Q.Find the values of
Concept understanding — Trigonometric Functions and Their Graphs
Trigonometric ratios were first defined only for acute angles inside a right triangle. This concept develops the full picture: trigonometric functions defined for any angle (or any real number), the rules that govern their signs and symmetry, and how they behave graphically.
1. From ratios to functions — the coordinate definition. Place an angle θ in standard position at the origin, initial side along the positive x-axis. Let P(x,y) be any point (other than the origin) on the terminal side, and r=OP=x2+y2. Then
sinθ=ry,cosθ=rx,tanθ=xy (x=0),cotθ=yx (y=0),cosecθ=yr (y=0),secθ=xr (x=0).
This matches the right-triangle ratios exactly when θ is acute, but now works for any θ. Since ∣x∣,∣y∣≤r, we always have −1≤sinθ≤1 and −1≤cosθ≤1. The value obtained does not depend on which point P is chosen on the terminal side (similar triangles give the same ratio).
Taking P on the unit circle x2+y2=1 makes r=1, so cosθ=x and sinθ=y directly — the point P is (cosθ,sinθ). This gives the exact values at the quadrantal angles:
| θ | 0∘ | 90∘ | 180∘ | 270∘ | 360∘ |
|---|---|---|---|---|---|
| cosθ | 1 | 0 | −1 | 0 | 1 |
| sinθ | 0 | 1 | 0 | −1 | 0 |
from which sinθ=0⟺θ=nπ and cosθ=0⟺θ=(2n+1)π/2 for integer n; tanθ is undefined exactly where cosθ=0. Also, any two angles differing by a whole multiple of 360∘ (2π) give identical values for every trigonometric function.
2. Signs — the ASTC rule. Since x,y change sign across the four quadrants while r>0 always, each function's sign is fixed by the quadrant of θ:
| Quadrant | Positive | Negative |
|---|---|---|
| I (x>0,y>0) | all six | — |
| II (x<0,y>0) | sin,cosec | cos,sec,tan,cot |
| III (x<0,y<0) | tan,cot | sin,cosec,cos,sec |
| IV (x>0,y<0) | cos,sec | sin,cosec,tan,cot |
Remembered by the mnemonic 'All Students Take Chocolate' (quadrants I, II, III, IV in order: All, Sine, Tangent, Cosine, each together with its reciprocal). Given one function's value and the quadrant, the Pythagorean identity sin2θ+cos2θ=1 (or 1+tan2θ=sec2θ) fixes the paired ratio up to a sign, and ASTC picks the correct sign; the remaining four functions then follow from the quotient identities (tan=sin/cos, cot=cos/sin) and reciprocal identities (cosec=1/sin, sec=1/cos).
3. Extending to real numbers — the wrapping function. For applications beyond geometry (waves, oscillations, calculus), trigonometric functions are extended to any real number t, not just an angle. Starting at A(1,0) on the unit circle, wrap an arc of length ∣t∣ around the circle — anticlockwise if t>0, clockwise if t<0 — to reach a point B(x,y). Since the circle has radius 1, the arc length equals the subtended angle θ in radians, so we simply define sint=sinθ=y and cost=cosθ=x. Every property already established for angles (bounds, signs, periodicity) carries over unchanged to real-number inputs.
4. Allied angles. Two angles are allied if their sum or difference is an integer multiple of π/2: so −θ, π/2±θ, π±θ, 3π/2±θ, 2π±θ are all allied to θ. Reflecting P(a,b) across the x-axis (to find the ratios of −θ) gives P′(a,−b), so sin(−θ)=−sinθ and cos(−θ)=cosθ (and hence tan(−θ)=−tanθ, etc.) — these two negative-angle facts are also exactly why cosine is even and sine is odd (point 6 below). A quarter-turn rotation similarly gives sin(90∘+θ)=cosθ, cos(90∘+θ)=−sinθ. Every allied-angle case (for 0<θ<π/2) is summarised in one table:
| −θ | 2π−θ | 2π+θ | π−θ | π+θ | 23π−θ | 23π+θ | 2π−θ | 2π+θ | |
|---|---|---|---|---|---|---|---|---|---|
| sine | −sinθ | cosθ | cosθ | sinθ | −sinθ | −cosθ | −cosθ | −sinθ | sinθ |
| cosine | cosθ | sinθ | −sinθ | −cosθ | −cosθ | −sinθ | sinθ | cosθ | cosθ |
| tangent | −tanθ | cotθ | −cotθ | −tanθ | tanθ | cotθ | −cotθ | −tanθ | tanθ |
(cosecant/secant/cotangent follow as reciprocals). Two rules rebuild this table instead of memorising it: (a) keep vs. co-change — an even multiple of π/2 away from θ (−θ,π±θ,2π±θ) keeps the same function name; an odd multiple (π/2±θ,3π/2±θ) switches to the co-function (sine↔cosine, tan↔cot, sec↔cosec); (b) the sign is then read off ASTC by checking which quadrant the allied angle itself falls into (treating θ as a small acute angle).
The practical technique: reduce any angle by (i) stripping off whole multiples of 360∘/2π (never changes the value), then (ii) writing what remains as 180∘±(acute) or 90∘±(acute) etc., and applying the table — e.g. sin150∘=sin(180∘−30∘)=sin30∘=21, or tan315∘=tan(360∘−45∘)=−tan45∘=−1.
5. Periodicity. f is periodic with period p (the smallest such positive number) if f(x+p)=f(x) for all x. Since a full 2π rotation returns the terminal side to itself, sin,cos,cosec,sec all have period 2π. But tan and cot repeat twice as fast, with period π, because a half rotation already sends both x and y to −x,−y, leaving the ratio y/x unchanged.
6. Graphs of sinx and cosx, and odd/even symmetry. The graph of y=sinx is a wave bounded between −1 and 1, repeating every 2π, rising on (−π/2,π/2) and falling on (π/2,3π/2), crossing zero at every multiple of π. The graph of y=cosx has the identical shape, just shifted left by π/2, since cosx=sin(x+π/2). A function f is even if f(−x)=f(x) (graph symmetric about the y-axis) and odd if f(−x)=−f(x) (graph symmetric about the origin). Because cos(−x)=cosx and sin(−x)=−sinx: cosine (and secant) are even, while sine, tangent, cosecant, and cotangent are all odd. To classify a combination like f(x)=sin2x−2cos2x−cosx: compute f(−x) using the negative-angle identities and compare to f(x) and −f(x) — here f(−x)=f(x) exactly (every term is a sin2, cos2, or bare cos, all unaffected by x→−x), so f is even; but f(x)=sinx+cosx gives f(−x)=−sinx+cosx, which is neither f(x) nor −f(x), so this sum is neither odd nor even. Not every function built from an odd piece and an even piece is itself odd or even — the test must be applied to the whole combination, term by term.
Reduce each angle to a first-quadrant reference angle by stripping off full rotations (360∘ or 2π) and applying the correct allied-angle identity, then fix the sign from the ASTC quadrant rule.
- (i) 480∘→120∘ (ii) −1110∘→ use −sin(1110∘), 1110∘→30∘ (iii) 300∘→−60∘ equiv. (iv) 1050∘→330∘ (v) 660∘→300∘ (vi) 19π/3→π/3 (vii) −11π/3→π/3 with a sign flip.
(i) 23 (ii) −21 (iii) 21 (iv) −31 (v) −31 (vi) 3 (vii) 23.
Every angle here reduces, after stripping off whole rotations of 360∘ (2π), to an allied angle of a standard acute angle; the table of §3.4.3 (or the negative-angle identity for a negative angle) then gives the value directly.
Step 1. Part (i): reduce 480∘. 480∘=360∘+120∘, and adding a full rotation never changes a trigonometric value, so sin480∘=sin120∘=sin(180∘−60∘)=sin60∘=23.
Step 2. Part (ii): reduce −1110∘. First use sin(−θ)=−sinθ: sin(−1110∘)=−sin1110∘. Since 1110∘=3×360∘+30∘, sin1110∘=sin30∘=21. So sin(−1110∘)=−21.
Step 3. Part (iii): reduce 300∘. 300∘=360∘−60∘, so cos300∘=cos(−60∘)=cos60∘=21 (using cos(−θ)=cosθ, equally read off the table's 2π−θ column).
Step 4. Part (iv): reduce 1050∘. 1050∘=2×360∘+330∘, so tan1050∘=tan330∘=tan(360∘−30∘)=−tan30∘=−31.
Step 5. Part (v): reduce 660∘. 660∘=360∘+300∘, so cot660∘=cot300∘=cot(360∘−60∘)=−cot60∘=−31.
Step 6. Part (vi): reduce 319π. 319π=6π+3π; stripping off 6π (three full rotations) leaves tan319π=tan3π=3.
Step 7. Part (vii): reduce −311π. sin(−311π)=−sin311π. Now 311π=4π−3π, so, dropping the 4π (two full rotations), sin311π=sin(−3π)=−sin3π=−23. Hence sin(−311π)=−(−23)=23.
(i) 23 (ii) −21 (iii) 21 (iv) −31 (v) −31 (vi) 3 (vii) 23.
Angle reduction via full-rotation stripping + allied-angle identities
- Stripping off a rotation that is not a whole multiple of 360∘ (2\pi), leaving an angle that isn't actually equivalent
- Getting the sign wrong by not checking which quadrant the reduced angle actually falls in before reading off ASTC
- CBSE 2026Set ANNUAL1 markMCQQ.Which of the following is not true?(a) tanθ=25(b) sinθ=−43(c) secθ=41(d) cosθ=−1
›Reveal solutionSolution
tanθ and sinθ can take many values, and cosθ=−1 is achievable at θ=180∘; but secθ=1/cosθ always satisfies ∣secθ∣≥1, so secθ=1/4 is impossible.
Check each option:
- tanθ=25: tangent is unbounded (ranges over all reals), so this is achievable — true.
- sinθ=−3/4: sine ranges over [−1,1], and −3/4 lies in this range — true, achievable.
- secθ=1/4: since secθ=1/cosθ and −1≤cosθ≤1 (with cosθ=0), we get ∣secθ∣≥1 always. So secθ=1/4 (magnitude <1) is impossible — this is NOT true.
- cosθ=−1: achieved at θ=180∘ — true.
✓Final answerThe correct option is (c) secθ=41 (not true, since ∣secθ∣≥1 always).
- CBSE 2023Set ANNUAL1 markMCQQ.Which of the following is not true?(a) tanθ=25(b) sinθ=−43(c) secθ=41(d) cosθ=−1
›Reveal solutionSolution
tanθ and cotθ can be any real number, and sinθ,cosθ range in [−1,1], but secθ and cscθ can never lie strictly between −1 and 1.
Check each option:
- tanθ=25: tangent is unbounded, this is achievable. Valid.
- sinθ=−43: within [−1,1]. Valid.
- secθ=41: since secθ=1/cosθ and ∣cosθ∣≤1, we always have ∣secθ∣≥1. A value of 41 has magnitude less than 1, so it is impossible.
- cosθ=−1: within [−1,1], attained at θ=180∘. Valid.
Only secθ=41 is impossible.
✓Final answersecθ=41 is NOT true.
- CBSE 2022Set ANNUAL1 markMCQQ.The value of tan90° is:(a) 23(b) 0(c) 1(d) ∞
›Reveal solutionSolution
tan90°=cos90°sin90°=01, which is undefined; we say it tends to infinity.
tanθ=cosθsinθ. At θ=90°, sin90°=1 and cos90°=0.
Dividing by zero is undefined, and as θ→90° from either side, tanθ grows without bound -- so tan90° is conventionally described as ∞ (undefined).
✓Final answerThe correct option is (d) ∞.
- CBSE 2019Set ANNUAL1 markMCQQ.Which of the following is not a periodic function with period 2π?(a) tanx(b) cosx(c) sinx(d) cosecx
›Reveal solutionSolution
sinx, cosx, and cosecx have fundamental period exactly 2π; tanx's fundamental period is π, so it is the odd one out among the listed periods.
The fundamental (smallest positive) period of each function:
- sinx: period 2π.
- cosx: period 2π.
- cosecx=1/sinx: period 2π (same as sinx).
- tanx: period π, since tan(x+π)=tanx — it repeats twice as fast as the others.
So among the options, tanx is the function whose characteristic period is not 2π.
✓Final answerThe correct option is (a) tanx.
- CBSE 2019Set ANNUAL1 markMCQQ.The minimum and the maximum values of ∣cosx∣−2 are respectively:(a) 0 and 2(b) −2 and 0(c) −2 and −1(d) −1 and 1
›Reveal solutionSolution
∣cosx∣ ranges over [0,1], so ∣cosx∣−2 ranges over [−2,−1]: minimum −2, maximum −1.
For any real x, −1≤cosx≤1, so 0≤∣cosx∣≤1.
Subtracting 2 from every part of this inequality: 0−2≤∣cosx∣−2≤1−2, i.e. −2≤∣cosx∣−2≤−1.
The minimum value −2 occurs when ∣cosx∣=0 (e.g. x=π/2), and the maximum value −1 occurs when ∣cosx∣=1 (e.g. x=0).
✓Final answerThe correct option is (c) −2 and −1.
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