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Precalculus / LEVEL 2 · DIFFICULTY 2/5

Trigonometric Relationships

Combine rotations and solve nonright triangles.

3 stages · 24 practice problems · two 6-question assessment forms

STAGE PC 2.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Adding Angles

Useful preparation: Trigonometry from Coordinates

Goal: Understand and apply adding angles.

Before you begin: Trigonometry from Coordinates

ONE ROTATION, TWO COORDINATES

From the unit circle to a wave

Parent relation: x2+y2=1\displaystyle x^{2} + y^{2} = 1. The radius is 1. Turn the radius and follow the horizontal cosine and vertical sine coordinates.

Trace one complete rotation
0%
xy10π2π

θ=0\displaystyle \theta = 0 rad; cos⁡ θ=1\displaystyle \operatorname{cos}\, \theta = 1; sin⁡ θ=0\displaystyle \operatorname{sin}\, \theta = 0

The solid purple curve records sin⁡ θ\displaystyle \operatorname{sin}\, \theta as the angle increases from 0 to 2 π\displaystyle 2\, \pi.

The dashed orange curve records cos⁡ θ\displaystyle \operatorname{cos}\, \theta. Both repeat after one complete turn.

Understand the idea

Combining rotations mixes horizontal and vertical components. The sine of a sum therefore contains two cross products, one from each rotation’s vertical contribution.

sin⁡ (A+B)=sin⁡ A cos⁡ B+cos⁡ A sin⁡ B\displaystyle \operatorname{sin}\, \left(A + B\right) = \operatorname{sin}\, A\, \operatorname{cos}\, B + \operatorname{cos}\, A\, \operatorname{sin}\, B

Choose and carry out a method

Apply sin⁡ (A+B)=sin⁡ A cos⁡ B+cos⁡ A sin⁡ B\displaystyle \operatorname{sin}\, \left(A + B\right) = \operatorname{sin}\, A\, \operatorname{cos}\, B + \operatorname{cos}\, A\, \operatorname{sin}\, B. Substitute exact fractions and use a common denominator before combining.

Check the reasoning

The result must lie from −1 to 1. Two acute angles can sum to an obtuse angle, so do not infer the combined quadrant merely from one angle.

WORKED EXAMPLE 1

A and B are acute angles with sin⁡ A=610\displaystyle \operatorname{sin}\, A = \frac{6}{10}, cos⁡ A=810\displaystyle \operatorname{cos}\, A = \frac{8}{10}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

  1. Use the addition identity, which mixes one sine and one cosine in each product.
  2. sin⁡ (A+B)=(610) (45)+(810) (35)\displaystyle \operatorname{sin}\, \left(A + B\right) = \left(\frac{6}{10}\right)\, \left(\frac{4}{5}\right) + \left(\frac{8}{10}\right)\, \left(\frac{3}{5}\right).
  3. The result is 2425\displaystyle \frac{24}{25}; adding the two sine values alone is not valid.

2425\displaystyle \frac{24}{25}

WORKED EXAMPLE 2

A and B are acute angles with sin⁡ A=817\displaystyle \operatorname{sin}\, A = \frac{8}{17}, cos⁡ A=1517\displaystyle \operatorname{cos}\, A = \frac{15}{17}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

  1. Use the addition identity, which mixes one sine and one cosine in each product.
  2. sin⁡ (A+B)=(817) (45)+(1517) (35)\displaystyle \operatorname{sin}\, \left(A + B\right) = \left(\frac{8}{17}\right)\, \left(\frac{4}{5}\right) + \left(\frac{15}{17}\right)\, \left(\frac{3}{5}\right).
  3. The result is 7785\displaystyle \frac{77}{85}; adding the two sine values alone is not valid.

7785\displaystyle \frac{77}{85}

Common pitfalls

Possible mix-up: sin⁡ (A+B)=sin⁡ A+sin⁡ B\displaystyle \operatorname{sin}\, \left(A + B\right) = \operatorname{sin}\, A + \operatorname{sin}\, B.

Trigonometric functions do not distribute over angle addition.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

How does the addition formula specialize when B is zero?

Preview the eight practice prompts
  1. Problem 1

    A and B are acute angles with sin⁡ A=1237\displaystyle \operatorname{sin}\, A = \frac{12}{37}, cos⁡ A=3537\displaystyle \operatorname{cos}\, A = \frac{35}{37}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

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  2. Problem 2

    A and B are acute angles with sin⁡ A=1450\displaystyle \operatorname{sin}\, A = \frac{14}{50}, cos⁡ A=4850\displaystyle \operatorname{cos}\, A = \frac{48}{50}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

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  3. Problem 3

    A and B are acute angles with sin⁡ A=1665\displaystyle \operatorname{sin}\, A = \frac{16}{65}, cos⁡ A=6365\displaystyle \operatorname{cos}\, A = \frac{63}{65}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

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  4. Problem 4

    A and B are acute angles with sin⁡ A=1882\displaystyle \operatorname{sin}\, A = \frac{18}{82}, cos⁡ A=8082\displaystyle \operatorname{cos}\, A = \frac{80}{82}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

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  5. Problem 5

    A and B are acute angles with sin⁡ A=20101\displaystyle \operatorname{sin}\, A = \frac{20}{101}, cos⁡ A=99101\displaystyle \operatorname{cos}\, A = \frac{99}{101}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

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  6. Problem 6

    A and B are acute angles with sin⁡ A=22122\displaystyle \operatorname{sin}\, A = \frac{22}{122}, cos⁡ A=120122\displaystyle \operatorname{cos}\, A = \frac{120}{122}, sin⁡ B=35\displaystyle \operatorname{sin}\, B = \frac{3}{5} and cos⁡ B=45\displaystyle \operatorname{cos}\, B = \frac{4}{5}. Find sin⁡ (A+B)\displaystyle \operatorname{sin}\, \left(A + B\right).

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  7. Problem 7New context

    Two successive rotations have sine/cosine pairs (24145,143145)\displaystyle \left(\frac{\frac{24}{145{,}143}}{145}\right) and (35,45)\displaystyle \left(\frac{3}{5}, \frac{4}{5}\right). Find the sine of their combined rotation.

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  8. Problem 8New context

    Two successive rotations have sine/cosine pairs (26170,168170)\displaystyle \left(\frac{\frac{26}{170{,}168}}{170}\right) and (35,45)\displaystyle \left(\frac{3}{5}, \frac{4}{5}\right). Find the sine of their combined rotation.

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Open stage PC 2.1 in the student workspace →

STAGE PC 2.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Double-Angle Identities

Useful preparation: Adding Angles

Goal: Understand and apply double-angle identities.

Before you begin: Adding Angles

ONE ROTATION, TWO COORDINATES

From the unit circle to a wave

Parent relation: x2+y2=1\displaystyle x^{2} + y^{2} = 1. The radius is 1. Turn the radius and follow the horizontal cosine and vertical sine coordinates.

Trace one complete rotation
0%
xy10π2π

θ=0\displaystyle \theta = 0 rad; cos⁡ θ=1\displaystyle \operatorname{cos}\, \theta = 1; sin⁡ θ=0\displaystyle \operatorname{sin}\, \theta = 0

The solid purple curve records sin⁡ θ\displaystyle \operatorname{sin}\, \theta as the angle increases from 0 to 2 π\displaystyle 2\, \pi.

The dashed orange curve records cos⁡ θ\displaystyle \operatorname{cos}\, \theta. Both repeat after one complete turn.

Understand the idea

Setting the two angles equal in an addition formula creates identities for twice an angle. The cosine identity may be expressed using both sine and cosine or using just one of them.

cos⁡ (2 θ)=cos⁡2 θ−sin⁡2 θ=1−2 sin⁡2 θ\displaystyle \operatorname{cos}\, \left(2\, \theta\right) = \operatorname{cos}^{2}\, \theta - \operatorname{sin}^{2}\, \theta = 1 - 2\, \operatorname{sin}^{2}\, \theta

Choose and carry out a method

Use cos⁡ (2 θ)=cos⁡2 θ−sin⁡2 θ\displaystyle \operatorname{cos}\, \left(2\, \theta\right) = \operatorname{cos}^{2}\, \theta - \operatorname{sin}^{2}\, \theta when both values are given. Square exact fractions and subtract carefully.

Check the reasoning

Substitute θ=0\displaystyle \theta = 0 as a quick identity check. Doubling a rotation changes the direction; it does not scale a unit vector’s length.

WORKED EXAMPLE 1

An acute angle θ\displaystyle \theta has sin⁡ θ=610\displaystyle \operatorname{sin}\, \theta = \frac{6}{10} and cos⁡ θ=810\displaystyle \operatorname{cos}\, \theta = \frac{8}{10}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

  1. The double-angle identity follows by setting both angles equal in the addition identity.
  2. cos⁡ (2 θ)=cos⁡2 θ−sin⁡2 θ=82−62102\displaystyle \operatorname{cos}\, \left(2\, \theta\right) = \operatorname{cos}^{2}\, \theta - \operatorname{sin}^{2}\, \theta = \frac{8^{2} - 6^{2}}{10^{2}}.
  3. The exact value is 725\displaystyle \frac{7}{25}. Doubling an angle does not double its cosine.

725\displaystyle \frac{7}{25}

WORKED EXAMPLE 2

An acute angle θ\displaystyle \theta has sin⁡ θ=817\displaystyle \operatorname{sin}\, \theta = \frac{8}{17} and cos⁡ θ=1517\displaystyle \operatorname{cos}\, \theta = \frac{15}{17}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

  1. The double-angle identity follows by setting both angles equal in the addition identity.
  2. cos⁡ (2 θ)=cos⁡2 θ−sin⁡2 θ=152−82172\displaystyle \operatorname{cos}\, \left(2\, \theta\right) = \operatorname{cos}^{2}\, \theta - \operatorname{sin}^{2}\, \theta = \frac{15^{2} - 8^{2}}{17^{2}}.
  3. The exact value is 161289\displaystyle \frac{161}{289}. Doubling an angle does not double its cosine.

161289\displaystyle \frac{161}{289}

Common pitfalls

Possible mix-up: cos⁡ (2 θ)=2 cos⁡ θ\displaystyle \operatorname{cos}\, \left(2\, \theta\right) = 2\, \operatorname{cos}\, \theta.

The double-angle formula includes squared values and a subtraction.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Derive the version involving only sine from the Pythagorean identity.

Preview the eight practice prompts
  1. Problem 1

    An acute angle θ\displaystyle \theta has sin⁡ θ=1237\displaystyle \operatorname{sin}\, \theta = \frac{12}{37} and cos⁡ θ=3537\displaystyle \operatorname{cos}\, \theta = \frac{35}{37}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

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  2. Problem 2

    An acute angle θ\displaystyle \theta has sin⁡ θ=1450\displaystyle \operatorname{sin}\, \theta = \frac{14}{50} and cos⁡ θ=4850\displaystyle \operatorname{cos}\, \theta = \frac{48}{50}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

    Practise this exact problem →
  3. Problem 3

    An acute angle θ\displaystyle \theta has sin⁡ θ=1665\displaystyle \operatorname{sin}\, \theta = \frac{16}{65} and cos⁡ θ=6365\displaystyle \operatorname{cos}\, \theta = \frac{63}{65}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

    Practise this exact problem →
  4. Problem 4

    An acute angle θ\displaystyle \theta has sin⁡ θ=1882\displaystyle \operatorname{sin}\, \theta = \frac{18}{82} and cos⁡ θ=8082\displaystyle \operatorname{cos}\, \theta = \frac{80}{82}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

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  5. Problem 5

    An acute angle θ\displaystyle \theta has sin⁡ θ=20101\displaystyle \operatorname{sin}\, \theta = \frac{20}{101} and cos⁡ θ=99101\displaystyle \operatorname{cos}\, \theta = \frac{99}{101}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

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  6. Problem 6

    An acute angle θ\displaystyle \theta has sin⁡ θ=22122\displaystyle \operatorname{sin}\, \theta = \frac{22}{122} and cos⁡ θ=120122\displaystyle \operatorname{cos}\, \theta = \frac{120}{122}. Find cos⁡ (2 θ)\displaystyle \operatorname{cos}\, \left(2\, \theta\right).

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  7. Problem 7New context

    A robot repeats a turn whose sine and cosine are 24145\displaystyle \frac{24}{145} and 143145\displaystyle \frac{143}{145}. What is the cosine of its total turn?

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  8. Problem 8New context

    A robot repeats a turn whose sine and cosine are 26170\displaystyle \frac{26}{170} and 168170\displaystyle \frac{168}{170}. What is the cosine of its total turn?

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Open stage PC 2.2 in the student workspace →

STAGE PC 2.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

The Cosine Rule

Useful preparation: Double-Angle Identities

Goal: Understand and apply the cosine rule.

Before you begin: Double-Angle Identities

Understand the idea

The Pythagorean theorem describes perpendicular sides. For a general included angle, a projection term adjusts the squared distance between the endpoints of two sides.

c2=a2+b2−2 ab cos⁡ C\displaystyle c^{2} = a^{2} + b^{2} - 2\, ab\, \operatorname{cos}\, C

Choose and carry out a method

Match the included angle to the two adjacent known sides. Compute c2=a2+b2−2 ab cos⁡ C\displaystyle c^{2} = a^{2} + b^{2} - 2\, ab\, \operatorname{cos}\, C and stop at c2\displaystyle c^{2} when the question asks for squared distance.

Check the reasoning

At C=90∘\displaystyle C = 90^{\circ}, the correction is zero. At an acute included angle, the opposite side’s square is smaller than a2+b2\displaystyle a^{2} + b^{2}.

WORKED EXAMPLE 1

Two sides of a triangle have lengths 4 and 6, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

  1. The cosine rule accounts for the included angle between two known sides.
  2. c2=42+62−2⋅4⋅6⋅(12)\displaystyle c^{2} = 4^{2} + 6^{2} - 2 \cdot 4 \cdot 6 \cdot \left(\frac{1}{2}\right).
  3. c2=28\displaystyle c^{2} = 28. The question asks for the square, so no square root is needed.

28

WORKED EXAMPLE 2

Two sides of a triangle have lengths 5 and 7, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

  1. The cosine rule accounts for the included angle between two known sides.
  2. c2=52+72−2⋅5⋅7⋅(12)\displaystyle c^{2} = 5^{2} + 7^{2} - 2 \cdot 5 \cdot 7 \cdot \left(\frac{1}{2}\right).
  3. c2=39\displaystyle c^{2} = 39. The question asks for the square, so no square root is needed.

39

Common pitfalls

Possible mix-up: Use a nonincluded angle with the two given sides.

The cosine rule pairs the included angle with its adjacent sides.

Possible mix-up: A correct numerical answer alone explains the method.

State the governing relationship and check the conditions described above.

Explain it to yourself

Explain why the formula reduces to the Pythagorean theorem for a right angle.

Preview the eight practice prompts
  1. Problem 1

    Two sides of a triangle have lengths 7 and 9, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

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  2. Problem 2

    Two sides of a triangle have lengths 8 and 10, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

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  3. Problem 3

    Two sides of a triangle have lengths 9 and 11, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

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  4. Problem 4

    Two sides of a triangle have lengths 10 and 12, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

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  5. Problem 5

    Two sides of a triangle have lengths 11 and 13, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

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  6. Problem 6

    Two sides of a triangle have lengths 12 and 14, with included angle 60∘\displaystyle 60^{\circ}. Find the square of the opposite side length.

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  7. Problem 7New context

    Two straight paths of lengths 13 m and 15 m leave the same junction at 60∘\displaystyle 60^{\circ}. What is the squared distance between their endpoints, in m2\displaystyle m^{2}?

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  8. Problem 8New context

    Two straight paths of lengths 14 m and 16 m leave the same junction at 60∘\displaystyle 60^{\circ}. What is the squared distance between their endpoints, in m2\displaystyle m^{2}?

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