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Lesson 4 of 5

Trigonometric Functions · Lesson 4 of 5

Mixed Applications and Half-Angle Reasoning

“Combine identities with quadrant information, exact algebra and sign-sensitive half-angle reasoning.”

Learning Objectives

• Recover missing function values before evaluating a sum of two angles. • Combine product-to-sum and sum-to-product in a single proof. • Use a double-angle relationship to obtain an exact tangent value through a quadratic equation. • Derive half-angle values and select signs from the interval of the halved angle. • Simplify shifted squared terms by coordinating double-angle and sum identities.

Connecting tools rather than applying one formula

A mixed problem may give only part of the information needed by the formula you intend to use. The task is to build that missing information carefully. Keep track of what is given, what must first be recovered and which identities can connect the two.

The examples here use a common strategy: determine values and signs, select an identity, substitute exact quantities, then check the result. In proof problems, recognise which transformation exposes a common factor or cancellation. A longer problem often becomes manageable when each of these decisions is made separately.

An angle sum with two different quadrants to track

An addition formula involves both sine and cosine of both angles. If only one value for each angle is given, first recover the missing partner. The quadrants determine which square-root signs are compatible with the angles.

Example — Combining two quadrant-II angles

Problem
If sin x = 3/5 and cos y = −12/13, with x and y both in quadrant II, find sin(x + y).

  1. 1.For x, cos²x = 1 − 9/25 = 16/25. Quadrant II has negative cosine, so cos x = −4/5.
  2. 2.For y, sin²y = 1 − 144/169 = 25/169. Quadrant II has positive sine, so sin y = 5/13.
  3. 3.Use sin(x + y) = sin x cos y + cos x sin y. Substitute (3/5)(−12/13) + (−4/5)(5/13).
  4. 4.The terms are −36/65 and −20/65, giving sin(x + y) = −56/65.
  5. 5.The answer lies between −1 and 1, as a sine output must. Its negative sign also agrees with the combined angle’s position for these particular inputs.

Knowing that two angles are both in quadrant II does not by itself locate their sum in one fixed quadrant; their sum lies between π and 2π. The actual values matter. The formula handles that information exactly, while the general interval still confirms that the sine of their sum is negative.

A proof that changes representation twice

Some expressions contain products that do not share an obvious factor. Transforming the products into sums can reveal cancellation. The remaining difference can then be turned back into a product that matches the required result.

Example — Products, cancellation, then a new product

Problem
Prove cos2x cos(x/2) − cos3x cos(9x/2) = sin5x sin(5x/2).

  1. 1.Apply 2cos A cos B = cos(A + B) + cos(A − B) to each product. The left side becomes one half of [cos(5x/2) + cos(3x/2) − cos(15x/2) − cos(−3x/2)].
  2. 2.Cosine is unchanged by a negative input, so cos(−3x/2) = cos(3x/2). These two terms cancel.
  3. 3.The result is (1/2)[cos(5x/2) − cos(15x/2)]. Apply the cosine difference-to-product identity.
  4. 4.The half-sum is 5x and the half-difference is −5x/2, so the expression becomes (1/2)[−2sin5x sin(−5x/2)].
  5. 5.Use sin(−5x/2) = −sin(5x/2). The two negative signs cancel, yielding sin5x sin(5x/2). All transformations used identities valid for every real x.

Notice that the order of subtraction was retained throughout. If you swap two terms in a sine difference, its sign reverses. Writing the half-sum and half-difference explicitly is a reliable way to avoid losing that sign in a long proof.

An exact value from a double-angle equation

The double-angle formula can work in reverse. When the function value at twice an angle is known, treat the unknown value at the original angle as a variable. Solving the resulting algebraic equation gives candidates; the angle’s interval selects the valid one.

Example — Finding tan(π/8)

Problem
Find tan(π/8) exactly using a double-angle identity.

  1. 1.Let t = tan(π/8). Twice the angle is π/4, whose tangent is 1. Since π/8 is in quadrant I and below π/4, the required tangent is defined and positive.
  2. 2.The double-angle formula gives 1 = 2t/(1 − t²). The denominator is nonzero because tan(π/4) exists.
  3. 3.Multiply by 1 − t²: 1 − t² = 2t. Rearrange to t² + 2t − 1 = 0.
  4. 4.Complete the square: (t + 1)² = 2. Hence t = −1 + √2 or t = −1 − √2.
  5. 5.The second candidate is negative and cannot be tan(π/8). Thus tan(π/8) = √2 − 1. This value is less than 1, consistent with an angle smaller than π/4 in quadrant I.
Algebra gives candidates; the angle selects the value

A quadratic or squared identity can produce two roots. Do not automatically choose a positive root without checking the actual function and angle. Here tangent is positive in quadrant I; in a different quadrant the negative candidate might be relevant.

Deriving half-angle relationships

A half-angle problem is a double-angle problem read backward. Write the known angle x as twice u, where u = x/2. The double-angle cosine identities then describe the squares of sine and cosine at u.

From cos2u = 1 − 2sin²u, subtract cos2u from 1 and divide by 2 to obtain sin²u = (1 − cos2u)/2. From cos2u = 2cos²u − 1, add 1 and divide by 2 to obtain cos²u = (1 + cos2u)/2. Replace 2u by x in both expressions.

Half-angle squared valuesLaTeX
These hold for every real x. They determine magnitudes, while the interval of x/2 determines the signs of the unsquared values.

Taking square roots gives two possible signs for each function. Use the location of x/2 before selecting them. If only a quadrant is supplied, the following problems use its standard representative interval within 0 < x < 2π. If extra full turns were allowed, the actual interval would be needed because halving can change the sine and cosine signs.

Sign-sensitive half-angle valuesLaTeX
Each sign is chosen independently to match the sine or cosine sign in the quadrant of x/2. The square roots themselves are nonnegative.
Original standard intervalHalved intervalSign of sin(x/2)Sign of cos(x/2)
0 < x < π/20 < x/2 < π/4++
π/2 < x < ππ/4 < x/2 < π/2++
π < x < 3π/2π/2 < x/2 < 3π/4+−
3π/2 < x < 2π3π/4 < x/2 < π+−
horizontalverticalOangle xangle x/2(1, 0)(0, 1)(−1, 0)(0, −1)
Halving an angle in quadrant III— For a representative x between π and 3π/2, the halved angle lies in quadrant II: its sine is positive and cosine negative.
Example — Three half-angle values from a tangent

Problem
If tan x = 3/4 and π < x < 3π/2, find sin(x/2), cos(x/2) and tan(x/2).

  1. 1.First locate the half-angle: π/2 < x/2 < 3π/4. It lies in quadrant II, so its sine is positive and cosine negative.
  2. 2.To recover cos x, use sec²x = 1 + tan²x = 1 + 9/16 = 25/16. Since x lies in quadrant III, sec x = −5/4 and cos x = −4/5.
  3. 3.The sine square is [1 − (−4/5)]/2 = (9/5)/2 = 9/10. Choose the positive value: sin(x/2) = 3/√10.
  4. 4.The cosine square is [1 + (−4/5)]/2 = (1/5)/2 = 1/10. Choose the negative value: cos(x/2) = −1/√10.
  5. 5.Therefore tan(x/2) = (3/√10)/(−1/√10) = −3. The squares sum to 9/10 + 1/10 = 1, and the quotient has the negative quadrant-II sign.
Example — A half-angle when the original angle is in quadrant II

Problem
Suppose cos x = −3/5 and π/2 < x < π. Find the sine, cosine and tangent of x/2.

  1. 1.Halving the interval gives π/4 < x/2 < π/2, which is in quadrant I. Both half-angle sine and cosine are positive.
  2. 2.sin²(x/2) = [1 − (−3/5)]/2 = 4/5, so sin(x/2) = 2/√5.
  3. 3.cos²(x/2) = [1 + (−3/5)]/2 = 1/5, so cos(x/2) = 1/√5.
  4. 4.Divide the two values to obtain tan(x/2) = 2. The original angle has negative cosine, but its half-angle has positive cosine; their quadrants are different.

Once the sine and cosine half-angle values are known, tangent is their quotient wherever the cosine value is nonzero. Squaring the quotient can determine its magnitude, but using the signed sine and cosine directly is often clearer. This avoids assuming that a square-root expression for tangent is always positive.

Do not halve a reduced angle without checking full turns

Angles x and x + 2π have identical sine and cosine, but their halves differ by π. Sine and cosine of those halves therefore have opposite signs. For a half-angle task, the actual interval for the original angle matters, not only its terminal side.

Shifted squares reveal a cancelling pattern

A sum of squared cosine terms at different angles may appear complicated. Convert each square using cos²t = (1 + cos2t)/2. This separates a constant part from a set of cosine terms that can often be grouped symmetrically.

Example — Three equally spaced squared terms

Problem
Prove cos²x + cos²(x + π/3) + cos²(x − π/3) = 3/2.

  1. 1.Replace each squared term by (1 + cosine of twice its angle)/2. The sum becomes [3 + cos2x + cos(2x + 2π/3) + cos(2x − 2π/3)]/2.
  2. 2.Group the last two cosine terms. Their half-sum is 2x and half-difference is 2π/3, so their sum is 2cos2x cos(2π/3).
  3. 3.Since cos(2π/3) = −1/2, the grouped terms equal −cos2x.
  4. 4.The variable terms cancel: [3 + cos2x − cos2x]/2 = 3/2. The equality holds for all real x.

The final constant is not a coincidence from a few numerical inputs. The identity explains how the changing terms compensate for one another. This is a useful distinction between checking examples and proving a statement for every input.

Quiz

Quick check

If sin x = 3/5 and cos y = −12/13, with both angles in quadrant II, what is sin(x + y)?

Quick check

Which root gives tan(π/8)?

Quick check

If π < x < 3π/2, where is x/2?

Quick check

If cos x = −4/5 and π < x < 3π/2, what is cos(x/2)?

Quick check

Which half-angle relationship is correct?

Quick check

What is cos²x + cos²(x + π/3) + cos²(x − π/3)?

Quick check

Why can a product-to-sum step help in a difference of products?

Practice Problems

Practice Problems
  1. If sin x = 5/13 and cos y = −3/5, with x and y in quadrant II, find sin(x + y) and cos(x + y) exactly.
  2. Prove cos2x cos(x/2) − cos3x cos(9x/2) = sin5x sin(5x/2), explicitly tracking the negative half-difference.
  3. Use tan2u to derive tan(π/8), and verify the chosen root by substitution into the resulting equation.
  4. If tan x = 3/4 and π < x < 3π/2, derive all three half-angle values and check their signs.
  5. If cos x = −1/3 and π < x < 3π/2, find sin(x/2), cos(x/2) and tan(x/2).
  6. If sin x = 1/4 and π/2 < x < π, first find cos x and then derive the half-angle values.
  7. Prove the three-term squared-cosine identity and explain why checking it only at x = 0 is insufficient.
  8. Compare the signs of sin(x/2) and sin((x + 2π)/2). Explain why terminal-side information alone does not determine half-angle signs.

Key Takeaways

Key Takeaways

• Recover missing values and signs before substituting into an angle-sum formula. • A proof may use product-to-sum, cancellation and sum-to-product in succession. • Double-angle equations give candidate exact values; interval information selects the appropriate root. • Half-angle squared values come from the cosine double-angle identities. • Choose sine and cosine signs using the interval of the halved angle, retaining information about full turns. • Converting shifted squares into double-angle cosines can reveal a sum that is constant for all inputs.

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Angle Combinations and Trigonometric Identities

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Chapter Summary and Practice