Signs of trigonometric functions by quadrants

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MODULE: Trigonometric Functions and Identities

[EXEC: MICRO_CORE]

✖️ 1. Determining signs using unit circle coordinates and the ASTC rule

📍 ASTC Rule and Unit Circle Signs

  • Quadrant I: All trig functions are positive (A = All).
  • Quadrant II: Only sin is positive (S = Students).
  • Quadrant III: Only tan is positive (T = Take).
  • Quadrant IV: Only cos is positive (C = Calculus).
  • On the unit circle, sin⁡(θ)=y\sin(\theta) = y and cos⁡(θ)=x\cos(\theta) = x, so signs follow coordinate signs.
  • tan⁡(θ)=yx\tan(\theta) = \frac{y}{x}, so it's positive when xx and yy have the same sign.

Example: At θ=150∘\theta = 150^\circ (Quadrant II), sin⁡(150∘)>0\sin(150^\circ) > 0, cos⁡(150∘)<0\cos(150^\circ) < 0, tan⁡(150∘)<0\tan(150^\circ) < 0.

💡 Memory hook: "All Students Take Calculus" counterclockwise from Quadrant I.

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1. Determining signs using unit circle coordinates and the ASTC rule

Determining signs using unit circle coordinates and the ASTC rule

For any angle θ\theta on the unit circle, the point (x,y)(x, y) satisfies cos⁡(θ)=x\cos(\theta) = x and sin⁡(θ)=y\sin(\theta) = y. The tangent function is defined as tan⁡(θ)=yx\tan(\theta) = \frac{y}{x} when x≠0x \neq 0.

The sign of each trigonometric function depends on which quadrant the terminal side of θ\theta lies in, determined by the signs of xx and yy.

Core Rules (ASTC mnemonic):

  • Quadrant I: All functions positive (x>0,y>0x > 0, y > 0)
  • Quadrant II: Sine positive only (x<0,y>0x < 0, y > 0)
  • Quadrant III: Tangent positive only (x<0,y<0x < 0, y < 0)
  • Quadrant IV: Cosine positive only (x>0,y<0x > 0, y < 0)

This pattern allows immediate determination of function signs without computation.

Example: For θ=150∘\theta = 150^\circ (Quadrant II), sin⁡(150∘)>0\sin(150^\circ) > 0, cos⁡(150∘)<0\cos(150^\circ) < 0, tan⁡(150∘)<0\tan(150^\circ) < 0.

TASK_1[0 / 3]
LVL_2
STRC: CLASSIFY

Determine the sign of cos⁡(225∘)\cos(225^\circ).

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✖️ 2. Even and odd properties of trigonometric functions

🔄 Even and Odd Function Properties

  • Cosine is even: cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x) (sign stays the same).
  • Sine is odd: sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x) (sign flips).
  • Tangent is odd: tan⁡(−x)=−tan⁡(x)\tan(-x) = -\tan(x) (sign flips).
  • Negative angles rotate clockwise instead of counterclockwise.
  • Use these to simplify expressions like sin⁡(−30∘)=−sin⁡(30∘)=−12\sin(-30^\circ) = -\sin(30^\circ) = -\frac{1}{2}.

Example: cos⁡(−45∘)=cos⁡(45∘)=22\cos(-45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}, but tan⁡(−45∘)=−tan⁡(45∘)=−1\tan(-45^\circ) = -\tan(45^\circ) = -1.

💡 Memory hook: Cosine is "even" like a mirror; sine and tangent "flip" for negatives.

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2. Even and odd properties of trigonometric functions

Even and odd properties of trigonometric functions

A function ff is even if f(−x)=f(x)f(-x) = f(x) for all xx in its domain, and odd if f(−x)=−f(x)f(-x) = -f(x). These symmetries determine how trigonometric functions behave under angle negation.

Cosine is even because reflecting across the yy-axis preserves the xx-coordinate. Sine and tangent are odd because reflection reverses the yy-coordinate and the ratio yx\frac{y}{x}.

Core Rules:

  • Cosine is even: cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x)
  • Sine is odd: sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x)
  • Tangent is odd: tan⁡(−x)=−tan⁡(x)\tan(-x) = -\tan(x)
  • Secant is even, cosecant and cotangent are odd

These properties simplify evaluation of negative angles by reducing them to positive angle calculations.

Example: sin⁡(−30∘)=−sin⁡(30∘)=−12\sin(-30^\circ) = -\sin(30^\circ) = -\frac{1}{2}, while cos⁡(−30∘)=cos⁡(30∘)=32\cos(-30^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2}.

TASK_1[0 / 3]
LVL_2
STRC: CLASSIFY

Which of the following expressions is equivalent to cos⁡(−45∘)\cos(-45^\circ)?

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✖️ 3. Finding exact values given one function and quadrant constraint

🧮 Finding All Trig Functions from One Value

  • Use Pythagorean identity: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to find the missing function.
  • The quadrant determines the sign of the result (use ASTC).
  • Once you have sin⁡\sin and cos⁡\cos, compute tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.
  • For reciprocals: csc⁡=1sin⁡\csc = \frac{1}{\sin}, sec⁡=1cos⁡\sec = \frac{1}{\cos}, cot⁡=1tan⁡\cot = \frac{1}{\tan}.

Example: If sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and θ\theta is in Quadrant II, then cos⁡(θ)=−1−(35)2=−45\cos(\theta) = -\sqrt{1 - (\frac{3}{5})^2} = -\frac{4}{5} (negative in QII), so tan⁡(θ)=3/5−4/5=−34\tan(\theta) = \frac{3/5}{-4/5} = -\frac{3}{4}.

💡 Memory hook: Pythagorean identity + quadrant sign = all six functions unlocked.

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3. Finding exact values given one function and quadrant constraint

Finding exact values given one function and quadrant constraint

Given one trigonometric function value and the quadrant of θ\theta, all other function values can be determined using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 and sign rules.

The quadrant constraint determines the sign of each function, while the identity provides magnitude relationships.

Core Rules:

  • Use sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to find the missing primary function
  • Apply quadrant-specific signs from ASTC to determine positive or negative values
  • Compute tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} and reciprocal functions as needed
  • Always verify that signs match the given quadrant

This process converts partial information into complete trigonometric data for the angle.

Example: Given sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and θ\theta in Quadrant II, then cos⁡(θ)=−1−(35)2=−45\cos(\theta) = -\sqrt{1 - (\frac{3}{5})^2} = -\frac{4}{5} (negative in QII), so tan⁡(θ)=3/5−4/5=−34\tan(\theta) = \frac{3/5}{-4/5} = -\frac{3}{4}.

TASK_1[0 / 3]
LVL_3
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Given cos⁡(θ)=4/5\cos(\theta) = 4/5 and θ\theta is in Quadrant IV. Find the exact value of sin⁡(θ)\sin(\theta).

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✖️ 4. Applications in 2D kinematics velocity vectors

🚀 Velocity Vectors and Quadrant Direction

  • A velocity vector (vx,vy)(v_x, v_y) has direction θ=tan⁡−1(vyvx)\theta = \tan^{-1}(\frac{v_y}{v_x}).
  • The signs of vxv_x and vyv_y determine which quadrant the vector points toward.
  • Positive vxv_x means moving right; negative means left.
  • Positive vyv_y means moving up; negative means down.
  • Always check the quadrant to adjust tan⁡−1\tan^{-1} output (calculators give principal values only).

Example: If vx=−3v_x = -3 m/s and vy=4v_y = 4 m/s, the vector is in Quadrant II, so θ=180∘−tan⁡−1(43)≈180∘−53∘=127∘\theta = 180^\circ - \tan^{-1}(\frac{4}{3}) \approx 180^\circ - 53^\circ = 127^\circ from the positive x-axis.

💡 Memory hook: Sign of components = quadrant; adjust angle accordingly for true direction.

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4. Applications in 2D kinematics velocity vectors

Applications in 2D kinematics velocity vectors

In two-dimensional motion, velocity vectors v⃗=(vx,vy)\vec{v} = (v_x, v_y) have components that correspond to unit circle coordinates. The direction angle θ\theta measured counterclockwise from the positive xx-axis satisfies vx=∣v⃗∣cos⁡(θ)v_x = |\vec{v}|\cos(\theta) and vy=∣v⃗∣sin⁡(θ)v_y = |\vec{v}|\sin(\theta).

Quadrant analysis determines the physical direction of motion based on component signs.

Core Rules:

  • Quadrant I: Motion northeast (vx>0,vy>0v_x > 0, v_y > 0)
  • Quadrant II: Motion northwest (vx<0,vy>0v_x < 0, v_y > 0)
  • Quadrant III: Motion southwest (vx<0,vy<0v_x < 0, v_y < 0)
  • Quadrant IV: Motion southeast (vx>0,vy<0v_x > 0, v_y < 0)

Correct quadrant identification ensures accurate angle calculation using θ=arctan⁡(vyvx)\theta = \arctan(\frac{v_y}{v_x}) with appropriate adjustments.

Example: A projectile with vx=−4v_x = -4 m/s and vy=3v_y = 3 m/s moves in Quadrant II at angle θ=180∘−arctan⁡(34)≈143.1∘\theta = 180^\circ - \arctan(\frac{3}{4}) \approx 143.1^\circ.

TASK_1[0 / 3]
LVL_2
STRC: CLASSIFY

A drone has a velocity vector with components vx=−15v_x = -15 m/s and vy=−8v_y = -8 m/s. Based on quadrant analysis, in which physical direction is the drone moving?

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