Trigonometric identities and equations flashcards

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Identity (symbol ≡)
A statement that is true for every value of the variable, e.g. sin²θ + cos²θ ≡ 1; not just for particular solutions.
sin²θ
(sin θ)², the square of the value of sin θ; not sin(θ²).
Principal value
The single angle the calculator's inverse function gives: sin⁻¹ and tan⁻¹ in −90° to 90°, cos⁻¹ in 0° to 180°.
sin θ and cos θ on the unit circle
A radius at angle θ meets the unit circle at P(cos θ, sin θ): cos θ is the x-coordinate and sin θ the y-coordinate.
Period of sin, cos and tan
sin and cos: 360° (2π); tan: 180° (π). sin(nθ + α) has period 360°/n.
tan θ in terms of sin θ and cos θ
tan θ ≡ sin θ / cos θ (cos θ ≠ 0). (No — learn it)
Pythagorean identity
sin²θ + cos²θ ≡ 1. (No — learn it)
Rearrangements of sin²θ + cos²θ ≡ 1
sin²θ ≡ 1 − cos²θ and cos²θ ≡ 1 − sin²θ. (No — derive it)
All solutions of sin θ = k (|k| ≤ 1)
θ = α and 180° − α, plus or minus multiples of 360° (radians: α, π − α, ± 2π). (No — derive it)
All solutions of cos θ = k (|k| ≤ 1)
θ = α and 360° − α (i.e. ±α), plus or minus multiples of 360°. (No — derive it)
All solutions of tan θ = k
θ = α plus or minus multiples of 180° (π in radians). (No — derive it)
Exact values at 30°, 45°, 60°
sin: ½, √2/2, √3/2; cos: √3/2, √2/2, ½; tan: √3/3, 1, √3. (No — learn it)
Which ratios are positive in each quadrant?
0°–90°: all; 90°–180°: sin; 180°–270°: tan; 270°–360°: cos.
Symmetry of y = sin θ used for solving
Symmetric about θ = 90°, so sin(180° − α) = sin α.
Symmetry of y = cos θ used for solving
Symmetric about θ = 0° (an even function), so cos(−α) = cos α.
Shape of y = tan θ
Increasing branches repeating every 180°, with vertical asymptotes at θ = ±90°, ±270°, … where cos θ = 0.
How many solutions does sin θ = k have in one period?
Two if −1 < k < 1; one if k = ±1; none if |k| > 1. Read it from where y = k cuts the curve.
Why is sin²θ + cos²θ = 1?
P(cos θ, sin θ) is on a circle of radius 1, so by Pythagoras x² + y² = 1.
Why can you divide a sin θ = b cos θ by cos θ?
If cos θ = 0 the equation would force sin θ = 0 too, which never happens at the same angle; so no solutions are lost.
Why must you not divide sin θ cos θ = k sin θ by sin θ?
It assumes sin θ ≠ 0 and loses the solutions where sin θ = 0. Factorise: sin θ(cos θ − k) = 0.
Why change the interval before solving sin(nθ + α) = k?
The bracket runs over n times the range, so it has about n times as many solutions; listing them in the new interval finds them all.
How do you turn 2 sin²x + 3 cos x = 0 into one function?
Replace sin²x with 1 − cos²x to get a quadratic in cos x.
Solve sin θ = 0.3 for 0° ≤ θ < 360°.
sin⁻¹ 0.3 = 17.5°; second solution 180° − 17.5° = 162.5°.
Solve tan θ = −3 for 0° ≤ θ < 360°.
tan⁻¹(−3) = −71.6°; add 180° and 360°: θ = 108.4°, 288.4°.
sin θ = 3/5 and θ is obtuse. Find cos θ and tan θ.
cos²θ = 1 − 9/25 = 16/25; obtuse so cos θ = −4/5, tan θ = −3/4.
Solve sin 2θ = ½ for 0° ≤ θ < 360°.
0° ≤ 2θ < 720°: 2θ = 30°, 150°, 390°, 510°, so θ = 15°, 75°, 195°, 255°.
Solve 2 cos²x = 3 sin x + 3 for 0° ≤ x < 360°.
2(1 − sin²x) = 3 sin x + 3 → 2 sin²x + 3 sin x + 1 = 0 → sin x = −½ or −1: x = 210°, 270°, 330°.
Solve tan x = 1 for 0 ≤ x < 2π.
x = π/4 and π/4 + π = 5π/4.
Prove (1 − cos²θ)/sin θ ≡ sin θ.
1 − cos²θ = sin²θ, and sin²θ / sin θ = sin θ.
Interval written with π
Answers in radians: calculator in radian mode; exact multiples of π or 3 s.f.
A quadratic gives sin θ = 2
Reject it with a reason: −1 ≤ sin θ ≤ 1, so there are no solutions from that factor.
Solving tan θ = k: second solution
α + 180°, not 180° − α (that is the rule for sine).
“Hence” after solving an equation in x
Reuse your unrounded values with x replaced by the new bracket; halve or convert them, then round.
Proving an identity
Start from one side and finish with exactly the other; never work on both sides at once.
Accuracy for angles
Degrees to 1 d.p. and radians to 3 s.f. unless the question says otherwise.

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