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3.3 Trigonometric Functions

🎓 Class 11 Mathematics CBSE Theory Ch 3 — Trigonometric Functions ⏱ ~15 min
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This MCQ module is based on: 3.3 Trigonometric Functions

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Targeting Class 11 level in Trigonometry, with Advanced difficulty.

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3.3 Trigonometric Functions

In earlier classes you defined sine and cosine for an acute angle in a right triangle. We now extend the definitions to any real-valued angle using the unit circle?.

Place the unit circle (centre O, radius 1) on the coordinate plane. Let \(A=(1,0)\). For any real number \(x\), wrap a length \(x\) anti-clockwise along the circle from \(A\) (or clockwise if \(x\) is negative). Call the landing point \(P(a,b)\). Then we define:

Definition: Sine and Cosine
\[\boxed{\;\cos x=a,\qquad \sin x=b\;}\] where \((a,b)\) are the coordinates of the point \(P\) on the unit circle that corresponds to the arc-length / angle \(x\).
X Y A(1,0) P(a,b) a = cos x b = sin x x O Unit circle: any real angle x ↔ unique point P(cos x, sin x).
Fig 3.6: Definition of sin x and cos x on the unit circle.

Since \(P\) lies on the unit circle, \(a^2+b^2=1\), giving the most fundamental identity:

Pythagorean identity
\[\sin^2 x+\cos^2 x=1\quad\text{for every real } x.\]

The four quadrantal angles \(0,\ \pi/2,\ \pi,\ 3\pi/2\) put \(P\) at \((1,0)\), \((0,1)\), \((-1,0)\), \((0,-1)\) respectively, so we read off:

x0π/2π3π/22π
cos x10−101
sin x010−10
Periodicity
A full revolution corresponds to \(2\pi\), so the same point is reached again. Therefore for every integer \(n\) and every real \(x\): \[\cos(2n\pi+x)=\cos x,\qquad \sin(2n\pi+x)=\sin x.\] The functions \(\sin\) and \(\cos\) repeat every \(2\pi\); we say they are periodic with period \(2\pi\).

The other four trigonometric functions

The remaining four are defined as ratios, with the natural restrictions on the denominator:

Definitions (with domain restrictions)
\(\displaystyle \operatorname{cosec} x=\dfrac{1}{\sin x}\), \(x\neq n\pi\)
\(\displaystyle \sec x=\dfrac{1}{\cos x}\), \(x\neq (2n+1)\dfrac{\pi}{2}\)
\(\displaystyle \tan x=\dfrac{\sin x}{\cos x}\), \(x\neq (2n+1)\dfrac{\pi}{2}\)
\(\displaystyle \cot x=\dfrac{\cos x}{\sin x}\), \(x\neq n\pi\)
where \(n\) is any integer.

Dividing the Pythagorean identity \(\sin^2 x+\cos^2 x=1\) by \(\cos^2 x\) and \(\sin^2 x\) respectively gives the two consequences:

\[1+\tan^2 x=\sec^2 x,\qquad 1+\cot^2 x=\operatorname{cosec}^2 x.\]

Values at the standard angles

Combining the right-triangle values with the unit-circle convention:

x →0π/6π/4π/3π/2π3π/22π
sin x01/21/√2√3/210−10
cos x1√3/21/√21/20−101
tan x01/√31√3n.d.0n.d.0

("n.d." = not defined.) The reciprocal functions cosec, sec, cot follow by inversion.

3.3.1 Sign of trigonometric functions

Whether each of the six functions is positive or negative depends only on which quadrant the angle's terminal side lies in. Recall: \(\cos x\) is the x-coordinate and \(\sin x\) is the y-coordinate of \(P\); the others are derived from these.

X Y (1,0) (0,1) (−1,0) (0,−1) I: All + II: sin, cosec + III: tan, cot + IV: cos, sec + P(cos x, sin x)
Fig 3.7: Sign of trig functions by quadrant. Mnemonic: All Students Take Calculus (I → II → III → IV).
FunctionQuadrant I (0 to π/2)II (π/2 to π)III (π to 3π/2)IV (3π/2 to 2π)
sin x, cosec x++−−
cos x, sec x+−−+
tan x, cot x+−+−

3.3.2 Domain and Range of Trigonometric Functions

The domain is the set of inputs for which the function is defined; the range is the set of attainable outputs.

FunctionDomainRange
sin xℝ[−1, 1]
cos xℝ[−1, 1]
tan xℝ \ {(2n+1)π/2}ℝ
cot xℝ \ {nπ}ℝ
sec xℝ \ {(2n+1)π/2}ℝ \ (−1, 1)
cosec xℝ \ {nπ}ℝ \ (−1, 1)

The behaviour in each quadrant — whether each function increases or decreases — can be read directly off the unit circle. For \(\sin x\) starting at the positive x-axis and tracing anti-clockwise:

QuadrantI (0 → π/2)II (π/2 → π)III (π → 3π/2)IV (3π/2 → 2π)
sin x↑ 0 to 1↓ 1 to 0↓ 0 to −1↑ −1 to 0
cos x↓ 1 to 0↓ 0 to −1↑ −1 to 0↑ 0 to 1
tan x↑ 0 to ∞↑ −∞ to 0↑ 0 to ∞↑ −∞ to 0
cot x↓ ∞ to 0↓ 0 to −∞↓ ∞ to 0↓ 0 to −∞

Graphs of the six trigonometric functions

X Y 0 −π/2 π/2 π −π 1 −1 y = sin x
Fig 3.8: y = sin x. Range [−1, 1]. Periodic with period 2π. Zeros at x = nπ.
0 1 −1 y = cos x
Fig 3.9: y = cos x. Same shape as sin x, shifted left by π/2. Zeros at x = (2n+1)π/2.
y = tan x −π/2 π/2 0
Fig 3.10: y = tan x. Vertical asymptotes at x = (2n+1)π/2. Period π.
y = cot x −π/2 π/2 −π π
Fig 3.11: y = cot x. Vertical asymptotes at x = nπ. Period π.
1 −1 y = sec x
Fig 3.12: y = sec x. Asymptotes at odd multiples of π/2. Range (−∞, −1] ∪ [1, ∞). Period 2π.
y = cosec x 1 −1
Fig 3.13: y = cosec x. Asymptotes at multiples of π. Range (−∞, −1] ∪ [1, ∞). Period 2π.
Interactive: Unit-Circle Trig Explorer

Drag the slider to move the point P around the unit circle. Watch all six trig values update live, and see how cos x and sin x are the projections of P onto the axes.

60°
cos x sin x x y
sin x = 0.8660  |  cos x = 0.5000  |  tan x = 1.7321
cosec x = 1.1547  |  sec x = 2.0000  |  cot x = 0.5774

Worked Examples

Example 6. If \(\cos x=-\dfrac{3}{5}\) and \(x\) lies in the third quadrant, find the values of the other five trigonometric functions.
\(\sec x=\dfrac{1}{\cos x}=-\dfrac{5}{3}\). Now \(\sin^2 x=1-\cos^2 x=1-\dfrac{9}{25}=\dfrac{16}{25}\), so \(\sin x=\pm\dfrac{4}{5}\). In the third quadrant \(\sin x\) is negative, hence \(\sin x=-\dfrac{4}{5}\) and \(\operatorname{cosec} x=-\dfrac{5}{4}\). \(\tan x=\dfrac{\sin x}{\cos x}=\dfrac{-4/5}{-3/5}=\dfrac{4}{3}\), \(\cot x=\dfrac{3}{4}\).
Example 7. If \(\cot x=-\dfrac{5}{12}\) and \(x\) lies in the second quadrant, find the other five trigonometric functions.
\(\tan x=\dfrac{1}{\cot x}=-\dfrac{12}{5}\). Then \(\sec^2 x=1+\tan^2 x=1+\dfrac{144}{25}=\dfrac{169}{25}\), so \(\sec x=\pm\dfrac{13}{5}\). In the second quadrant \(\sec x\) is negative, so \(\sec x=-\dfrac{13}{5}\) and \(\cos x=-\dfrac{5}{13}\). Now \(\sin x=\tan x\cdot\cos x=\left(-\dfrac{12}{5}\right)\left(-\dfrac{5}{13}\right)=\dfrac{12}{13}\) and \(\operatorname{cosec} x=\dfrac{13}{12}\).
Example 8. Find the value of \(\sin\dfrac{31\pi}{3}\).
Using periodicity \(\sin(2n\pi+x)=\sin x\): \[\sin\dfrac{31\pi}{3}=\sin\!\left(10\pi+\dfrac{\pi}{3}\right)=\sin\dfrac{\pi}{3}=\dfrac{\sqrt 3}{2}.\]
Example 9. Find the value of \(\cos(-1710°)\).
Cos is even and periodic with period 360°. So \[\cos(-1710°)=\cos 1710°=\cos(1710°-5\times 360°)=\cos(1710°-1800°)=\cos(-90°)=\cos 90°=0.\]
Activity: ASTC Sign Detective
L4 Analyse
Materials: Calculator (or graphing tool), graph paper.
Predict: If you double an angle that lies in Quadrant II, in which quadrant does the new angle lie? Will sin double-angle have the same sign as the original sin?
  1. Pick five angles, one from each of the regions \([0,\pi/2)\), \([\pi/2,\pi)\), \([\pi, 3\pi/2)\), \([3\pi/2, 2\pi)\), and one negative angle.
  2. For each, compute sin, cos, tan and identify the sign without a calculator using ASTC. Then verify with a calculator.
  3. Tabulate "predicted sign" vs "actual sign" — they should match for all five.
  4. Now double each angle. Re-classify the doubled angle's quadrant and predict its signs.
  5. Identify whether the sign of sin doubles, flips, or stays — and explain why using the unit circle.
Doubling an angle in Q2 (\(\pi/2 < x < \pi\)) gives \(2x\in(\pi, 2\pi)\) — so the new angle is in Q3 or Q4. Sin x was positive in Q2, but sin 2x is negative. The sign change is forced by the geometry: doubling rotates P past the negative y-axis. The exercise builds intuition for why "double angle" formulas like \(\sin 2x=2\sin x\cos x\) (next part) involve sign-changing factors.

Competency-Based Questions

Scenario: A Ferris wheel of radius 8 m rotates anti-clockwise once every 60 seconds. The lowest point is 1 m above the ground. A passenger boards at the bottom (call this position the start, \(t=0\)). Their height \(h(t)\) above the ground at time \(t\) seconds satisfies \(h(t)=9-8\cos\!\left(\dfrac{\pi t}{30}\right)\).
Q1. The maximum height attained is:
L3 Apply
  • (a) 1 m
  • (b) 8 m
  • (c) 9 m
  • (d) 17 m
Answer: (d) 17 m. Max of \(-\cos\theta\) is +1, so \(h_{\max}=9-8(-1)=17\) m. (Reached at \(t=30\) s when \(\theta=\pi\).)
Q2. (Fill in the blank) The angle \(\theta=\dfrac{\pi t}{30}\) lies in the third quadrant for \(t\) in the interval ____ to ____ seconds.
L4 Analyse
Answer: Q3 means \(\pi<\theta<3\pi/2\), i.e. \(\pi<\pi t/30<3\pi/2\), i.e. \(30
Q3. (True/False) "\(\tan\theta\) is undefined at \(t=15\) s." Justify.
L5 Evaluate
True. At \(t=15\), \(\theta=\pi/2\). Since \(\cos(\pi/2)=0\), the ratio \(\tan\theta=\sin\theta/\cos\theta\) is undefined. Geometrically: at \(t=15\) the passenger is at the 3-o'clock position; the radius is horizontal, so the "tangent of the angle" — which is the slope of OP — is infinite.
Q4. Which of the six trig values for \(\theta=\dfrac{2\pi}{3}\) (i.e. \(t=20\) s) is positive? Choose all that apply.
L3 Apply
  • (a) sin θ
  • (b) cos θ
  • (c) tan θ
  • (d) cosec θ
Answer: (a) and (d). \(2\pi/3\) is in Quadrant II (between \(\pi/2\) and \(\pi\)). By ASTC, only sin and its reciprocal cosec are positive; cos, sec, tan, cot are negative.
Q5. Design: write a function \(g(t)\) describing the horizontal east–west displacement (in metres) of the passenger from the wheel's centre, in terms of \(t\). Verify it returns 0 at \(t=0\) and \(t=30\), and reaches +8 at \(t=15\). Specify what trig function you used and explain why.
L6 Create
Solution: The vertical position from centre is \(-8\cos(\pi t/30)\); the horizontal is \(8\sin(\pi t/30)\). So \(g(t)=8\sin(\pi t/30)\). Check: \(g(0)=8\sin 0=0\) ✓; \(g(15)=8\sin(\pi/2)=8\) ✓; \(g(30)=8\sin\pi=0\) ✓. Sin was used because the horizontal coordinate of P on a unit circle (with the ferris-wheel parametrisation starting at the bottom) is \(\sin\theta\), where \(\theta\) is the angle measured from the downward vertical.

Assertion–Reason Questions

Assertion (A): The range of \(\sec x\) is \((-\infty,-1]\cup[1,\infty)\).
Reason (R): \(\sec x=1/\cos x\) and \(|\cos x|\le 1\) for every real \(x\) where \(\cos x\ne 0\).
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Since \(0<|\cos x|\le 1\), \(|\sec x|=1/|\cos x|\ge 1\). So sec x cannot lie in \((-1,1)\). R provides the precise reason for A.
Assertion (A): \(\tan x\) is positive in the third quadrant.
Reason (R): In Q3 both \(\sin x\) and \(\cos x\) are negative; the ratio of two negatives is positive.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). ASTC: only T (and cot) are positive in Q3. R is the algebraic reason — sign of a quotient is the product of the signs.
Assertion (A): \(\sin\) is a periodic function with period \(\pi\).
Reason (R): \(\sin(\pi+x)=-\sin x\), which is not equal to \(\sin x\) in general.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (d). A is false — sin has period 2π, not π. R is true and is in fact the proof that π is NOT a period of sin.

Frequently Asked Questions

How are sin and cos defined on the unit circle?
For an angle x measured anti-clockwise from the positive x-axis, the corresponding point P on the unit circle has coordinates (cos x, sin x). So cos x is the x-coordinate of P and sin x is its y-coordinate.
In which quadrants is sin x positive?
Sin x is positive in the first quadrant (0 < x < π/2) and the second quadrant (π/2 < x < π), and negative in the third (π < x < 3π/2) and fourth (3π/2 < x < 2π) quadrants.
What is the domain and range of tan x?
Domain: all real numbers except odd multiples of π/2 (i.e. x ≠ (2n+1)π/2). Range: all real numbers.
Why is sin x periodic with period 2π?
Adding 2π (one full revolution) to an angle returns to the same point on the unit circle, so the y-coordinate (and hence sin) is unchanged. Thus sin(x + 2π) = sin x for all real x.
Where does the graph of tan x have asymptotes?
The graph of y = tan x has vertical asymptotes at every odd multiple of π/2 — i.e. at x = ±π/2, ±3π/2, ±5π/2, … because cos x = 0 there and tan x = sin x / cos x is undefined.
What is the ASTC rule for trigonometric signs?
In the four quadrants taken in order Q1→Q2→Q3→Q4: All trig functions are positive in Q1; only Sin (and cosec) in Q2; only Tan (and cot) in Q3; only Cos (and sec) in Q4. Mnemonic: 'All Students Take Calculus'.
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