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Exact values using compound angles

Find exact trigonometric values with compound angle formulae, signed triangles and quadrant checks. Apply them to line angles, triangle geometry and algebraic relationships.

Before you startCompound angle formulae, exact values and quadrant signs

01 / Keep the ratios exact

You can combine angles without calculating either angle.

If you know sine and cosine for two angles, the compound angle formulae give the exact sine and cosine of their sum or difference. There is usually no reason to convert those angles to rounded calculator values first.

Known signed ratios → compound angle formula → exact result

The model gives three choices for each angle. Select a sum or difference and compare the gold endpoint with its exact coordinates. A sum of an obtuse angle and an acute angle can cross into the third quadrant.

Combine exact signed ratiosExplore

A coordinatesB coordinatesCombine

Blue: A. Green: B. Gold: the combined angle, measured from the positive x-axis.

cos(A + B) = −220/221; sin(A + B) = −21/221; tan(A + B) = 21/220.

Cosine: (−12 × 15 − 5 × 8)/(13 × 17). Sine: (5 × 15 + (−12) × 8)/(13 × 17).

Choose the signs from the quadrants before substituting. The displayed ratios are exact; you do not need decimal angles.

02 / Split an unfamiliar angle into familiar ones

Choose a sum or difference of angles with known exact values.

Find cos 105° exactlyWorked example

105° = 60° + 45°

Use cosine addition.

cos 105° = cos 60° cos 45° − sin 60° sin 45°

Substitute 1/2, √2/2, √3/2 and √2/2.

cos 105° = (√2 − √6)/4

The result is negative, as expected in quadrant two.

Find sin 105° exactlyWorked example

sin 105° = sin 60° cos 45° + cos 60° sin 45°

The sine addition has a plus.

sin 105° = (√6 + √2)/4

The result is positive and less than 1.

Watch a 60° direction turn through 45°

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

03 / Recover a missing ratio with its sign

Pythagoras gives the magnitude; the quadrant gives the sign.

sin A = 5/13 and A is obtuse. Find cos A.Worked example

cos² A = 1 − 25/169 = 144/169

Apply sin² A + cos² A = 1.

cos A = −12/13

An obtuse angle lies in quadrant two, where cosine is negative.

tan B = −7/24 and B is in quadrant fourWorked example

Use a triangle with horizontal component 24 and vertical component −7

Its hypotenuse is 25.

cos B = 24/25 and sin B = −7/25

The denominator is a positive length; the coordinates carry the signs.

Writing cos A = √(1 − sin² A) without a sign choice silently assumes nonnegative cosine. The given interval is part of the information, not an optional extra.

04 / Combine two signed triangles

Work with a common denominator and check the resulting quadrant.

sin A = 5/13, A obtuse; sin B = 8/17, B acute. Find sin(A + B) and cos(A + B).Worked example

cos A = −12/13 and cos B = 15/17

Pythagoras with the stated quadrant signs.

sin(A + B) = (5 × 15 − 12 × 8)/221 = −21/221

The vertical coordinate is negative.

cos(A + B) = (−12 × 15 − 5 × 8)/221 = −220/221

The horizontal coordinate is negative.

A + B lies in quadrant three

Here 90° < A + B < 270°; the two negative coordinates locate it more precisely.

As a consistency check, 21² + 220² = 221². That checks the combined sine/cosine pair has unit length, although it cannot by itself detect every possible sign error.

05 / Changing plus to minus changes both patterns

Use the difference formula afresh.

For the same A and B, find sin(A − B) and cos(A − B)Worked example

sin(A − B) = (5 × 15 + 12 × 8)/221

Subtracting cos A sin B adds a positive quantity here.

sin(A − B) = 171/221

The vertical coordinate is positive.

cos(A − B) = (−12 × 15 + 5 × 8)/221 = −140/221

The horizontal coordinate is negative.

A − B is in quadrant two

The original bounds give 0° < A − B < 180°.

A minus inside the angle is not a command to negate the whole answer for the sum. Keep the signs of the original ratios separate from the signs in the formula.

06 / Use the combined sine and cosine for other ratios

Check the denominator before dividing.

Use cos 105° and sin 105° to find tan 105° and sec 105°Worked example

tan 105° = (√6 + √2)/(√2 − √6)

Divide sine by cosine.

tan 105° = −(2 + √3)

Multiply numerator and denominator by √2 + √6.

sec 105° = 4/(√2 − √6) = −(√2 + √6)

Rationalise again.

For a combined angle with cosine zero, tangent and secant are undefined. If the sine is zero, cosecant and cotangent are undefined. Do not replace an undefined ratio by zero.

07 / Find the acute angle between two lines

The gradient is the tangent of an inclination angle.

For two nonvertical lines with gradients m₁ and m₂, use tangent subtraction on their inclination angles. When 1 + m₁m₂ ≠ 0, the acute angle φ between the lines satisfies:

tan φ = |(m₂ − m₁)/(1 + m₁m₂)|

Find the acute angle between y = x/2 + 7 and y = −x/3 + 2Worked example

m₁ = −1/3, m₂ = 1/2

The intercepts do not affect the angle.

tan φ = |(1/2 + 1/3)/(1 − 1/6)| = 1

Use the acute angle, not an arbitrary tangent solution.

φ = 45°

The supplementary angle is 135°.

If 1 + m₁m₂ = 0, the lines are perpendicular: φ = 90°. If the gradients match, φ = 0°. For a vertical line, use its 90° inclination directly rather than an undefined gradient.

08 / Connect compound angles with the sine rule

Find the third angle before comparing opposite sides.

A triangle has acute angles U and V with tan U = 1/2 and tan V = 1/3. The side opposite the third angle is 10. Find the other two sides.Worked example

sin U = 1/√5, cos U = 2/√5; sin V = 1/√10, cos V = 3/√10

Both signed triangles are positive.

tan(U + V) = (1/2 + 1/3)/(1 − 1/6) = 1

Since 0° < U + V < 180°, their sum is 45°.

The third angle is 135°, with sine √2/2

Triangle angles total 180°.

u = 10 sin U / sin 135° = 2√10

Use the side opposite U.

v = 10 sin V / sin 135° = 2√5

Use the side opposite V.

09 / Use an algebraic relationship between trig terms

Translate the relationship before expanding.

The numbers sin θ, cos θ and 2sin(θ + π/4) form an arithmetic progression, with θ acute. Find tan θ.Worked example

2cos θ = sin θ + 2sin(θ + π/4)

Twice the middle term equals the sum of the outer terms.

(2 − √2)cos θ = (1 + √2)sin θ

Expand and collect terms.

tan θ = (2 − √2)/(1 + √2) = 3√2 − 4

Division by cos θ is valid because θ is acute.

The result is positive, consistent with an acute θ. Without the interval, division might need a separate check for cos θ = 0.

10 / Use magnitude, signs and the original information

Exact answers should still make geometric sense.

Sine and cosine must lie between −1 and 1.
Their squared values must add to 1 for the same angle.
Use the given quadrants when recovering a missing ratio.
A tangent value alone does not specify a unique angle.
Keep radicals and fractions through the calculation; round only a requested final angle.

Does tan(A + B) = 1 prove A + B = 45°?Worked example

Not without an interval

A calculation gives sin C = 21/221 and cos C = −220/221. Does the unit-length check prove the signs are correct?

Hint

Squaring removes signs.

Worked solution

No. The pair passes sin² C + cos² C = 1, but quadrant two and quadrant three pairs can have the same magnitudes. Check the original angle information and signed expansion.

12 / Recap

Exact ratios and interval information work together.

Split special angles into known sums or differences.
Recover missing ratios using Pythagoras and quadrant signs.
Substitute signed fractions before simplifying.
Use a nonzero combined sine or cosine to form reciprocal ratios.
Apply tangent subtraction to line inclinations, with perpendicular and vertical cases handled separately.
Use the sine rule and algebraic relationships after identifying the right angle expression.

Section 1 of 12 · Keep the ratios exact