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Exact trigonometric values in radians

Find exact sine, cosine and tangent values in radians using special triangles and the unit circle. Practise reference angles, quadrant signs and extra turns with worked solutions.

Before you startRadian measure, right-triangle trigonometry and surds

01 / Coordinates give the signs

The angle can change quadrant; the reference triangle stays familiar.

On the circle of radius 1 centred at the origin, the point reached by angle θ has coordinates (cos θ, sin θ). Sine is the vertical coordinate and cosine is the horizontal coordinate. Their signs follow the point’s position.

P = (cos θ, sin θ)
tan θ = sin θ / cos θ, when cos θ ≠ 0

Choose an angle in the model. The reference triangle provides the magnitudes; the quadrant supplies the signs. At an axis, use the point’s coordinates directly.

Read sine and cosine from coordinatesExplore
Exact unit-circle coordinatesAt 5pi/6, the unit-circle point is minus root3 over2, one half. Cosine is negative and sine positive.xy1−11−1P = (−√3/2, 1/2)

θ = 5π/6 rad. Quadrant II.

cos θ = −√3/2; sin θ = 1/2; tan θ = −√3/3.

Reference angle to the x-axis: π/6.

Blue horizontal coordinate: cosine. Green vertical coordinate: sine. Gold radius: length 1. The signed coordinates, not the positive triangle lengths, give the ratios.

Watch a reference triangle cross quadrants

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Derive the first-quadrant values

Two triangles are enough for the standard exact angles.

An isosceles right triangleWorked example

Legs 1 and 1 give hypotenuse √2

Pythagoras: c² =1² +1² =2.

The two acute angles are π/4

They share the remaining π/2 equally.

sin(π/4) = cos(π/4) =1/√2 =√2/2

Rationalising changes the form, not the value. tan(π/4) =1.

Bisect an equilateral triangle of side 2Worked example

One right triangle has sides 1, √3, 2

Its acute angles are π/6 and π/3.

sin(π/6) =1/2; cos(π/6) =√3/2

The shorter leg is opposite π/6.

sin(π/3) =√3/2; cos(π/3) =1/2

Swapping the acute angle swaps opposite and adjacent.

tan(π/6) =1/√3 =√3/3; tan(π/3) =√3

Tangent is opposite divided by adjacent.

03 / Find the reference angle

Measure the small angle to the horizontal axis.

For a terminal ray inside a quadrant, the reference angle α lies between 0 and π/2. First remove any complete turns to place θ in 0 ≤ θ < 2π. Then use its position.

Quadrant I: α = θ
Quadrant II: α = π − θ
Quadrant III: α = θ − π
Quadrant IV: α = 2π − θ

Three angles with reference π/6Worked example

5π/6: π −5π/6 =π/6

Quadrant II.

7π/6: 7π/6 −π =π/6

Quadrant III.

11π/6: 2π −11π/6 =π/6

Quadrant IV.

The reference angle finds the magnitude only. It does not mean the sine, cosine or tangent is always positive.

04 / Apply the quadrant signs

Use horizontal and vertical position, then divide for tangent.

Find sin(5π/6), cos(5π/6) and tan(5π/6)Worked example

Reference angle =π/6; point lies in Quadrant II

x is negative, y is positive.

sin(5π/6) =1/2; cos(5π/6) =−√3/2

Use the π/6 magnitudes with the correct signs.

tan(5π/6) =(1/2)/(−√3/2) =−√3/3

Tangent is negative because the coordinate signs differ.

In Quadrant III both sine and cosine are negative, so tangent is positive. In Quadrant IV cosine is positive and sine is negative, so tangent is negative. In Quadrant I all three are positive.

05 / Axis angles and undefined tangent

Do not invent a value when the denominator is zero.

θ =0: (cos θ, sin θ) =(1,0)
θ =π/2: (0,1)
θ =π: (−1,0)
θ =3π/2: (0,−1)
θ =2π: (1,0)

Tangent is 0 at 0 and π, because the numerator is 0 and cosine is nonzero. It is undefined at π/2 and 3π/2, because the denominator is 0. “Infinity” is not a value of the tangent function at these inputs.

tan θ is undefined when θ = π/2 + kπ,
where k is any integer.

Boundary angles lie on the axes, rather than inside a quadrant. Read their exact coordinates instead of assigning a quadrant sign rule.

06 / Negative and multiple-turn angles

Reduce the angle before reading the circle.

Exact values beyond the first turnWorked example

sin(13π/6) =sin(π/6) =1/2

Subtract 2π; sine repeats after a full turn.

cos(−3π/4) =cos(5π/4) =−√2/2

Add 2π; the terminal ray lies in Quadrant III.

tan(11π/4) =tan(3π/4) =−1

Subtract 2π. Alternatively, tangent repeats after π.

sin(−π/6) =−1/2; cos(−π/6) =√3/2

Reflection in the horizontal axis reverses y and preserves x.

For any θ, sin(θ +2π) =sin θ and cos(θ +2π) =cos θ. Also sin(−θ) =−sin θ and cos(−θ) =cos θ. Tangent repeats after π wherever it is defined.

07 / Combine exact values

Keep surds and fractions until the calculation is finished.

Evaluate 2sin(2π/3) −cos(5π/4)Worked example

sin(2π/3) =√3/2

Quadrant II makes sine positive.

cos(5π/4) =−√2/2

Quadrant III makes cosine negative.

2(√3/2) −(−√2/2) =√3 +√2/2

Subtracting a negative contributes a positive term.

Evaluate tan(7π/6) cos(π/3)Worked example

tan(7π/6) =√3/3; cos(π/3) =1/2

Tangent is positive in Quadrant III.

Product =√3/6

Do not round the two factors first.

√3 +√2/2 cannot be combined into a single simpler surd by adding the numbers under the square roots.

08 / Use exact radians in linked triangles

Name the side opposite the angle before choosing a ratio.

Two right triangles share side AB of length 6. In triangle ABC the right angle is at B and ∠BAC =π/6. In triangle ABD the right angle is at D and ∠BAD =π/3. The second triangle therefore uses AB as its hypotenuse.

Find BC, AC, AD and BDWorked example

BC/AB =tan(π/6), so BC =6/√3 =2√3

AB is adjacent in the first triangle.

AB/AC =cos(π/6), so AC =6/(√3/2) =4√3

AC is the first triangle’s hypotenuse.

AD/AB =cos(π/3), so AD =3

AB is now the hypotenuse, not the adjacent leg.

BD/AB =sin(π/3), so BD =3√3

Check AD² +BD² =9 +27 =36 =AB².

The same named side can play a different role in each triangle. Label the right angle and reference angle again before substituting an exact value.

09 / Your turn

Give exact values and explain any undefined expression.

01 · Quadrant II

Find sin(3π/4), cos(3π/4) and tan(3π/4).

Hint

Reference angle π/4; x<0, y>0.

Worked solution

√2/2, −√2/2, −1 respectively.

02 · Quadrant III

Find sin(4π/3), cos(4π/3) and tan(4π/3).

Hint

Reference angle π/3; both coordinates are negative.

Worked solution

−√3/2, −1/2, √3 respectively.

03 · Quadrant IV

Find sin(11π/6), cos(11π/6) and tan(11π/6).

Hint

Reference angle π/6; only cosine is positive.

Worked solution

−1/2, √3/2, −√3/3 respectively.

04 · Clockwise

Evaluate sin(−π/3) and cos(−π/3).

Hint

Reflect the π/3 point in the x-axis.

Worked solution

−√3/2 and 1/2. Sine changes sign; cosine does not.

05 · Extra turns

Evaluate cos(17π/6) and sin(−11π/6).

Hint

Reduce each by whole multiples of 2π.

Worked solution

cos(17π/6) =cos(5π/6) =−√3/2. sin(−11π/6) =sin(π/6) =1/2.

06 · Tangent period

Evaluate tan(9π/4) and tan(−2π/3).

Hint

Tangent repeats after π, where defined.

Worked solution

tan(π/4) =1 and tan(π/3) =√3.

07 · Axis values

Evaluate sin(3π/2), cos(2π) and tan(π). Explain why tan(5π/2) is undefined.

Hint

Use the axis coordinates and tan =sin/cos.

Worked solution

−1, 1 and 0. At 5π/2 the point is (0,1), so the quotient would divide by 0.

08 · Exact expression

Evaluate 2cos(2π/3) +3sin(π/6).

Hint

Cosine is negative in Quadrant II.

Worked solution

2(−1/2) +3(1/2) =1/2.

09 · A surd product

Evaluate sin(5π/4) cos(7π/6).

Hint

Both factors are negative.

Worked solution

(−√2/2)(−√3/2) =√6/4.

10 · Reference-angle reasoning

An angle lies in Quadrant II and has reference angle π/3. Find the angle in 0 ≤ θ < 2π and its tangent.

Hint

Subtract the reference angle fromπ.

Worked solution

θ =2π/3; tan θ =−√3.

11 · Right triangle

A right triangle has hypotenuse 10 and an acute angle π/6. Find the opposite and adjacent lengths exactly.

Hint

Use sine and cosine with the hypotenuse.

Worked solution

Opposite =10(1/2) =5. Adjacent =10(√3/2) =5√3. Check25 +75 =100.

12 · Spot the sign error

A student says cos(5π/6) =cos(π/6) because both have reference angle π/6. Correct the statement.

Hint

The reference angle fixes the magnitude, not the sign.

Worked solution

cos(5π/6) =−cos(π/6) =−√3/2. The point has a negative horizontal coordinate.

10 / Recap

Find the terminal ray, reference angle and signs in that order.

  • Unit-circle coordinates are (cos θ, sin θ).
  • Special triangles give exact magnitudes at π/6, π/4 and π/3.
  • Reference angles do not supply the quadrant signs.
  • Axis values come directly from the coordinates.
  • Tangent is undefined when cosine is 0.
  • Remove complete turns and keep exact surds until the end.

Review radians and degrees →

Section 1 of 10 · Coordinates give the signs