Precalculus · F.TF.C.9

Combining a sine and a cosine into one wave

A sine wave plus a cosine wave of the same period: is the sum another wave, and if so, how tall is it and where does it peak?

You need

Sine and tangent of a sum or differencePrecalculus Amplitude and reflectionPrecalculus

Where you are 13 in Trigonometric identities

Trigonometry7

Trigonometric identities13

▴ 8 above Product to sum and sum to product formulas Sine and tangent of a sum or difference

Combining a sine and a cosine into one wave

Double angle formulas Power reducing and half angle formulas

Why this exists

Expand 5\sin(x + 1) with the sine sum formula and you get 5\cos 1\,\sin x + 5\sin 1\,\cos x. A shifted sine wave is a mix of \sin x and \cos x.

Mixes like that also turn up on their own. A weight on a spring that starts away from rest and already moving, two sound sources a little apart, two parts of an alternating current circuit: each produces a sine term plus a cosine term of the same period. The questions are the ones asked of any wave. How far does it swing? When is it largest? When does it reach a given value? None of them is easy to read from 3\sin x + 2\cos x, because two terms change at once.

Running the sum formula backward turns the mix into one wave, R\sin(x + \alpha). A single wave answers all three questions at sight: its Amplitude and reflection is R, and its peaks and its solutions come from one sine.

Definition

Let a and b be real numbers, not both 0. The sum a\sin x + b\cos x is a linear combination of \sin x and \cos x: each is multiplied by a constant and the results are added. Writing it as a single sinusoid means finding a number R > 0 and an angle \alpha with

a\sin x + b\cos x = R\sin(x + \alpha)

for every real x. The two numbers are fixed by

\begin{aligned} R &= \sqrt{a^2 + b^2} \\ \cos\alpha &= \frac{a}{R}, \quad \sin\alpha = \frac{b}{R} \end{aligned}

Both conditions on \alpha are needed. Together they pick out one angle in each turn of 2\pi; either one alone allows two.

The same sum can also be written in cosine form, with the same R:

a\sin x + b\cos x = R\cos(x - \beta)

where \displaystyle \cos\beta = \frac{b}{R} and \displaystyle \sin\beta = \frac{a}{R}. The roles of a and b swap, because in the expansion of \cos(x - \beta) the term \cos x is the one that carries \cos\beta.

Intuition

Treat the two coefficients as a point (a, b) in the plane. Its distance from the origin is \displaystyle \sqrt{a^2 + b^2} = R. A point at distance R and at angle \alpha from the positive x-axis has coordinates (R\cos\alpha, R\sin\alpha) (the The unit circle scaled by R), so \alpha is exactly the direction of (a, b). Converting the sum is describing one point by its distance and direction instead of by its coordinates. Figure 2, in Example 2, draws it.

On a graph, the two terms are waves whose peaks fall in different places. Where one is high the other is partway down, so the sum is again a wave of the same period. It is taller than either part, since \displaystyle \sqrt{a^2 + b^2} exceeds both |a| and |b| when neither is 0, but shorter than the two heights stacked, |a| + |b|, because the two peaks never line up.

Figure 1The sum of the two dashed waves is the solid wave, a single sine of amplitude 2. Its peak at \displaystyle \frac{2\pi}{3} is where neither part peaks.

Results

Single sine. For a and b not both 0, with \displaystyle R = \sqrt{a^2 + b^2},

a\sin x + b\cos x = R\sin(x + \alpha)

where \displaystyle \cos\alpha = \frac{a}{R} and \displaystyle \sin\alpha = \frac{b}{R}.

Single cosine. With the same R, a\sin x + b\cos x = R\cos(x - \beta), where \displaystyle \cos\beta = \frac{b}{R} and \displaystyle \sin\beta = \frac{a}{R}. The angle \displaystyle \beta = \frac{\pi}{2} - \alpha always works.

Reading the wave. The amplitude is R. The greatest value R occurs where \displaystyle x + \alpha = \frac{\pi}{2} + 2\pi k, which is x = \beta + 2\pi k, and the least value -R where \displaystyle x + \alpha = \frac{3\pi}{2} + 2\pi k, for every integer k.

Faster waves. For k \neq 0, a\sin(kx) + b\cos(kx) = R\sin(kx + \alpha) with the same R and \alpha. Its period is \displaystyle \frac{2\pi}{|k|}.

Solving. The equation a\sin x + b\cos x = c becomes \displaystyle \sin(x + \alpha) = \frac{c}{R}. It has solutions exactly when |c| \le R.

Finding \alpha. If a > 0, \displaystyle \alpha = \tan^{-1}\frac{b}{a}. If a < 0, \displaystyle \alpha = \tan^{-1}\frac{b}{a} + \pi. If a = 0, \displaystyle \alpha = \frac{\pi}{2} when b > 0 and \displaystyle \alpha = -\frac{\pi}{2} when b < 0.

Why it is true

Expand the single sine with the sum formula and group the terms:

\begin{aligned} R\sin(x + \alpha) &= R\sin x\cos\alpha + R\cos x\sin\alpha \\ &= (R\cos\alpha)\sin x + (R\sin\alpha)\cos x \end{aligned}

If R\cos\alpha = a and R\sin\alpha = b, this is a\sin x + b\cos x for every x. Those two conditions are also forced: putting \displaystyle x = \frac{\pi}{2} in an identity a\sin x + b\cos x = R\sin(x + \alpha) gives a = R\cos\alpha, and putting x = 0 gives b = R\sin\alpha.

Squaring the two conditions and adding gives R^2(\cos^2\alpha + \sin^2\alpha) = a^2 + b^2, so \displaystyle R = \sqrt{a^2 + b^2}, which is positive because a and b are not both 0. Then \displaystyle \left(\frac{a}{R}\right)^2 + \left(\frac{b}{R}\right)^2 = 1, so the point \displaystyle \left(\frac{a}{R}, \frac{b}{R}\right) lies on the unit circle and is (\cos\alpha, \sin\alpha) for an angle \alpha that is unique up to adding multiples of 2\pi. That proves the single sine form exists, and that \alpha needs both its cosine and its sine.

The cosine form is the same argument with R\cos(x - \beta) = (R\sin\beta)\sin x + (R\cos\beta)\cos x. By the Cofunction identities, \displaystyle \cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha = \frac{b}{R} and \displaystyle \sin\left(\frac{\pi}{2} - \alpha\right) = \cos\alpha = \frac{a}{R}, so \displaystyle \beta = \frac{\pi}{2} - \alpha fits.

For the arctangent rule: when a \neq 0, \displaystyle \tan\alpha = \frac{b}{a}. The inverse tangent returns an angle in \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right), where cosine is positive, so it is the right angle exactly when a > 0. When a < 0, adding \pi changes the sign of both the sine and the cosine and leaves the tangent alone, which moves the angle to the side where \cos\alpha < 0. For faster waves, replace x by kx throughout: an identity true for every input is true for the input kx.

Worked examples

Example 1

A first combination

Write \displaystyle \sqrt{3}\sin x - \cos x as a single sine. Give its amplitude and the first positive x where it is greatest.

Solution

Here \displaystyle a = \sqrt{3} and b = -1, so \displaystyle R = \sqrt{3 + 1} = 2. Then \displaystyle \cos\alpha = \frac{\sqrt{3}}{2} and \displaystyle \sin\alpha = -\frac{1}{2}. A positive cosine and a negative sine put \alpha in Quadrant IV, and \displaystyle \alpha = -\frac{\pi}{6}. So

\sqrt{3}\sin x - \cos x = 2\sin\left(x - \frac{\pi}{6}\right)

The amplitude is 2. The sine is 1 when \displaystyle x - \frac{\pi}{6} = \frac{\pi}{2} + 2\pi k, so \displaystyle x = \frac{2\pi}{3} + 2\pi k, and the first positive one is \displaystyle x = \frac{2\pi}{3}.

Check. At x = 0 the left side is -1 and the right side is \displaystyle 2\sin\left(-\frac{\pi}{6}\right) = -1. At \displaystyle x = \frac{2\pi}{3} the left side is \displaystyle \sqrt{3}\cdot\frac{\sqrt{3}}{2} - \left(-\frac{1}{2}\right) = 2. Figure 1 shows the two parts and their sum.

Example 2

When the arctangent points the wrong way

Write \displaystyle -3\sin x + 3\sqrt{3}\cos x in the form R\sin(x + \alpha).

Solution

\displaystyle R = \sqrt{9 + 27} = 6, so \displaystyle \cos\alpha = -\frac{1}{2} and \displaystyle \sin\alpha = \frac{\sqrt{3}}{2}. A negative cosine and a positive sine put \alpha in Quadrant II, and \displaystyle \alpha = \frac{2\pi}{3}:

-3\sin x + 3\sqrt{3}\cos x = 6\sin\left(x + \frac{2\pi}{3}\right)

A calculator given only \displaystyle \tan\alpha = \frac{b}{a} = -\sqrt{3} returns \displaystyle \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3}, whose cosine \displaystyle \frac{1}{2} has the wrong sign. The tangent cannot tell \displaystyle (-3, 3\sqrt{3}) from \displaystyle (3, -3\sqrt{3}): both lie on one line through the origin (Figure 2). Using \displaystyle -\frac{\pi}{3} would give \displaystyle 6\sin\left(x - \frac{\pi}{3}\right), which is \displaystyle -3\sqrt{3} at x = 0, while the original sum is \displaystyle 3\sqrt{3} there. Since a < 0, the rule adds \pi: \displaystyle -\frac{\pi}{3} + \pi = \frac{2\pi}{3}.

Figure 2The two points lie on one line through the origin, so they have the same ratio \displaystyle \frac{b}{a} and the same tangent. Only the signs of a and b say which one is meant.

Example 3

Both forms, with no special angle

Write 4\sin x - 7\cos x in the form R\cos(x - \beta) and in the form R\sin(x + \alpha), with angles in radians to four decimal places. Where is it greatest?

Solution

\displaystyle R = \sqrt{16 + 49} = \sqrt{65} \approx 8.0623.

Cosine form: \displaystyle \cos\beta = \frac{-7}{\sqrt{65}} and \displaystyle \sin\beta = \frac{4}{\sqrt{65}}, so \beta is in Quadrant II. Its reference angle is the acute angle with tangent \displaystyle \frac{4}{7}, \displaystyle \tan^{-1}\frac{4}{7} \approx 0.5191, so \beta \approx \pi - 0.5191 \approx 2.6224 and

4\sin x - 7\cos x \approx \sqrt{65}\cos(x - 2.6224)

Sine form: \displaystyle \cos\alpha = \frac{4}{\sqrt{65}} and \displaystyle \sin\alpha = \frac{-7}{\sqrt{65}}, Quadrant IV, so \displaystyle \alpha = -\tan^{-1}\frac{7}{4} \approx -1.0517 and \displaystyle 4\sin x - 7\cos x \approx \sqrt{65}\sin(x - 1.0517). The two agree: \displaystyle \frac{\pi}{2} - \alpha \approx 1.5708 + 1.0517 = 2.6225, which is \beta up to rounding.

The cosine form shows the peak directly. \cos(x - \beta) = 1 when x - \beta = 2\pi k, so the greatest value \displaystyle \sqrt{65} \approx 8.0623 occurs at x \approx 2.6224 + 2\pi k.

Check. At x = \beta, \displaystyle \sin x = \frac{4}{\sqrt{65}} \approx 0.4961 and \displaystyle \cos x = \frac{-7}{\sqrt{65}} \approx -0.8682, so 4\sin x - 7\cos x is about 1.9846 + 6.0777 = 8.0623.

Example 4

Ragnheidur’s faster wave

Ragnheidur graphs f(x) = 2\sin(3x) - 2\cos(3x). Write it as a single sine, and give its period, its phase shift and the first positive x where it is greatest.

Solution

The coefficients are a = 2 and b = -2, so \displaystyle R = \sqrt{8} = 2\sqrt{2}, \displaystyle \cos\alpha = \frac{\sqrt{2}}{2} and \displaystyle \sin\alpha = -\frac{\sqrt{2}}{2}: Quadrant IV, \displaystyle \alpha = -\frac{\pi}{4}. The identity holds with 3x in place of x, so

f(x) = 2\sqrt{2}\sin\left(3x - \frac{\pi}{4}\right)

The period is \displaystyle \frac{2\pi}{3}. The angle \displaystyle -\frac{\pi}{4} is not the slide: factor out the 3 first, \displaystyle 3x - \frac{\pi}{4} = 3\left(x - \frac{\pi}{12}\right), so the graph is \displaystyle y = 2\sqrt{2}\sin(3x) moved \displaystyle \frac{\pi}{12} to the right.

f is greatest, \displaystyle 2\sqrt{2}, when \displaystyle 3x - \frac{\pi}{4} = \frac{\pi}{2} + 2\pi k, that is \displaystyle x = \frac{\pi}{4} + \frac{2\pi k}{3}. The first positive one is \displaystyle x = \frac{\pi}{4}. Check: \displaystyle f\left(\frac{\pi}{4}\right) = 2\sin\frac{3\pi}{4} - 2\cos\frac{3\pi}{4} = \sqrt{2} + \sqrt{2} = 2\sqrt{2}.

Example 5

Thi solves an equation

Thi solves \sin x - \cos x = 1 for 0 \le x < 2\pi.

Solution

Two different functions of x sit on the left, and one sine is easier to undo. Here \displaystyle R = \sqrt{2}, \displaystyle \cos\alpha = \frac{\sqrt{2}}{2} and \displaystyle \sin\alpha = -\frac{\sqrt{2}}{2}, so \displaystyle \alpha = -\frac{\pi}{4} and the equation becomes

\begin{aligned} \sqrt{2}\sin\left(x - \frac{\pi}{4}\right) &= 1 \\ \sin\left(x - \frac{\pi}{4}\right) &= \frac{\sqrt{2}}{2} \end{aligned}

This is a basic sine equation in \displaystyle u = x - \frac{\pi}{4}. As x runs over [0, 2\pi), u runs over \displaystyle \left[-\frac{\pi}{4}, \frac{7\pi}{4}\right), and there \displaystyle \sin u = \frac{\sqrt{2}}{2} only at \displaystyle u = \frac{\pi}{4} and \displaystyle u = \frac{3\pi}{4}. Adding \displaystyle \frac{\pi}{4} back, \displaystyle x = \frac{\pi}{2} or x = \pi.

Check. \displaystyle \sin\frac{\pi}{2} - \cos\frac{\pi}{2} = 1 - 0 = 1 and \sin\pi - \cos\pi = 0 - (-1) = 1.

The single sine also shows when there is no solution: \sin x - \cos x never exceeds \displaystyle \sqrt{2} \approx 1.414, so \displaystyle \sin x - \cos x = \frac{3}{2} has none.

Example 6

Hoa’s hammock

Hoa’s hammock sways sideways. Its distance from rest in feet, t seconds after she starts timing, is d(t) = 1.2\sin(\pi t) + 0.5\cos(\pi t), positive on one side and negative on the other. How far from rest does it swing, and when does it first get there? In the first 2 seconds, when is d(t) = -0.65?

Solution

\displaystyle R = \sqrt{1.44 + 0.25} = \sqrt{1.69} = 1.3. Then \displaystyle \cos\alpha = \frac{1.2}{1.3} = \frac{12}{13} and \displaystyle \sin\alpha = \frac{0.5}{1.3} = \frac{5}{13}, both positive, so \displaystyle \alpha = \tan^{-1}\frac{5}{12} \approx 0.3948 and

d(t) \approx 1.3\sin(\pi t + 0.3948)

with period \displaystyle \frac{2\pi}{\pi} = 2 seconds. The hammock swings 1.3 feet from rest. It first reaches d = 1.3 when \displaystyle \pi t + \alpha = \frac{\pi}{2}, at \displaystyle t = \frac{1}{2} - \frac{\alpha}{\pi} \approx 0.5 - 0.1257 \approx 0.374 seconds.

d(t) = -0.65 means \displaystyle \sin(\pi t + \alpha) = \frac{-0.65}{1.3} = -\frac{1}{2}. For 0 \le t < 2 the angle \pi t + \alpha runs over [\alpha, 2\pi + \alpha) \approx [0.395, 6.678). The sine is \displaystyle -\frac{1}{2} there at \displaystyle \frac{7\pi}{6} \approx 3.665 and \displaystyle \frac{11\pi}{6} \approx 5.760; the next, \displaystyle \frac{7\pi}{6} + 2\pi \approx 9.948, is too large. Solving \displaystyle \pi t + \alpha = \frac{7\pi}{6} and \displaystyle \pi t + \alpha = \frac{11\pi}{6} gives \displaystyle t = \frac{7}{6} - \frac{\alpha}{\pi} \approx 1.041 and \displaystyle t = \frac{11}{6} - \frac{\alpha}{\pi} \approx 1.708 seconds.

Check. d(1.041) = 1.2\sin(3.2704) + 0.5\cos(3.2704), which is about -0.1541 - 0.4959 = -0.650.

Figure 3One full sway of the hammock. The farthest point and the two crossings of d = -0.65 are read from the single sine.

Boundaries

Both terms need the same input. a\sin(kx) + b\cos(kx) combines; \sin(2x) + \cos(3x) does not, and is not a single sinusoid at all. Its period is found in Finding the period from the equation, Example 6. Even \sin(2x) + \cos x is out of reach (Exercise 22).
a = b = 0 gives the zero function, with R = 0 and no angle \alpha. Every other pair works.
\alpha is fixed only up to adding 2\pi k for an integer k: \displaystyle 6\sin\left(x + \frac{2\pi}{3}\right) and \displaystyle 6\sin\left(x - \frac{4\pi}{3}\right) are the same function. The same holds for \beta.
\displaystyle \tan^{-1}\frac{b}{a} is \alpha only when a > 0. When a < 0 add \pi; when a = 0 there is no ratio and \displaystyle \alpha = \pm\frac{\pi}{2}.
a\sin x + b\cos x = c has solutions only when |c| \le R.

Where this goes next

Equations that do not fit one sine, such as 2\sin x\cos x = \cos x, are solved by splitting them into factors in Solving trig equations by factoring, quadratics and identities. In Simple harmonic motion, the simplest model of a weight on a spring, with no friction, gives its position as \displaystyle x_0\cos(\omega t) + \frac{v_0}{\omega}\sin(\omega t), where x_0 is where it starts and v_0 its starting velocity; this lesson turns that mix into an amplitude and a phase.

Exercises

Writing one sine

Write each as R\sin(x + \alpha) with R > 0. Give \alpha exactly when it is a special angle, otherwise in radians to four decimal places.

\sin x + \cos x
\displaystyle 5\sqrt{3}\sin x + 5\cos x
-4\sin x - 4\cos x
8\sin x - 15\cos x
-12\sin x + 9\cos x

The cosine form

Write each as R\cos(x - \beta) with R > 0.

\displaystyle \cos x - \sqrt{3}\sin x
6\sin x + 6\cos x
-2\sin x - 3\cos x, with \beta to four decimal places

Amplitude, period and extremes

Give the amplitude and the greatest and least values of 7\sin x + 24\cos x.
For \displaystyle g(x) = \sin(2x) + \sqrt{3}\cos(2x), give the period, the greatest value, and every x where it occurs.
For h(t) = 4\sin(\pi t) - 4\cos(\pi t), give the period, and the greatest and least values with where each occurs for 0 \le t < 2.
Find the least value of \displaystyle -\sqrt{2}\sin x + \sqrt{2}\cos x and the first positive x where it occurs.

Solving equations

Solve \sin x + \cos x = -1 for 0 \le x < 2\pi.
Solve \displaystyle \sqrt{3}\sin x + \cos x = \sqrt{2} for 0 \le x < 2\pi.
Solve 2\sin x + 5\cos x = 3 for 0 \le x < 2\pi, to four decimal places.
Give every solution of \displaystyle \sin x - \sqrt{3}\cos x = 2.
For which numbers c does 5\sin x - 12\cos x = c have a solution? Does it have one for c = 14?

Waves that add

Ragnheidur adds two waves of amplitude 3, y = 3\sin t and \displaystyle y = 3\sin\left(t + \frac{\pi}{3}\right). Write the sum as one sine and give its amplitude.
Thi’s model of a vibrating reed is f(t) = 0.8\sin(4t) + 0.6\cos(4t). Give the amplitude, the period, and the first t > 0 where f is greatest, to four decimal places.

Reasoning

Explain why \displaystyle \tan^{-1}\frac{b}{a} gives a wrong \alpha for -\sin x - \cos x, and find the right one.
Show that the amplitude of a\sin x + b\cos x is never more than |a| + |b|, and say when it equals |a| + |b|.
Hoa claims \sin(2x) + \cos x = p\sin x + q\cos x for some constants p and q. Show that no such constants exist.
With b \neq 0, show that a\sin x + b\cos x is greatest at an x with \displaystyle \tan x = \frac{a}{b}, and explain why not every such x gives the greatest value.

Answers

\displaystyle \sqrt{2}\sin\left(x + \frac{\pi}{4}\right). \displaystyle R = \sqrt{2}, \displaystyle \cos\alpha = \sin\alpha = \frac{\sqrt{2}}{2}, Quadrant I.
\displaystyle 10\sin\left(x + \frac{\pi}{6}\right). \displaystyle R = \sqrt{75 + 25} = 10, \displaystyle \cos\alpha = \frac{\sqrt{3}}{2}, \displaystyle \sin\alpha = \frac{1}{2}.
\displaystyle 4\sqrt{2}\sin\left(x - \frac{3\pi}{4}\right). \displaystyle R = \sqrt{32} = 4\sqrt{2}, \displaystyle \cos\alpha = \sin\alpha = -\frac{\sqrt{2}}{2}, Quadrant III, so \displaystyle \alpha = -\frac{3\pi}{4} (or \displaystyle \frac{5\pi}{4}).
17\sin(x - 1.0808). \displaystyle R = \sqrt{64 + 225} = 17, \displaystyle \cos\alpha = \frac{8}{17}, \displaystyle \sin\alpha = -\frac{15}{17}, Quadrant IV, \displaystyle \alpha = -\tan^{-1}\frac{15}{8} \approx -1.0808.
15\sin(x + 2.4981). \displaystyle R = \sqrt{144 + 81} = 15, \displaystyle \cos\alpha = -\frac{4}{5}, \displaystyle \sin\alpha = \frac{3}{5}, Quadrant II, \displaystyle \alpha = \pi - \tan^{-1}\frac{3}{4} \approx 2.4981.
\displaystyle 2\cos\left(x + \frac{\pi}{3}\right). Here \displaystyle a = -\sqrt{3}, b = 1, R = 2; \displaystyle \cos\beta = \frac{1}{2}, \displaystyle \sin\beta = -\frac{\sqrt{3}}{2}, so \displaystyle \beta = -\frac{\pi}{3} and \displaystyle x - \beta = x + \frac{\pi}{3}.
\displaystyle 6\sqrt{2}\cos\left(x - \frac{\pi}{4}\right). \displaystyle R = 6\sqrt{2}, \displaystyle \cos\beta = \sin\beta = \frac{\sqrt{2}}{2}.
\displaystyle \sqrt{13}\cos(x - 3.7296). \displaystyle \cos\beta = -\frac{3}{\sqrt{13}}, \displaystyle \sin\beta = -\frac{2}{\sqrt{13}}, Quadrant III, \displaystyle \beta = \pi + \tan^{-1}\frac{2}{3} \approx 3.7296 (equally -2.5536).
Amplitude 25, greatest 25, least -25. \displaystyle R = \sqrt{49 + 576} = \sqrt{625} = 25.
\displaystyle g(x) = 2\sin\left(2x + \frac{\pi}{3}\right), period \pi, greatest value 2 at \displaystyle x = \frac{\pi}{12} + \pi k for every integer k, from \displaystyle 2x + \frac{\pi}{3} = \frac{\pi}{2} + 2\pi k.
\displaystyle h(t) = 4\sqrt{2}\sin\left(\pi t - \frac{\pi}{4}\right), period 2. Greatest \displaystyle 4\sqrt{2} at \displaystyle t = \frac{3}{4} (from \displaystyle \pi t - \frac{\pi}{4} = \frac{\pi}{2}), least \displaystyle -4\sqrt{2} at \displaystyle t = \frac{7}{4} (from \displaystyle \pi t - \frac{\pi}{4} = \frac{3\pi}{2}).
Least value -2 at \displaystyle x = \frac{3\pi}{4}. The sum is \displaystyle 2\sin\left(x + \frac{3\pi}{4}\right) (\displaystyle \cos\alpha = -\frac{\sqrt{2}}{2}, \displaystyle \sin\alpha = \frac{\sqrt{2}}{2}), least when \displaystyle x + \frac{3\pi}{4} = \frac{3\pi}{2}.
x = \pi or \displaystyle x = \frac{3\pi}{2}. \displaystyle \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) = -1, so \displaystyle \sin\left(x + \frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} with \displaystyle x + \frac{\pi}{4} in \displaystyle \left[\frac{\pi}{4}, \frac{9\pi}{4}\right): \displaystyle \frac{5\pi}{4} or \displaystyle \frac{7\pi}{4}.
\displaystyle x = \frac{\pi}{12} or \displaystyle x = \frac{7\pi}{12}. \displaystyle 2\sin\left(x + \frac{\pi}{6}\right) = \sqrt{2}, so \displaystyle x + \frac{\pi}{6} = \frac{\pi}{4} or \displaystyle \frac{3\pi}{4}.
x \approx 1.3604 or x \approx 5.6838. \displaystyle \sqrt{29}\sin(x + 1.1903) = 3 with \displaystyle \alpha = \tan^{-1}\frac{5}{2} \approx 1.1903; \displaystyle \sin^{-1}\frac{3}{\sqrt{29}} \approx 0.5909, so x + \alpha \approx \pi - 0.5909 = 2.5507 or 0.5909 + 2\pi \approx 6.8741, then subtract \alpha.
\displaystyle x = \frac{5\pi}{6} + 2\pi k for every integer k. The left side is \displaystyle 2\sin\left(x - \frac{\pi}{3}\right), so \displaystyle \sin\left(x - \frac{\pi}{3}\right) = 1 and \displaystyle x - \frac{\pi}{3} = \frac{\pi}{2} + 2\pi k.
Exactly when -13 \le c \le 13, since \displaystyle R = \sqrt{25 + 144} = 13 and the left side takes every value from -13 to 13. For c = 14 there is no solution.
\displaystyle 3\sqrt{3}\sin\left(t + \frac{\pi}{6}\right), amplitude \displaystyle 3\sqrt{3} \approx 5.196, not 6. Expanding, the sum is \displaystyle \frac{9}{2}\sin t + \frac{3\sqrt{3}}{2}\cos t; \displaystyle R = \sqrt{\frac{81}{4} + \frac{27}{4}} = \sqrt{27}, \displaystyle \cos\alpha = \frac{\sqrt{3}}{2}, \displaystyle \sin\alpha = \frac{1}{2}.
Amplitude 1, period \displaystyle \frac{\pi}{2}, first greatest at t \approx 0.2318. f(t) = \sin(4t + 0.6435) with \displaystyle \alpha = \tan^{-1}\frac{0.6}{0.8} \approx 0.6435; \displaystyle 4t + \alpha = \frac{\pi}{2} gives \displaystyle t = \frac{\pi/2 - 0.6435}{4}.
\displaystyle \frac{b}{a} = \frac{-1}{-1} = 1 and \displaystyle \tan^{-1} 1 = \frac{\pi}{4}, but \displaystyle \cos\frac{\pi}{4} > 0 while \displaystyle \frac{a}{R} = -\frac{\sqrt{2}}{2} < 0. The point (-1, -1) is in Quadrant III, so \displaystyle \alpha = \frac{\pi}{4} + \pi = \frac{5\pi}{4}: the sum is \displaystyle \sqrt{2}\sin\left(x + \frac{5\pi}{4}\right).
(|a| + |b|)^2 = a^2 + b^2 + 2|a||b|, which is at least a^2 + b^2 = R^2, and both R and |a| + |b| are not negative, so R \le |a| + |b|. Equality needs 2|a||b| = 0, that is a = 0 or b = 0, when the sum is already a single sine or cosine.
At x = 0 the identity would give 1 = q, and at \displaystyle x = \frac{\pi}{2} it would give 0 = p, so the right side would be \cos x. But at \displaystyle x = \frac{\pi}{4} the left side is \displaystyle 1 + \frac{\sqrt{2}}{2} and \displaystyle \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}. So no constants work.
The greatest value is at x = \beta, where \displaystyle \cos\beta = \frac{b}{R} \neq 0 and \displaystyle \sin\beta = \frac{a}{R}, so \displaystyle \tan\beta = \frac{a}{b}. But tangent repeats every \pi, so x = \beta + \pi has the same tangent; there \cos(x - \beta) = -1 and the sum is least.