Q.Find the range of the function 2cosx−11.
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🔒 Start your 14-day free trial to unlock the full solution →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 |
Since cosx ranges over [−1,1], the denominator 2cosx−1 ranges over [−3,1] excluding 0; analyzing the reciprocal on each piece gives the full ran …
Letting u=2cosx−1∈[−3,1]∖{0} and studying y=1/u on the positive and negative parts of that interval gives the full range.
Since cosx∈[−1,1], we get u=2cosx−1∈[−3,1]. But u=0 (i.e. cosx=21) must be excluded since it makes the denominator zero. So u∈[−3,0)∪(0,1].
Let y=u1.
For u∈(0,1]: as u→0+, y→+∞; at u=1, y=1. So y∈[1,∞).
…
- 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. …
- 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. …
- 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.
…
- 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). …
- 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.
…
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