Radians and the Unit Circle: Give Every Angle a Place

CBSE Mathematics · 2026–27 · Class 11

Learn with Rajesh Sir

A triangle is a good beginning. A turning radius takes you further: through a quarter-turn, past a straight angle and even backwards. See how one moving point gives sine and cosine a meaning for every real angle.

By the end, you can…
  • Convert signed angles between degrees and radians.
  • Use arc length and sector area with angles in radians.
  • Read all six functions from a circle point and explain their signs.
  • Reduce angles using symmetry and complete turns.
  • Relate a chord and its half-angle to a right triangle.
About 5 minutes
TOPIC 01

An angle records a turn

Imagine a radius initially pointing right. Its starting ray is the initial side; its final ray is the terminal side. The angle records the turn from one to the other. In the coordinate plane, an angle in standard position starts on the positive xx-axis with its vertex at the origin. This gives everyone the same starting point for describing a rotation.

Counterclockwise turns are positive; clockwise turns are negative. Thus 90∘90^\circ turns upward, while −90∘-90^\circ turns downward. A turn may also be larger than a complete revolution. The angles 30∘30^\circ, 390∘390^\circ and −330∘-330^\circ finish on the same ray. They are coterminal (समान अंतिम भुजा वाले): their differences are complete turns.

For any integer nn, adding 360∘n360^\circ n changes the number of revolutions but leaves the terminal ray unchanged. Keep two ideas separate: an oriented angle includes the direction and amount of rotation; the terminal point tells you where the radius finally arrives. A complete-turn adjustment preserves the trigonometric functions, but it does not make the two recorded rotations literally the same turn.

When reducing a large or negative angle, choose a convenient representative first. Then identify the quadrant or axis. This is more dependable than trying to remember the sign of a function from the size of a number alone. Axis angles belong to the axes rather than to either neighbouring open quadrant.

Work it through

Try a clockwise turn. For −225∘-225^\circ, add one complete turn:

−225∘+360∘=135∘. -225^\circ+360^\circ=135^\circ.

The radius finishes in Quadrant II. Its horizontal coordinate is negative and its vertical coordinate is positive. The original turn was clockwise; its final ray is the same as that of 135∘135^\circ.

Pause. Think. Try.

1. Which angle is coterminal with −45∘-45^\circ?
A hint, pleaseAdd one complete turn.
Read the answer and explanation

315∘315^\circ

−45∘+360∘=315∘-45^\circ+360^\circ=315^\circ. Both rotations finish below the positive xx-axis.
2. An angle in standard position is −120∘-120^\circ. Where does its terminal ray lie?
A hint, pleaseUse −120∘+360∘-120^\circ+360^\circ.
Read the answer and explanation

Quadrant III

The coterminal positive angle is 240∘240^\circ, between 180∘180^\circ and 270∘270^\circ, so the terminal ray is in Quadrant III.
About 5 minutes
TOPIC 02

Radians compare an arc with its radius

A degree divides a complete turn into 360 equal parts. A radian measures the same turn by comparing an arc length with the radius. If an arc of length ss is cut from a circle of radius rr, the central angle has radian measure θ=s/r\theta=s/r. One radian is the angle for which the arc length equals the radius. Since both lengths use the same unit, their ratio is a number.

A full circumference is 2πr2\pi r, so a complete turn is 2π2\pi radians. A straight angle is therefore π\pi radians, giving the exact conversion:

180∘=π rad,θrad=θdegπ180,θdeg=θrad180π.180^\circ=\pi\text{ rad},\qquad \theta_{\rm rad}=\theta_{\rm deg}\frac{\pi}{180},\qquad \theta_{\rm deg}=\theta_{\rm rad}\frac{180}{\pi}.

The number π\pi is not itself an angle unit. It is a real number, approximately 3.14159; π\pi radians and π\pi degrees are very different turns. In trigonometric-function work an argument such as sin⁡x\sin x, without a degree symbol, is normally read in radians. Keep the degree symbol whenever degrees are intended.

Negative angles and angles beyond one revolution convert by the same multiplication. There is no need for a second rule. First keep fractions exact, simplify, and only then use a decimal if it serves the question. That habit avoids rounding errors and makes familiar angles such as π/6\pi/6, π/4\pi/4 and π/3\pi/3 easy to recognise.

Work it through

Two conversions.

−135∘π180∘=−3π4 rad,13π6180∘π=390∘.-135^\circ\frac{\pi}{180^\circ}=-\frac{3\pi}{4}\text{ rad},\qquad \frac{13\pi}{6}\frac{180^\circ}{\pi}=390^\circ.

The second angle makes one complete turn and another 30∘30^\circ. Its coterminal angle in [0,2π)[0,2\pi) is π/6\pi/6.

Pause. Think. Try.

1. Convert 150∘150^\circ to radians.
A hint, pleaseMultiply by π/180\pi/180.
Read the answer and explanation

5π/65\pi/6

150×π/180=5π/6150\times\pi/180=5\pi/6.
2. What angle in degrees is 7π/47\pi/4 radians?
A hint, pleaseMultiply by 180/π180/\pi.
Read the answer and explanation

315∘315^\circ

(7π/4)(180∘/π)=7×45∘=315∘(7\pi/4)(180^\circ/\pi)=7\times45^\circ=315^\circ.
About 5 minutes
TOPIC 03

Use radians to measure a curved distance

Two walkers can turn through the same angle while travelling different distances: the walker on a larger circle follows a longer arc. The radian definition separates these quantities cleanly. For a circle of radius r>0r>0 and a nonnegative central angle θ\theta in radians, the corresponding arc length is

s=rθ.s=r\theta.

The area of a sector is the same fraction of the circle’s area as its angle is of a complete turn. Since a complete turn is 2π2\pi radians,

A=θ2π πr2=12r2θ.A=\frac{\theta}{2\pi}\,\pi r^2=\frac12r^2\theta.

These forms already contain the degree-to-radian conversion. Do not substitute a bare degree value into them. Either convert the angle first or use the equivalent fraction θdeg/360\theta_{\rm deg}/360 of a circumference or circle. The formulas should give the same answer whichever method you choose.

Distinguish an arc, which follows the curved circumference, from a chord, which is the straight segment between its endpoints. Except at the zero-length case, a chord is shorter than its corresponding minor arc. A photograph or sketch can make them look similar; the mathematical quantities are different.

For ordinary length and area questions, use a nonnegative angle magnitude. A clockwise angular displacement can be signed, but a distance travelled and an ordinary sector area are not negative. Also say whether the question concerns one chosen sector, a minor arc or accumulated travel over several revolutions. Stating the model prevents the angle from doing two different jobs in the same calculation.

θ radiansrs = rθArea = ½r²θRadians connect angle, arc and sector.
A sector with central angle θ radians has arc length rθ and area ½r²θ.

Work it through

A circular garden. A sector has radius 12 cm12\text{ cm} and angle 45∘=π/445^\circ=\pi/4 radians.

s=12π4=3π cm,A=12(12)2π4=18π cm2.s=12\frac{\pi}{4}=3\pi\text{ cm},\qquad A=\frac12(12)^2\frac{\pi}{4}=18\pi\text{ cm}^2.

The units also check the result: radius times angle gives a length; radius squared times angle gives an area.

Pause. Think. Try.

1. A radius is 8 cm8\text{ cm} and the central angle is π/3\pi/3 radians. Find the arc length.
A hint, pleaseUse s=rθs=r\theta, with the angle already in radians.
Read the answer and explanation

8π/3 cm8\pi/3\text{ cm}

s=8(π/3)=8π/3 cms=8(\pi/3)=8\pi/3\text{ cm}.
2. A sector has radius 6 cm6\text{ cm} and angle π/2\pi/2 radians. What is its area?
A hint, pleaseUse A=12r2θA=\tfrac12r^2\theta.
Read the answer and explanation

9π cm29\pi\text{ cm}^2

A=12×36×π/2=9π cm2A=\tfrac12\times36\times\pi/2=9\pi\text{ cm}^2. It is a quarter-circle.
About 5 minutes
TOPIC 04

A moving point defines the six functions

Draw a circle of radius 1 centred at the origin. Let a radius turn through an angle θ\theta, measured from the positive xx-axis, and meet the circle at P(x,y)P(x,y). Define

cos⁡θ=x,sin⁡θ=y.\cos\theta=x,\qquad\sin\theta=y.

For an acute angle, dropping a perpendicular from PP gives the familiar right triangle. Its hypotenuse is 1, so adjacent divided by hypotenuse is xx and opposite divided by hypotenuse is yy. The circle definition agrees with the triangle definition while extending it to every real angle. At an axis the triangle can flatten, but the circle point still exists.

The remaining functions are ratios or reciprocals of these coordinates:

tan⁡θ=yx,cot⁡θ=xy,sec⁡θ=1x,cosec⁡θ=1y.\tan\theta=\frac{y}{x},\quad\cot\theta=\frac{x}{y},\quad\sec\theta=\frac1x,\quad\operatorname{cosec}\theta=\frac1y.

A zero denominator means undefined. At θ=π/2\theta=\pi/2, for example, P=(0,1)P=(0,1). Thus sine is 1 and cosine is 0; tangent and secant are undefined. This is not an instruction to replace a denominator by a small number and call the answer infinity.

The circle equation also explains a permanent identity: x2+y2=1x^2+y^2=1 becomes cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1. It remains true in every quadrant because squared coordinates remove signs. Move the point through a complete turn and watch which quantities vary continuously, which change sign, and which become undefined at an axis. Check exact-angle positions before interpreting decimal readouts between them.

Work it through

A point on the circle. If P=(−3/5,4/5)P=(-3/5,4/5), then cos⁡θ=−3/5\cos\theta=-3/5, sin⁡θ=4/5\sin\theta=4/5, tan⁡θ=−4/3\tan\theta=-4/3, cot⁡θ=−3/4\cot\theta=-3/4, sec⁡θ=−5/3\sec\theta=-5/3, and cosec⁡θ=5/4\operatorname{cosec}\theta=5/4. The check 9/25+16/25=19/25+16/25=1 confirms that the point is on the unit circle.

Explore this diagram with JavaScript enabled. The worked example above remains available to read and print.

Pause. Think. Try.

1. The unit-circle point is (5/13,−12/13)(5/13,-12/13). What is tan⁡θ\tan\theta?
A hint, pleaseTangent is vertical coordinate divided by horizontal coordinate.
Read the answer and explanation

−12/5-12/5

tan⁡θ=(−12/13)/(5/13)=−12/5\tan\theta=(-12/13)/(5/13)=-12/5.
2. Which pair is undefined at θ=π/2\theta=\pi/2?
A hint, pleaseThe circle point is (0,1)(0,1).
Read the answer and explanation

tan⁡θ\tan\theta and sec⁡θ\sec\theta

Tangent and secant divide by the horizontal coordinate 0. Cotangent is 0 and cosecant is 1.
About 5 minutes
TOPIC 05

Find the sign before calculating a value

The signs of trigonometric functions follow from coordinates. In Quadrant I, both xx and yy are positive. In Quadrant II, x<0x<0 and y>0y>0. In Quadrant III, both are negative. In Quadrant IV, x>0x>0 and y<0y<0. Since cosine is xx, sine is yy, and tangent is y/xy/x, the sign table needs no separate guesswork.

Open quadrantSine/cosecantCosine/secantTangent/cotangent
I+++
II+−−
III−−+
IV−+−

A reference angle is the acute angle between a non-axis terminal ray and the nearest part of the xx-axis. It supplies the magnitudes of sine and cosine; the quadrant supplies their signs. For 150∘150^\circ, the reference angle is 30∘30^\circ, so sine has magnitude 1/2 and cosine has magnitude 3/2\sqrt3/2. Quadrant II makes sine positive and cosine negative.

When a problem gives one ratio, first find a missing magnitude using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, then use the stated quadrant to choose the sign. A square root alone produces a nonnegative magnitude, not the automatically correct signed coordinate. Try predicting the signs first, then calculate the fractions; separating those two decisions makes the reasoning easier to check. Finally form any quotient from the signed values, and check that the denominator is nonzero.

At the axes, evaluate the actual coordinates instead of forcing the angle into a quadrant row. This small distinction accounts for zeros and undefined values without adding exceptions to a memorised sign slogan.

Work it through

One ratio and a quadrant. If sin⁡θ=−5/13\sin\theta=-5/13 and θ\theta is in Quadrant III, then cos⁡2θ=1−25/169=144/169\cos^2\theta=1-25/169=144/169. The cosine magnitude is 12/13, but Quadrant III requires a negative horizontal coordinate. Thus cos⁡θ=−12/13\cos\theta=-12/13 and tan⁡θ=5/12\tan\theta=5/12.

Pause. Think. Try.

1. What is cos⁡150∘\cos150^\circ?
A hint, pleaseThe reference angle is 30° and cosine is negative in Quadrant II.
Read the answer and explanation

−3/2-\sqrt3/2

cos⁡150∘=−cos⁡30∘=−3/2\cos150^\circ=-\cos30^\circ=-\sqrt3/2.
2. If cos⁡θ=−4/5\cos\theta=-4/5 and θ\theta is in Quadrant II, what is sin⁡θ\sin\theta?
A hint, pleaseFind the squared sine, then choose its sign from the quadrant.
Read the answer and explanation

3/53/5

sin⁡2θ=1−16/25=9/25\sin^2\theta=1-16/25=9/25. Sine is positive in Quadrant II, giving 3/5.
About 5 minutes
TOPIC 06

Reduce angles by seeing the symmetry

A reflection in the xx-axis changes (x,y)(x,y) to (x,−y)(x,-y). On the unit circle, the angle −θ-\theta is exactly this reflection of θ\theta. Therefore cosine keeps its value while sine changes sign:

cos⁡(−θ)=cos⁡θ,sin⁡(−θ)=−sin⁡θ.\cos(-\theta)=\cos\theta,\qquad\sin(-\theta)=-\sin\theta.

Cosine is an even function and sine is an odd function. Their quotient shows that tangent is odd wherever it is defined. These are statements about inputs and outputs, not instructions to delete every minus sign in an expression.

A half-turn moves (x,y)(x,y) to (−x,−y)(-x,-y), while a full turn returns it to (x,y)(x,y). Hence

sin⁡(θ+π)=−sin⁡θ,cos⁡(θ+π)=−cos⁡θ,tan⁡(θ+π)=tan⁡θ,\sin(\theta+\pi)=-\sin\theta,\quad\cos(\theta+\pi)=-\cos\theta,\quad\tan(\theta+\pi)=\tan\theta,sin⁡(θ+2π)=sin⁡θ,cos⁡(θ+2π)=cos⁡θ.\sin(\theta+2\pi)=\sin\theta,\qquad\cos(\theta+2\pi)=\cos\theta.

Reflection in the yy-axis gives sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta)=\sin\theta and cos⁡(π−θ)=−cos⁡θ\cos(\pi-\theta)=-\cos\theta. Exchanging the coordinates through complementary acute angles gives sin⁡(π/2−θ)=cos⁡θ\sin(\pi/2-\theta)=\cos\theta and cos⁡(π/2−θ)=sin⁡θ\cos(\pi/2-\theta)=\sin\theta. These relationships can be checked on the circle before using them in algebra.

A dependable reduction has three steps: remove complete turns, identify a reference angle, and apply the sign. Write the equality used at each step, especially when a negative input and a negative quadrant coordinate occur together. A diagram turns a long list of allied-angle formulas into a few simple movements.

Work it through

A large negative input.

sin⁡(−13π6)=sin⁡(−π6)=−sin⁡π6=−12.\sin\left(-\frac{13\pi}{6}\right)=\sin\left(-\frac{\pi}{6}\right)=-\sin\frac{\pi}{6}=-\frac12.

For comparison, cos⁡(−13π/6)=cos⁡(π/6)=3/2\cos(-13\pi/6)=\cos(\pi/6)=\sqrt3/2, because cosine is even and a full turn does not change its value.

Pause. Think. Try.

1. Simplify sin⁡(π−θ)\sin(\pi-\theta).
A hint, pleaseReflect the circle point in the vertical axis.
Read the answer and explanation

sin⁡θ\sin\theta

The vertical coordinate is unchanged by reflection in the yy-axis, so sine keeps its value.
2. What is tan⁡(7π/6)\tan(7\pi/6)?
A hint, pleaseRemove a half-turn; tangent has period π.
Read the answer and explanation

1/31/\sqrt3

tan⁡(7π/6)=tan⁡(π+π/6)=tan⁡(π/6)=1/3\tan(7\pi/6)=\tan(\pi+\pi/6)=\tan(\pi/6)=1/\sqrt3.
About 5 minutes
TOPIC 07

A chord connects the circle to a right triangle

Choose two points AA and BB on a circle of radius RR. The straight segment ABAB is their chord; the curved path along the circle is an arc. Join both endpoints to the centre OO. If the chosen central angle is ϕ\phi, triangle OABOAB has equal sides OA=OB=ROA=OB=R.

For the minor-angle construction 0<ϕ<π0<\phi<\pi, draw the perpendicular from OO to the chord and call its foot MM. In an isosceles triangle this line bisects both the chord and its vertex angle. Thus AM=AB/2AM=AB/2, ∠AOM=ϕ/2\angle AOM=\phi/2, and triangle OAMOAM is right-angled. Applying the sine ratio gives

sin⁡ϕ2=AMR,AM=Rsin⁡ϕ2,AB=2Rsin⁡ϕ2.\sin\frac\phi2=\frac{AM}{R},\qquad AM=R\sin\frac\phi2,\qquad AB=2R\sin\frac\phi2.

The half-angle appears because the perpendicular divides the central angle in two. It does not mean that the radius or the original angle has disappeared. Keep ϕ\phi, ϕ/2\phi/2, the right-angle mark and both equal radii visible as you move the endpoints. A chord calculation should be traceable directly to that labelled triangle.

At ϕ=π\phi=\pi, the chord is a diameter of length 2R2R; the half-triangle has flattened and should not be drawn as a nondegenerate right triangle. The same chord length for a chosen angle 0≤ϕ≤2π0\le\phi\le2\pi is 2Rsin⁡(ϕ/2)2R\sin(\phi/2); at a full turn the endpoints coincide. For a general signed angle, the nonnegative length is 2R∣sin⁡(ϕ/2)∣2R|\sin(\phi/2)|. For a minor central angle 0≤ϕ≤π0\le\phi\le\pi, the corresponding minor-arc length is RϕR\phi, with ϕ\phi in radians. Comparing the two formulas reinforces the difference between a curve and its straight shortcut.

Work it through

A 60° chord. Let R=10 cmR=10\text{ cm} and ϕ=60∘\phi=60^\circ. Each half of the chord is 10sin⁡30∘=5 cm10\sin30^\circ=5\text{ cm}, so AB=10 cmAB=10\text{ cm}. Its minor arc is 10(π/3)=10π/3 cm10(\pi/3)=10\pi/3\text{ cm}, which is longer than the chord.

Explore this diagram with JavaScript enabled. The worked example above remains available to read and print.

Pause. Think. Try.

1. A circle has radius 5 cm5\text{ cm}. A chord subtends 120∘120^\circ at the centre. Find its length.
A hint, pleaseUse the sine of half the central angle.
Read the answer and explanation

53 cm5\sqrt3\text{ cm}

AB=2(5)sin⁡60∘=10(3/2)=53 cmAB=2(5)\sin60^\circ=10(\sqrt3/2)=5\sqrt3\text{ cm}.
2. For a central angle ϕ\phi, why does the chord formula use ϕ/2\phi/2?
A hint, pleaseLook at the two congruent right triangles.
Read the answer and explanation

The perpendicular from the centre bisects the isosceles triangle.

The perpendicular from the centre bisects the chord and the vertex angle. The sine ratio in either right triangle uses ϕ/2\phi/2.
PUT IT TOGETHER

Your chapter checkpoint

Mix the ideas. Try each question before opening its explanation.

1. Reduce 19π/619\pi/6 to a coterminal angle in [0,2π)[0,2\pi).
A hint, pleaseSubtract 2π=12π/62\pi=12\pi/6.
Read the answer and explanation

7π/67\pi/6

19π/6−12π/6=7π/619\pi/6-12\pi/6=7\pi/6.
2. A unit-circle point has sin⁡θ=3/5\sin\theta=3/5 in Quadrant II. What is sec⁡θ\sec\theta?
A hint, pleaseFind the signed cosine before taking its reciprocal.
Read the answer and explanation

−5/4-5/4

cos⁡θ=−4/5\cos\theta=-4/5, so sec⁡θ=1/cos⁡θ=−5/4\sec\theta=1/\cos\theta=-5/4.
3. A radius is 9 cm9\text{ cm} and an arc length is 6 cm6\text{ cm}. Find the central angle in radians.
A hint, pleaseRadian measure is arc length divided by radius.
Read the answer and explanation

2/32/3

θ=s/r=6/9=2/3 rad\theta=s/r=6/9=2/3\text{ rad}.
4. Which identity holds for every real θ\theta?
A hint, pleaseChoose an identity whose expressions are defined for every real input.
Read the answer and explanation

sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1

The unit-circle equation proves the Pythagorean identity for all real angles. Reciprocal formulas need nonzero denominators.

Write the reasoning

Make a labelled sketch or a few lines of working. Then compare each step.

Convert, locate and evaluate

Convert −510∘-510^\circ to radians and find its sine and cosine.
Compare your solution

−510∘=−17π/6-510^\circ=-17\pi/6. Adding 4π4\pi gives the coterminal angle 7π/6=210∘7\pi/6=210^\circ, in Quadrant III, with reference angle π/6\pi/6. Hence sin⁡(−17π/6)=−1/2\sin(-17\pi/6)=-1/2 and cos⁡(−17π/6)=−3/2\cos(-17\pi/6)=-\sqrt3/2. The signs match two negative coordinates.

An arc and a sector

A circle has radius 14 cm14\text{ cm}. Find the arc length and sector area for a central angle of 150∘150^\circ. Leave answers in terms of π\pi.
Compare your solution

150∘=5π/6150^\circ=5\pi/6. Arc length: s=14(5π/6)=35π/3 cms=14(5\pi/6)=35\pi/3\text{ cm}. Sector area: A=12(14)2(5π/6)=245π/3 cm2A=\tfrac12(14)^2(5\pi/6)=245\pi/3\text{ cm}^2. Both equal 150/360=5/12150/360=5/12 of their corresponding full-circle quantities.

Recover all six functions

An angle ends in Quadrant IV and tan⁡θ=−5/12\tan\theta=-5/12. Find the other five functions.
Compare your solution

Use a right-triangle magnitude pattern 5:12:13. In Quadrant IV, cosine is positive and sine is negative. Thus sin⁡θ=−5/13\sin\theta=-5/13, cos⁡θ=12/13\cos\theta=12/13, cot⁡θ=−12/5\cot\theta=-12/5, sec⁡θ=13/12\sec\theta=13/12, and cosec⁡θ=−13/5\operatorname{cosec}\theta=-13/5. Check 25/169+144/169=125/169+144/169=1 and the negative quotient.

Class 11 milestone

Your Class 11 angle foundation is in place

You can move from an angle to a position, a sign and a ratio. Continue to the six moving graphs, or use compound angles to calculate new exact values. Looking for the Class 10 route? Return to heights and distances.