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.
Understand · explore · practise
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
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.
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.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Derive the first-quadrant values
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.
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
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
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
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
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
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
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.
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
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.
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
Find sin(3π/4), cos(3π/4) and tan(3π/4).
Reference angle π/4; x<0, y>0.
√2/2, −√2/2, −1 respectively.
Find sin(4π/3), cos(4π/3) and tan(4π/3).
Reference angle π/3; both coordinates are negative.
−√3/2, −1/2, √3 respectively.
Find sin(11π/6), cos(11π/6) and tan(11π/6).
Reference angle π/6; only cosine is positive.
−1/2, √3/2, −√3/3 respectively.
Evaluate sin(−π/3) and cos(−π/3).
Reflect the π/3 point in the x-axis.
−√3/2 and 1/2. Sine changes sign; cosine does not.
Evaluate cos(17π/6) and sin(−11π/6).
Reduce each by whole multiples of 2π.
cos(17π/6) =cos(5π/6) =−√3/2. sin(−11π/6) =sin(π/6) =1/2.
Evaluate tan(9π/4) and tan(−2π/3).
Tangent repeats after π, where defined.
tan(π/4) =1 and tan(π/3) =√3.
Evaluate sin(3π/2), cos(2π) and tan(π). Explain why tan(5π/2) is undefined.
Use the axis coordinates and tan =sin/cos.
−1, 1 and 0. At 5π/2 the point is (0,1), so the quotient would divide by 0.
Evaluate 2cos(2π/3) +3sin(π/6).
Cosine is negative in Quadrant II.
2(−1/2) +3(1/2) =1/2.
Evaluate sin(5π/4) cos(7π/6).
Both factors are negative.
(−√2/2)(−√3/2) =√6/4.
An angle lies in Quadrant II and has reference angle π/3. Find the angle in 0 ≤ θ < 2π and its tangent.
Subtract the reference angle fromπ.
θ =2π/3; tan θ =−√3.
A right triangle has hypotenuse 10 and an acute angle π/6. Find the opposite and adjacent lengths exactly.
Use sine and cosine with the hypotenuse.
Opposite =10(1/2) =5. Adjacent =10(√3/2) =5√3. Check25 +75 =100.
A student says cos(5π/6) =cos(π/6) because both have reference angle π/6. Correct the statement.
The reference angle fixes the magnitude, not the sign.
cos(5π/6) =−cos(π/6) =−√3/2. The point has a negative horizontal coordinate.
10 / Recap
Section 1 of 10 · Coordinates give the signs