Precalculus by Richard Wright

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Watch out for false prophets. They come to you in sheep’s clothing, but inwardly they are ferocious wolves. By their fruit you will recognize them. Matthew‬ ‭7‬:‭15‬-‭16‬a ‭NIV‬‬

5-06 Multiple Angle Formulas

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.5.1

Skittles houses with snowy roofs
Figure 1: Skittles houses with snowy roofs, Poulsbo, Washington. credit (wikimedia/nick hoke)

A roof is planned with a steepness of 3/12 meaning that the angle formed with horizontal is \(\tan θ = \frac{3}{12}\). To make snow slide off easier, the designer wants to double the angle of the roof. What would be the new steepness? The formulas for this lesson will be used to solve problems like these.

Double-Angle Formulas

The double-angle formulas are derived from the sum formulas from lesson 5-05. Let u = v in the sum formulas and simplify.

Double-Angle Formulas

Example 1: Evaluate Trigonometric Functions

A roof is planned with a steepness of 3/12 meaning that the angle formed with horizontal is \(\tan θ = \frac{3}{12}\). To make snow slide off easier, the designer wants to double the angle of the roof. a) What would be the new steepness? b) Also find sin 2θ and c) cos 2θ.

Solution
  1. By steepness, the problem is asking for what is tan 2θ? Use the double-angle formula for tangent.

    $$\tan 2θ = \frac{2 \tan θ}{1 - \tan^2 θ}$$

    $$= \frac{2\left(\frac{3}{12}\right)}{1 - \left(\frac{3}{12}\right)^2}$$

    $$= \frac{2\left(\frac{1}{4}\right)}{1 - \left(\frac{1}{4}\right)^2}$$

    $$= \frac{\frac{1}{2}}{1 - \frac{1}{16}}$$

    $$= \frac{\frac{1}{2}}{\frac{15}{16}}$$

    $$= \frac{1}{2} ⋅ \frac{16}{15}$$

    $$= \frac{8}{15}$$

  2. Tangent gives the ratio opposite/adjacent in right triangles. Draw the triangle and use the Pythagorean theorem to find the hypotenuse.

    Figure 2

    $$a^2 + b^2 = c^2$$

    $$12^2 + 3^2 = hyp^2$$

    $$153 = hyp^2$$

    $$hyp = 3\sqrt{17}$$

    So \(\sin θ = \frac{3}{3\sqrt{17}} = \frac{\sqrt{17}}{17}\) and \(\cos θ = \frac{12}{3\sqrt{17}} = \frac{4\sqrt{17}}{17}\). Now use the double angle formula for sine.

    $$\sin 2θ = 2 \sin θ \cos θ$$

    $$= 2\left(\frac{\sqrt{17}}{17}\right)\left(\frac{4\sqrt{17}}{17}\right)$$

    $$= \frac{136}{289} = \frac{8}{17}$$

  3. Cosine has three options for the double-angle formula. Choose one that is convenient.

    $$\cos 2θ = 2 \cos^2 θ - 1$$

    $$= 2 \left(\frac{4\sqrt{17}}{17}\right)^2 - 1$$

    $$= 2 \left(\frac{16}{17}\right) - 1$$

    $$= \frac{32}{17} - 1$$

    $$= \frac{15}{17}$$

Try It 1

If \(\sin α = \frac{24}{25}\), find sin 2α, cos 2α, and tan 2α.

Answer

\(\frac{336}{625}\), \(\frac{527}{625}\), \(\frac{336}{527}\)

Example 2: Solve a Trigonometric Equation

Find all solutions on the interval [0, 2π) of \(\sin x + \sin 2x = 0\).

Solution

Start by using the double-angle formula for sine.

$$\sin x + 2 \sin x \cos x = 0$$

Factor out sin x.

$$\sin x \left(1 + 2\cos x\right) = 0$$

Set each factor equal to zero and solve.

\(\sin x = 0\) \(1 + 2\cos x = 0\)
\(\cos x = -\frac{1}{2}\)
\(x = 0, π\) \(x = \frac{2π}{3}, \frac{4π}{3}\)
Try It 2

Solve on the interval [0, 2π) for \(\sin^2 x + \cos 2x = 0\).

Answer

\(\frac{π}{2}\), \(\frac{3π}{2}\)

Example 3: Derive a New Identity

Derive a triple-angle formula for sin 3x.

Solution

Start by rewriting the sin 3x as sin (2x + x) so a sum formula can be used.

$$\sin 3x = \sin \left(2x + x\right)$$

$$= \sin 2x \cos x + \cos 2x \sin x$$

Use double-angle formulas

$$= \left(\textcolor{blue}{2 \sin x \cos x}\right)\cos x + \left(\textcolor{blue}{2 \cos^2 x - 1}\right)\sin x$$

Multiply and distribute.

$$= 2 \sin x \cos^2 x + 2 \sin x \cos^2 x - \sin x$$

Combine like terms.

$$= 4 \sin x \cos^2 x - \sin x$$

Use a Pythagorean identity to change \(\cos^2 x\) to \(1 - \sin^2 x\) so that only sine is used.

$$= 4 \sin x\left(\textcolor{red}{1 - \sin^2 x}\right) - \sin x$$

Distribute.

$$= 4 \sin x - 4 \sin^3 x - \sin x$$

$$= 3 \sin x - 4 \sin^3 x$$

Try It 3

Derive a quadruple angle formula for cos 4x.

Answer

\(8\cos^4 x - 8\cos^2 x + 1\)

Power-Reducing Formulas

The power-reducing formulas change a squared trigonometric expression to no exponent. These are derived from taking the double-angle formulas for cosine and solving for the sin2 u or cos2 u. Tangent is then \(\frac{\sin u}{\cos u}\).

Power-Reducing Formulas

Example 4: Derive a Power-Reducing Formula of Higher Degree

Rewrite \(\sin^4 x\) as a sum of 1st powers of cosine.

Solution

Rewrite sin4 x in terms of sin2 x.

$$\sin^4 x = \sin^2 x \sin^2 x$$

Substitute the power-reducing formula for sine.

$$= \left(\frac{1 - \cos 2x}{2}\right)\left(\frac{1 - \cos 2x}{2}\right)$$

Multiply.

$$= \frac{1}{4}\left(1 - 2\cos 2x + \cos^2 2x\right)$$

Since a new squared term resulted, use the power-reducing formula for cosine with u = 2x.

$$= \frac{1}{4}\left(1 - 2\cos 2x + \left(\frac{1 + \cos 2\left(2x\right)}{2}\right)\right)$$

Get a common denominator.

$$= \frac{1}{4}\left(\frac{2}{2} - \frac{4\cos 2x}{2} + \frac{1 + \cos 4x}{2}\right)$$

Multiply the denominator and combine like terms.

$$= \frac{1}{8}\left(3 - 4\cos 2x + \cos 4x\right)$$

Try It 4

Rewrite cos3 x as a sum of 1st powers of cosine.

Answer

\(\frac{1}{2}\left(\cos x + \cos x \cos 2x\right)\)

Half-Angle Formulas

The half-angle formulas are derived from the power-reducing formulas. Let the angle be \(\frac{u}{2}\), then square root both sides.

Half-Angle Formulas

The ± sign means to choose the correct sign based on which quadrant the half-angle is in on the unit circle.

Example 5: Evaluate a Trigonometric Function

Find the exact value of sin 165°.

Solution

Notice that \(165° = \frac{330°}{2}\), a half-angle formula can be used with u = 330°.

$$\sin 165° = \sin \frac{330°}{2}$$

$$= ±\sqrt{\frac{1 - \cos 330°}{2}}$$

$$= ±\sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}$$

$$= ±\sqrt{\frac{\frac{2}{2} - \frac{\sqrt{3}}{2}}{2}}$$

$$= ±\sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}$$

$$= ±\sqrt{\frac{2 - \sqrt{3}}{4}}$$

$$= ±\frac{\sqrt{2 - \sqrt{3}}}{2}$$

Because 165° is in quadrant II and sine is positive in quadrant II, only the + should be used from the ±.

$$\sin 165° = \frac{\sqrt{2 - \sqrt{3}}}{2}$$

Try It 5

Find the exact value of \(\cos \frac{π}{8}\).

Answer

\(\frac{\sqrt{2+\sqrt{2}}}{2}\)

Example 6: Solve a Trigonometric Equation

Find all the solutions of \(\cos x = 2 \sin^2 \frac{x}{2}\).

Solution

The quickest way to solve this equation is to use a half-angle formula for \(\sin \frac{x}{2}\). The square on this term will cancel the square root in the half-angle formula.

$$\cos x = 2 \sin^2 \frac{x}{2}$$

$$\cos x = 2 \left(±\sqrt{\frac{1 - \cos x}{2}}\right)^2$$

$$\cos x = 2\left(\frac{1 - \cos x}{2}\right)$$

$$\cos x = 1 - \cos x$$

$$2 \cos x = 1$$

$$\cos x = \frac{1}{2}$$

$$x = \frac{2π}{3} + 2πn, \frac{5π}{3} + 2πn$$

Try It 6

Solve on the interval [0, 2π) for \(\cos x = \cos^2 \frac{x}{2}\).

Answer

0

Lesson Summary

Double-Angle Formulas

Power-Reducing Formulas

Half-Angle Formulas

The ± sign means to choose the correct sign based on which quadrant the half-angle is in on the unit circle.

Helpful videos about this lesson.

Practice Exercises (*Optional)

  1. Derive a power-reducing formula from a double-angle formula.
  2. If \(\tan θ = \frac{2}{3}\), find (a) sin 2θ, (b) cos 2θ, (c) tan 2θ.
  3. If \(\sin α = \frac{7}{25}\), find (a) sin 2α, (b) cos 2α, (c) tan 2α.
  4. *If \(\cos β = \frac{1}{2}\), find (a) sin 2β, (b) cos 2β, (c) tan 2β.
  5. Find all solutions on the interval [0, 2π).

  6. sin2 x + cos 2x = 0
  7. \(2 \sin x \cos x = \frac{\sqrt{2}}{2}\)
  8. sin 2x − cos x = 0
  9. Derive a new identity for the given expression.

  10. cos 3x
  11. sin 4x
  12. *cos 4x
  13. Rewrite the expression as a sum of 1st powers of cosine.

  14. cos3 x
  15. cos4 x
  16. tan4 x
  17. Find the exact value of the following expressions.

  18. *cos 15°
  19. tan 75°
  20. sin 105°
  21. Find all the solutions.

  22. \(\cos \frac{x}{2} = \cos x\)
  23. \(\tan \frac{x}{2} = \sin x\)
  24. *\(\cos^2 \frac{x}{2} = \cos x\)
  25. Mixed Review

  26. (5-05) Derive a formula for cos(x + π).
  27. (5-05) Solve \(\cos\left(x - \frac{3π}{2}\right) + \cos\left(x + \frac{3π}{2}\right) = 0\) on [0, 2π).
  28. (5-04) Find all the solutions of 2 tan x + 3 = 1.
  29. (5-03) Verify \(\frac{1 - \sin^2 x}{\cos^2 x} = 1\).
  30. (5-01) Simplify \(\frac{\tan^2 x}{\sec x + 1}\).

Answers

  1. \(\cos 2u = 2 \cos^2 u - 1\) → \(1 + \cos 2u = 2 \cos^2 u\) → \(\frac{1 + \cos 2u}{2} = \cos^2 u\)
  2. (a) \(\frac{12}{13}\); (b) \(\frac{5}{13}\); (c) \(\frac{12}{5}\)
  3. (a) \(\frac{336}{625}\); (b) \(\frac{527}{625}\); (c) \(\frac{336}{527}\)
  4. (a) \(\frac{\sqrt{3}}{2}\); (b) \(-\frac{1}{2}\); (c) \(-\sqrt{3}\)
  5. \(\frac{π}{2}, \frac{3π}{2}\)
  6. \(\frac{π}{8}\), \(\frac{3π}{8}\), \(\frac{9π}{8}\), \(\frac{11π}{8}\)
  7. \(\frac{π}{2}\), \(\frac{3π}{2}\), \(\frac{π}{6}\), \(\frac{5π}{6}\)
  8. cos 3x = 4 cos3 x − 3 cos x
  9. sin 4x = 8 cos3 x sin x − 4 cos x sin x
  10. cos 4x = 8 cos4 x − 8 cos2 x + 1
  11. \(\frac{1}{2}\left(\cos x + \cos 2x \cos x\right)\)
  12. \(\frac{1}{8}\left(3 + 4 \cos 2x + \cos 4x\right)\)
  13. \(\frac{3 - 4\cos 2x + \cos 4x}{3 + 4\cos 2x + \cos 4x}\)
  14. \(\frac{\sqrt{2+\sqrt{3}}}{2}\)
  15. \(2 + \sqrt{3}\)
  16. \(\frac{\sqrt{2 + \sqrt{3}}}{2}\)
  17. \(2πn, \frac{4π}{3} + 2πn\)
  18. \(\frac{π}{2}n\)
  19. n
  20. \(-\cos x\)
  21. All real numbers
  22. \(\frac{3π}{4} + πn\)
  23. \(\frac{1 - \sin^2 x}{\cos^2 x}\) \(= \frac{\cos^2 x}{\cos^2 x} = 1\)
  24. \(\sec x - 1\)