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Trigonometric maxima and minima

Find trigonometric maxima, minima and ranges using amplitude, double angles and completing the square. Check restricted intervals, reciprocal signs and whether bounds are attained.

Before you startThe R formula, trig graphs, inequalities and completing the square

01 / A global amplitude is only the starting point

The allowed interval decides whether a bound is attained.

The R formula gives the full-wave range. A question may restrict x to a short interval, so you must check whether the wave actually reaches its highest and lowest points there. Use the model to change both the interval and the expression.

f(x) = 3sin x + 4cos x
Global range: −5 ≤ f(x) ≤ 5

A maximum or minimum must be a value the function actually takes at an allowed input. A bound that is approached but never reached is not an attained extremum.

Extrema inside the windowExplore

ExpressionClosed interval

Here f(x) = 3sin x + 4cos x. Green marks minima, gold maxima. Readouts are rounded to four decimal places; angles are in degrees.

Range: 4.0000 ≤ y ≤ 4.9641.

Minimum at x = 0.0000°. Maximum at x = 30.0000°.

The global bounds −5 and 5 need not be reached in this window.

02 / Add the vertical offset to both amplitude bounds

The midline and amplitude have different roles.

Find the global range of 7 + 3sin x − 4cos x.Worked example

3sin x − 4cos x has amplitude 5

The signs change the phase, but not the amplitude.

−5 ≤ 3sin x − 4cos x ≤ 5

Over all real x, both bounds are attained.

2 ≤ 7 + 3sin x − 4cos x ≤ 12

Add the offset to the whole inequality.

For c + R sin(mx + α), with R ≥ 0 and nonzero m, the global range is [c − R, c + R]. If m = 0 the expression is constant and the full-wave range rule does not apply.

03 / Test the endpoints and any internal peaks or troughs

Do not replace an interval question with a full-turn answer.

Find the range of 3sin x + 4cos x for 0° ≤ x ≤ 30°.Worked example

Write f(x) = 5cos(x − β), with β = arctan(3/4)

Here β ≈ 36.87°.

The global peak occurs at x = β

It lies outside this interval.

The function increases throughout the interval

The cosine input moves towards zero from below.

f(0°) = 4; f(30°) = 3/2 + 2√3

The endpoints give the attained extrema.

4 ≤ f(x) ≤ 3/2 + 2√3

The upper bound is less than 5.

For a longer closed interval, include every internal input where the shifted sine or cosine equals ±1, then compare those values with both endpoints.

Watch the maximum enter an expanding interval

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

04 / An internal peak can beat both endpoints

The smallest endpoint is not necessarily a global trough.

Find the range of f(x) = 3sin x + 4cos x on 0° ≤ x ≤ 90°.Worked example

f(0°) = 4 and f(90°) = 3

Compare both included endpoints.

x = arctan(3/4) lies inside the interval

At this internal peak f(x) = 5.

There is no trough inside this interval

The next full-wave trough is 180° after the peak.

The range is 3 ≤ f(x) ≤ 5

The minimum occurs at 90°, the maximum at arctan(3/4).

05 / Distinguish a bound from an attained extremum

An open endpoint can remove a maximum or minimum.

For the same f, restrict x to 0° < x < β, where β = arctan(3/4). The function is strictly increasing on this interval.

4 < f(x) < 5

It approaches 4 near the excluded lower endpoint and 5 near the excluded upper endpoint. Neither value is attained, so there is no minimum and no maximum on this open interval. The lower and upper bounds are still 4 and 5.

Change just one endpointWorked example

On 0° ≤ x < β, the range is [4, 5)

There is a minimum 4, but no maximum.

On 0° < x ≤ β, the range is (4, 5]

There is a maximum 5, but no minimum.

06 / Reduce squared and product terms to one frequency

A double-angle identity can reveal the amplitude.

Find the global range of 6sin² x + 8sin x cos x.Worked example

6sin² x = 3 − 3cos 2x

Use sin² x = (1 − cos 2x)/2.

8sin x cos x = 4sin 2x

Use sin 2x = 2sin x cos x.

The expression is 3 + 4sin 2x − 3cos 2x

The same-frequency terms have amplitude 5.

The global range is [−2, 8]

Add the offset 3 to [−5, 5].

The cross-product term can be negative, so a negative minimum is consistent with the squared term being nonnegative.

07 / For a quadratic in one trig value, use its allowed input range

Completing the square identifies the vertex; endpoints still matter.

Find the global range of 4sin² x − 4sin x + 3.Worked example

Put u = sin x, with −1 ≤ u ≤ 1

This is a quadratic on a restricted u interval.

4u² − 4u + 3 = 4(u − 1/2)² + 2

The vertex u = 1/2 lies inside the allowed range.

Minimum = 2, when sin x = 1/2

The square is zero there.

At u = −1 the value is 11; at u = 1 it is 3

Check both endpoints for the maximum.

Range: 2 ≤ y ≤ 11

Every intermediate value is attained by continuity.

Do not assume that the largest sine always gives the largest quadratic expression. If the vertex lies outside the permitted trig-value range, it is not a candidate extremum.

08 / For a positive reciprocal, the order reverses

First prove the denominator stays away from zero.

Find the global range of 1/(8 + 3sin x + 4cos x).Worked example

3 ≤ 8 + 3sin x + 4cos x ≤ 13

The denominator is positive for every real x.

1/13 ≤ 1/(8 + 3sin x + 4cos x) ≤ 1/3

A larger positive denominator gives a smaller reciprocal.

Both bounds are attained

The original wave reaches both ±5.

Now put −2 in the numeratorWorked example

Multiply the reciprocal inequality by −2

The inequality directions reverse.

−2/3 ≤ −2/(8 + 3sin x + 4cos x) ≤ −2/13

Keep the lower, more negative number on the left.

09 / Zero crossings and negative denominators need their own checks

Never take reciprocals across a zero-containing interval.

For D = −8 + 3sin x + 4cos x, the denominator range is [−13, −3]. It is never zero, and the reciprocal range is [−1/3, −1/13]. The reciprocal function is decreasing on this negative interval too.

Why endpoints fail for 1/(3 + 3sin x + 4cos x)Worked example

The denominator ranges from −2 to 8

Its range includes zero.

It reaches zero at some inputs

The reciprocal is undefined at those inputs.

It can become arbitrarily large positive or negative near a zero

There is no finite global maximum or minimum.

Find the range of 2 + 3sin(0x + π/6).

Hint

The input does not vary with x.

Worked solution

The expression is constantly 2 + 3/2 = 7/2. Its range is the single value {7/2}, not [−1, 5].

11 / Recap

Find bounds and prove that the relevant values are reached.

Use amplitude and offset for a full nonconstant wave.
On a closed interval, compare endpoints with internal peaks and troughs.
Open endpoints may remove attainment.
Reduce double-angle expressions or complete a square in a bounded trig variable.
Check denominator signs and zeros before taking reciprocals.

Section 1 of 11 · A global amplitude is only the starting point