3/5 and cos β = 5/13. Find cos(α + β).

Show solution

1) sin(60° + 45°) = (√3/2)(√2/2) + (1/2)(√2/2) = (√6 + √2)/4.
2) tan(45° + 30°) = (1 + √3/3)/(1 − √3/3) = (3 + √3)/(3 − √3) = 2 + √3.
3) The pattern cos α cos β + sin α sin β is cos(α − β): cos 60° = 1/2.
4) cos α = 4/5 and sin β = 12/13 (acute angles, positive values): cos(α + β) = (4/5)(5/13) − (3/5)(12/13) = −16/65.

Example 7 (CSCA style): splitting the angle

α is an acute angle and cos(α + π/6) = 3/5. What is sin α?
A) (4√3 − 3)/10 B) (4√3 + 3)/10 C) (3√3 − 4)/10 D) (3√3 + 4)/10

Show solution

Do not look for α itself; write α = (α + π/6) − π/6. Since α + π/6 ∈ (π/6, 2π/3), its sine is positive: sin(α + π/6) = 4/5.
sin α = sin(α + π/6) cos(π/6) − cos(α + π/6) sin(π/6) = (4/5)(√3/2) − (3/5)(1/2) = (4√3 − 3)/10. Answer A.
Why the others tempt: B) uses the sum formula instead of the difference; C) and D) swap the roles of 3/5 and 4/5 (the cosine and sine of the helper angle).

sin 2α = 2 sin α cos α cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α tan 2α = 2 tan α / (1 − tan²α) cos²α = (1 + cos 2α)/2 sin²α = (1 − cos 2α)/2sin 2α = 2 sin α cos α cos 2α = cos²α − sin²α = 2cos²α − 1 = 1 − 2sin²α tan 2α = 2 tan α / (1 − tan²α) cos²α = (1 + cos 2α)/2 sin²α = (1 − cos 2α)/2

where:

αany angle (for tan 2α, tan α ≠ ±1)

The double-angle formulas (二倍角公式) are the sum formulas with β = α. Read from right to left, the formula for cos 2α gives the power-reduction formulas (降幂公式), which replace squares by the first power of cos 2α.

Example 8: double angles

1) tan α = 1/2. Find tan 2α, sin 2α and cos 2α.
2) sin 15° cos 15° 3) 1 − 2 sin² 75° 4) sin² 22.5°

Show solution

1) tan 2α = 1/(1 − 1/4) = 4/3; sin 2α = 2 tan α/(1 + tan²α) = 1/(5/4) = 4/5; cos 2α = (1 − tan²α)/(1 + tan²α) = (3/4)/(5/4) = 3/5 (check: (4/5)² + (3/5)² = 1).
2) ½ sin 30° = 1/4.
3) cos 150° = −√3/2.
4) (1 − cos 45°)/2 = (2 − √2)/4.

The auxiliary angle: a·sin x + b·cos x = R sin(x + φ)

A sum a sin x + b cos x is again a single sine wave. Take R = √(a² + b²) out of the bracket: the numbers a/R and b/R satisfy (a/R)² + (b/R)² = 1, so they are cos φ and sin φ of some angle φ, and by the sum formula R(cos φ sin x + sin φ cos x) = R sin(x + φ). Chinese textbooks call this the auxiliary-angle formula (辅助角公式).

a sin x + b cos x = R sin(x + φ), R = √(a² + b²), cos φ = a/R, sin φ = b/Ra sin x + b cos x = R sin(x + φ), R = √(a² + b²), cos φ = a/R, sin φ = b/R

where:

Rthe amplitude: maximum R, minimum −R
φthe auxiliary angle; its quadrant is chosen from the signs of a and b

The period and the graph of R sin(ωx + φ) are in «Graphs of trigonometric functions»; here we only need the rewriting and the range [−R, R].

Example 9: the auxiliary angle

1) Write sin x + √3 cos x as R sin(x + φ).
2) Write sin x − cos x in the same form.
3) Find the greatest and the least value of 3 sin x + 4 cos x.
4) Find the maximum of y = sin x cos x + cos² x.

Show solution

1) R = √(1 + 3) = 2: 2((1/2) sin x + (√3/2) cos x) = 2 sin(x + π/3).
2) R = √2, cos φ = 1/√2, sin φ = −1/√2 ⇒ φ = −π/4: √2 sin(x − π/4).
3) R = √(9 + 16) = 5: greatest 5, least −5.
4) Lower the degree first: y = ½ sin 2x + (1 + cos 2x)/2 = ½(sin 2x + cos 2x) + ½ = (√2/2) sin(2x + π/4) + ½. The maximum is (1 + √2)/2.

Interactive

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The wave a sin x + b cos x between the lines y = ±√(a² + b²). With a = 3 and b = 4 it touches ±5; change a and b, and the wave always stays between the two lines and touches both.

Example 10 (CSCA style): the maximum value

What is the maximum value of f(x) = sin x + cos(x + π/6)?
A) 2 B) 1 C) √3 D) √3/2

Show solution

Expand: cos(x + π/6) = (√3/2) cos x − (1/2) sin x, so f(x) = (1/2) sin x + (√3/2) cos x = sin(x + π/3). The maximum is 1, answer B.
Why the others tempt: A) adds the two maxima 1 + 1, but they are not reached at the same x; C) √3 comes from expanding with the wrong sign, (3/2) sin x + (√3/2) cos x with R = √3; D) √3/2 is just f(0).

How CSCA asks about this

Exact values: special angles after a reduction, or 15°, 75°, 105° through the sum and difference formulas.
One value given: sin α (or cos α, tan α) and the quadrant → the other functions, sin 2α, cos 2α, tan 2α; the trap is the sign.
tan α given: homogeneous fractions — divide by cos α or cos²α.
Angle splitting: α = (α + β) − β, 2α = (α + β) + (α − β); find the sign for the helper angle from its interval.
Max/min: lower the degree with the double-angle formulas, then use a sin + b cos = R sin(… + φ); the answer is ±R plus a constant.
Time-savers (75 s per item): test an identity option by putting α = 0 or α = π/6; before calculating, strike out options whose sign is wrong for the quadrant; «sin α is given» items often hide the 3-4-5 and 5-12-13 triangles.

Term	中文	Pinyin
radian	弧度	húdù
radian measure	弧度制	húdùzhì
unit circle	单位圆	dānwèiyuán
terminal side	终边	zhōngbiān
quadrant	象限	xiàngxiàn
sine / cosine / tangent	正弦 / 余弦 / 正切	zhèngxián / yúxián / zhèngqiē
trigonometric function	三角函数	sānjiǎo hánshù
arc length	弧长	húcháng
sector	扇形	shànxíng
identity	恒等式	héngděngshì
reduction formulas	诱导公式	yòudǎo gōngshì
sum and difference formulas	两角和与差公式	liǎngjiǎo hé yǔ chā gōngshì
double-angle formulas	二倍角公式	èrbèijiǎo gōngshì
power-reduction formulas	降幂公式	jiàngmì gōngshì
auxiliary-angle formula	辅助角公式	fǔzhùjiǎo gōngshì

Key terms in English and Chinese: you may take the test in either language. Chinese textbooks write the tangent as «tan».

Key points

180° = π rad; arc l = |α| r, sector S = ½ l r = ½ |α| r² (α in radians).
On the unit circle P = (cos α, sin α) and tan α = y/x; positive: I all, II sin, III tan, IV cos.
sin²α + cos²α = 1, tan α = sin α/cos α; after a square root the quadrant decides the sign.
Reduction: odd multiples of π/2 swap sin ↔ cos, even ones do not; the sign is that of the original function in its quadrant.
sin(α ± β) = sin α cos β ± cos α sin β, cos(α ± β) = cos α cos β ∓ sin α sin β; sin 2α = 2 sin α cos α, cos 2α = 1 − 2sin²α = 2cos²α − 1.
a sin x + b cos x = √(a² + b²) sin(x + φ): the values fill [−√(a² + b²), √(a² + b²)].

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