Precalculus

Double angle formulas

The sum formulas handle the sine and cosine of any two angles added together. What do they say when the two angles are the same, so that the angle is simply doubled?

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Sine and tangent of a sum or differencePrecalculus

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Power reducing and half angle formulasPrecalculus Solving trig equations by factoring, quadratics and identitiesPrecalculus Trigonometric integralsCalculus 2

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Double angle formulas

Power reducing and half angle formulas

Why this exists

The sum formulas (Cosine of a sum or difference and Sine and tangent of a sum or difference) take two angles a and b. The case that comes up more than any other is b = a: an angle doubled. A point on a wheel that turns twice as fast, a wave of twice the frequency, a launch angle that appears in a range formula as 2\theta: each needs \sin 2x or \cos 2x written in terms of \sin x and \cos x.

Doubling does not pass through the sine. At x = 90^\circ, \sin 2x = \sin 180^\circ = 0, while 2\sin x = 2. Setting b = a in the sum formulas gives the true rules, and the Pythagorean identity then gives cosine three useful shapes.

Definition

The double angle formulas are the identities

\begin{aligned} \sin 2x &= 2\sin x\cos x \\ \cos 2x &= \cos^2 x - \sin^2 x \\ &= 1 - 2\sin^2 x \\ &= 2\cos^2 x - 1 \end{aligned}

which hold for every real x (or angle in degrees), and

\tan 2x = \frac{2\tan x}{1 - \tan^2 x}

which holds whenever \tan x and \tan 2x both exist: x is not an odd multiple of \displaystyle \frac{\pi}{2}, and x is not \displaystyle \frac{\pi}{4} + k\cdot\frac{\pi}{2} for any integer k. They are identities in the sense of What an identity is. The three lines for \cos 2x are one identity in three forms: pick the form whose ingredients are known.

Intuition

Figure 1One triangle, two areas. With base OA = 1 the height is BF = \sin 2\theta. With base AB = 2\sin\theta the height is OM = \cos\theta. The two areas must agree.

Take an acute angle \theta and build an isosceles triangle with two sides of length 1 meeting at O with angle 2\theta. Its area can be found two ways.

Lay side OA along the ground as the base. The top vertex B is the terminal point of 2\theta on a circle of radius 1, so its height is BF = \sin 2\theta, and the area is \displaystyle \frac{1}{2}\sin 2\theta.

Now use AB as the base. The segment OM to the midpoint of AB splits the triangle into two right triangles, each with hypotenuse 1 and angle \theta at O. So OM = \cos\theta, MA = \sin\theta, the base is AB = 2\sin\theta, and the area (Area of triangles) is \displaystyle \frac{1}{2}(2\sin\theta)(\cos\theta). Setting the two areas equal gives \sin 2\theta = 2\sin\theta\cos\theta. The picture only covers 0^\circ < \theta < 90^\circ; the proof below covers every real number.

Results

Sine of a double angle. For all real x, \sin 2x = 2\sin x\cos x.

Cosine of a double angle, three forms. For all real x,

\begin{aligned} \cos 2x &= \cos^2 x - \sin^2 x \\ &= 1 - 2\sin^2 x = 2\cos^2 x - 1 \end{aligned}

Tangent of a double angle. When \tan x and \tan 2x exist, \tan 2x = \dfrac{2\tan x}{1 - \tan^2 x}, and then the denominator is not 0.

Triple angles. For all real x, \sin 3x = 3\sin x - 4\sin^3 x (Example 5) and \cos 3x = 4\cos^3 x - 3\cos x (Exercise 15).

Why it is true

Sine. The sine sum formula \sin(a + b) = \sin a\cos b + \cos a\sin b holds for every pair of real numbers, so it holds when a = b = x:

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

The two products are equal because multiplication is commutative.

Cosine. The cosine sum formula with a = b = x gives \cos 2x = \cos^2 x - \sin^2 x, as in Exercise 17 of Cosine of a sum or difference. The Pythagorean identity \sin^2 x + \cos^2 x = 1 (The Pythagorean identities) lets either square be traded for the other. Replacing \cos^2 x by 1 - \sin^2 x gives 1 - 2\sin^2 x; replacing \sin^2 x by 1 - \cos^2 x gives 2\cos^2 x - 1.

Tangent. Where \cos x \ne 0 and \cos 2x \ne 0, tangent is sine over cosine (Reciprocal and quotient identities), so

\tan 2x = \frac{2\sin x\cos x}{\cos^2 x - \sin^2 x}

Divide the top and bottom by \cos^2 x, which is not 0. The top becomes \displaystyle 2\cdot\frac{\sin x}{\cos x} = 2\tan x and the bottom becomes 1 - \tan^2 x. The same division shows

1 - \tan^2 x = \frac{\cos 2x}{\cos^2 x}

so the denominator is 0 exactly when \cos 2x = 0, that is, when \tan 2x does not exist. That happens when \tan x = \pm 1, at \displaystyle x = \frac{\pi}{4} + k\cdot\frac{\pi}{2} with k any integer.

Worked examples

Examples 1 to 3 belong to Day 1: the sine and the three cosines of a double angle, found from one value and a quadrant, and read backward. Examples 4 to 7 belong to Day 2: the tangent of a double angle, building larger multiples, and a use in physics.

Example 1

Both double angle values from one cosine

Nikola knows that \displaystyle \cos\theta = -\frac{48}{73} and that \theta is in Quadrant II. Find \sin 2\theta and \cos 2\theta, and the quadrant of 2\theta.

Solution

The sine formula needs \sin\theta as well. The Pythagorean identity gives it up to sign, and in Quadrant II sine is positive:

\sin\theta = \sqrt{1 - \tfrac{2304}{5329}} = \sqrt{\tfrac{3025}{5329}} = \frac{55}{73}

Then, with 73^2 = 5329,

\begin{aligned} \sin 2\theta &= 2\cdot\tfrac{55}{73}\cdot\left(-\tfrac{48}{73}\right) = -\frac{5280}{5329} \\ \cos 2\theta &= \tfrac{2304}{5329} - \tfrac{3025}{5329} = -\frac{721}{5329} \end{aligned}

The form \displaystyle 2\cos^2\theta - 1 = \frac{4608 - 5329}{5329} gives the same \displaystyle -\frac{721}{5329} without needing \sin\theta at all. Both values are negative, so 2\theta ends in Quadrant III.

Check. \theta lies between 90^\circ and 180^\circ, so 2\theta lies between 180^\circ and 360^\circ, where sine is negative, as found. Also 5280^2 + 721^2 = 28{,}398{,}241 = 5329^2.

Example 2

When the quadrant is missing

Slavica is told only that \displaystyle \sin\theta = \frac{1}{4}. Which of \cos 2\theta and \sin 2\theta can she find?

Solution

The form \cos 2\theta = 1 - 2\sin^2\theta uses nothing but \sin\theta, and squaring removes any doubt about sign:

\cos 2\theta = 1 - 2\cdot\tfrac{1}{16} = \frac{7}{8}

The sine formula is different. It needs \displaystyle \cos\theta = \pm\sqrt{1 - \frac{1}{16}} = \pm\frac{\sqrt{15}}{4}, and a sine of \displaystyle \frac{1}{4} happens in Quadrant I, where cosine is positive, and in Quadrant II, where it is negative. So

\sin 2\theta = 2\cdot\tfrac{1}{4}\cdot\left(\pm\tfrac{\sqrt{15}}{4}\right) = \pm\frac{\sqrt{15}}{8}

and Slavica cannot choose the sign without the quadrant. This is the reason for having three forms of \cos 2\theta: each one fits a different piece of information.

Example 3

Reading the formulas backward

Find the exact values of (a) \sin 22.5^\circ\cos 22.5^\circ and (b) \displaystyle 1 - 2\sin^2\frac{5\pi}{12}, and (c) write \cos^4 x - \sin^4 x as a single trig function.

Solution

(a) A product of the sine and cosine of the same angle is half of a double angle sine, since \displaystyle \sin x\cos x = \frac{1}{2}\sin 2x:

\sin 22.5^\circ\cos 22.5^\circ = \tfrac{1}{2}\sin 45^\circ = \frac{\sqrt{2}}{4}

(b) This is the form 1 - 2\sin^2 x with \displaystyle x = \frac{5\pi}{12}, so it equals \displaystyle \cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}.

\begin{aligned} \cos^4 x - \sin^4 x &= (\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x) \\ &= 1\cdot\cos 2x = \cos 2x \end{aligned}

for every real x. Check at x = 0: both sides are 1.

Example 4

The tangent of a double angle, and where it fails

Tomislav knows that \tan\theta = 3 with \theta in Quadrant III. Find \tan 2\theta, \sin 2\theta and \cos 2\theta. Then try the tangent formula for an angle \varphi with \tan\varphi = 1.

Solution

The tangent formula needs only \tan\theta:

\tan 2\theta = \frac{2\cdot 3}{1 - 3^2} = \frac{6}{-8} = -\frac{3}{4}

For the sine and cosine, a point on the terminal side in Quadrant III with \displaystyle \frac{y}{x} = 3 is (-1, -3), at distance \displaystyle \sqrt{10} from the origin. So \displaystyle \cos\theta = -\frac{1}{\sqrt{10}} and \displaystyle \sin\theta = -\frac{3}{\sqrt{10}}, and

\begin{aligned} \sin 2\theta &= 2\left(-\tfrac{3}{\sqrt{10}}\right)\left(-\tfrac{1}{\sqrt{10}}\right) = \frac{3}{5} \\ \cos 2\theta &= \tfrac{1}{10} - \tfrac{9}{10} = -\frac{4}{5} \end{aligned}

Check. \displaystyle \frac{3/5}{-4/5} = -\frac{3}{4}, agreeing with the tangent formula. Had \theta been in Quadrant I, both \sin\theta and \cos\theta would change sign, and their product would not. A tangent fixes \theta up to a multiple of 180^\circ, so it fixes 2\theta up to a multiple of 360^\circ: the quadrant of \theta was not needed after all.

For \tan\varphi = 1 the denominator is 1 - 1 = 0. That is the formula reporting that \tan 2\varphi does not exist: \varphi = 45^\circ + k\cdot 180^\circ, so 2\varphi = 90^\circ + k\cdot 360^\circ, where tangent is undefined.

Example 5

Sine of a triple angle

Write \sin 3x in terms of \sin x alone, and use it to find \sin 3x when \displaystyle \sin x = \frac{1}{3}.

Solution

Write 3x = 2x + x and use the sine sum formula, then the double angle formulas, choosing the cosine form that uses only \sin x:

\begin{aligned} \sin 3x &= \sin 2x\cos x + \cos 2x\sin x \\ &= 2\sin x\cos^2 x + (1 - 2\sin^2 x)\sin x \end{aligned}

The leftover \cos^2 x becomes 1 - \sin^2 x (the Pythagorean identity), and the distributive property collects the terms:

\begin{aligned} \sin 3x &= 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \\ &= 3\sin x - 4\sin^3 x \end{aligned}

With \displaystyle \sin x = \frac{1}{3}, \displaystyle \sin 3x = 1 - \frac{4}{27} = \frac{23}{27}, and no quadrant was needed. Check at \displaystyle x = \frac{\pi}{6}: \displaystyle \sin\frac{\pi}{2} = 1, and \displaystyle 3\cdot\frac{1}{2} - 4\cdot\frac{1}{8} = 1.

Example 6

An identity, and the tangent of an eighth of a turn

Show that \cot x - \tan x = 2\cot 2x wherever both sides are defined, and use it to find \displaystyle \tan\frac{\pi}{8} exactly.

Solution

Work on the left side alone (Verifying identities). Over the common denominator \sin x\cos x,

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

The top is \cos 2x, and the bottom is \displaystyle \frac{1}{2}\sin 2x by the sine formula read backward. So the left side is \displaystyle \frac{\cos 2x}{\frac{1}{2}\sin 2x} = 2\cot 2x. Both sides are defined exactly when \sin x\cos x \ne 0, which is when \sin 2x \ne 0: x not a multiple of \displaystyle \frac{\pi}{2}.

Now put \displaystyle x = \frac{\pi}{8} and write \displaystyle t = \tan\frac{\pi}{8}. Since \displaystyle \cot\frac{\pi}{4} = 1,

\frac{1}{t} - t = 2

Multiplying by t gives t^2 + 2t - 1 = 0, so \displaystyle t = -1 \pm \sqrt{2}. The angle \displaystyle \frac{\pi}{8} is acute, so its tangent is positive, and \displaystyle \tan\frac{\pi}{8} = \sqrt{2} - 1 \approx 0.4142.

Example 7

The best angle for a long kick

Tomislav kicks a soccer ball from level ground at 20 meters per second, at an angle \theta above the ground. In the model that ignores air resistance, after t seconds the ball is 20\cos\theta\cdot t meters downfield and -4.9t^2 + 20\sin\theta\cdot t meters up. Find the range R (the distance downfield where the ball lands) as a function of \theta, the angle with the longest range, and the angles that land the ball 30 meters away.

Solution

The height factors as t(-4.9t + 20\sin\theta), which is 0 at t = 0 (the kick) and at \displaystyle t = \frac{20\sin\theta}{4.9} (the landing). At that time the ball is downfield by

\begin{aligned} R &= 20\cos\theta\cdot\frac{20\sin\theta}{4.9} = \frac{400\sin\theta\cos\theta}{4.9} \\ &= \frac{400\sin 2\theta}{9.8} \end{aligned}

The last step multiplies top and bottom by 2 to make 2\sin\theta\cos\theta, which is \sin 2\theta. Now the range is easy to read. A sine is at most 1, and \sin 2\theta = 1 for 0^\circ < \theta < 90^\circ only at 2\theta = 90^\circ. So the longest kick is at \theta = 45^\circ, with \displaystyle R = \frac{400}{9.8} \approx 40.82 meters.

Figure 2Range of a ball launched at 20 meters per second, \displaystyle R = \frac{400}{9.8}\sin 2\theta. The curve is symmetric about 45^\circ, so two complementary angles reach every range below the maximum.

For R = 30, solve \displaystyle \sin 2\theta = \frac{30 \cdot 9.8}{400} = 0.735. The inverse sine gives 2\theta \approx 47.31^\circ, and sine takes the same value at 180^\circ - 47.31^\circ = 132.69^\circ. Both are between 0^\circ and 180^\circ, so \theta \approx 23.65^\circ or \theta \approx 66.35^\circ. The two angles add to 90^\circ, and that always happens: \sin 2(90^\circ - \theta) = \sin(180^\circ - 2\theta) = \sin 2\theta, so complementary angles give equal ranges in this model.

Boundaries

The sine and cosine formulas hold for every real number, in radians or degrees, with no exceptions.
The tangent formula needs \tan x to exist. At \displaystyle x = \frac{\pi}{2} it cannot be written, even though \tan 2x = \tan\pi = 0 exists; go back to \displaystyle \frac{\sin 2x}{\cos 2x} there. It also fails where \tan x = \pm 1, which is exactly where \tan 2x does not exist.
A double angle value found from \cos\theta or \sin\theta alone may still need the quadrant of \theta: \sin 2\theta does, and the form of \cos 2\theta built from the given function does not (Example 2).
Doubling does not pass through: \sin 2x = 2\sin x and \cos 2x = 2\cos x are false as identities, though each holds at a few values (Exercise 21).
The range formula \displaystyle R = \frac{v^2\sin 2\theta}{9.8} belongs to the model in Example 7, with v in meters per second, R in meters, and 9.8 meters per second squared as the pull of gravity near the ground: no air resistance, and launch and landing at the same height. A real ball in air falls short of it.

Where this goes next

Solving the cosine forms for \sin^2 x and \cos^2 x gives the Power reducing and half angle formulas, and putting \displaystyle \frac{x}{2} in place of x gives the half angle formulas there. The Product to sum and sum to product formulas generalize the step \displaystyle \sin x\cos x = \frac{1}{2}\sin 2x to two different angles. An equation such as \sin 2x = \cos x becomes \cos x(2\sin x - 1) = 0, which is the method of Solving trig equations by factoring, quadratics and identities. A wave such as \sin x\cos x is a single wave of half the height and twice the frequency, a first case of Combining a sine and a cosine into one wave. In calculus, these rewrites are how powers of sine and cosine are handled in Trigonometric integrals.

Exercises

Day 1: from one value and a quadrant

\displaystyle \sin\theta = -\frac{5}{13} with \theta in Quadrant IV. Find \sin 2\theta and \cos 2\theta.
\displaystyle \cos\theta = -\frac{2}{7} with \theta in Quadrant III. Find \sin 2\theta and \cos 2\theta.
\displaystyle \sin\theta = \frac{2}{3} with \theta in Quadrant II. Find \sin 2\theta and \cos 2\theta, and the quadrant of 2\theta.
Slavica knows only that \displaystyle \cos\theta = \frac{3}{10}. Find \cos 2\theta, and find both possible values of \sin 2\theta.
\sec\theta = -3 with \theta in Quadrant II. Find \sin 2\theta and \cos 2\theta.

Day 1: reading the formulas backward

Find the exact value of \displaystyle 2\sin\frac{5\pi}{12}\cos\frac{5\pi}{12}.
Find the exact value of \cos^2 105^\circ - \sin^2 105^\circ.
Find the exact value of \displaystyle 1 - 2\sin^2\frac{\pi}{12}.
Write each as a single trig function: (a) 2\cos^2 5x - 1; (b) 6\sin 2x\cos 2x.
Show that 4\sin x\cos x\cos 2x = \sin 4x.

Day 2: tangent of a double angle

\displaystyle \tan\theta = \frac{1}{2} with \theta acute. Find \tan 2\theta, \sin 2\theta and \cos 2\theta.
\displaystyle \cos\theta = -\frac{13}{85} with \theta in Quadrant III. Find \sin 2\theta, \cos 2\theta and \tan 2\theta.
\displaystyle \tan\theta = 2 - \sqrt{3} with \theta acute. Find \tan 2\theta, and from it find \theta.
Write \dfrac{2\tan 3x}{1 - \tan^2 3x} as a single tangent, and say for which x the two agree.

Day 2: building on the formulas

Show that \cos 3x = 4\cos^3 x - 3\cos x.
Write \cos 4x in terms of \cos x alone.
\displaystyle \sin x = \frac{1}{3} with x acute. Find \sin 4x exactly.
Show that \sin 2x = \dfrac{2\tan x}{1 + \tan^2 x} wherever \tan x exists.
Using the model of Example 7, Tomislav kicks at 15 meters per second and 30^\circ. Find the range to two decimal places, and the other angle with the same range.
In the same model, what launch speed lands a ball 50 meters away at the best angle? Give it to two decimal places.

Reasoning

Nikola writes \sin 2x = 2\sin x. Give one x where this is true and one where it is false, and explain why it is not an identity.
Explain why \cos 2\theta can be found from \sin\theta alone, but \sin 2\theta generally cannot.
Explain why the tangent formula cannot be used at \displaystyle x = \frac{\pi}{2} even though \tan\pi exists, and find every x where its denominator is 0.
Find the largest value of \sin x\cos x and every x where it is reached.
Use the double angle formula to explain why \sin 2\theta < 2\sin\theta for 0^\circ < \theta < 90^\circ, and what this says about the triangle in the first figure.

Answers

\displaystyle \sin 2\theta = -\frac{120}{169}, \displaystyle \cos 2\theta = \frac{119}{169}. In Quadrant IV, \displaystyle \cos\theta = \frac{12}{13}; then \displaystyle 2\left(-\frac{5}{13}\right)\frac{12}{13} and \displaystyle 1 - 2\cdot\frac{25}{169}.
\displaystyle \sin 2\theta = \frac{12\sqrt{5}}{49}, \displaystyle \cos 2\theta = -\frac{41}{49}. In Quadrant III, \displaystyle \sin\theta = -\frac{3\sqrt{5}}{7}; then \displaystyle 2\left(-\frac{3\sqrt{5}}{7}\right)\left(-\frac{2}{7}\right) and \displaystyle 2\cdot\frac{4}{49} - 1.
\displaystyle \sin 2\theta = -\frac{4\sqrt{5}}{9}, \displaystyle \cos 2\theta = \frac{1}{9}, so 2\theta is in Quadrant IV. Here \displaystyle \cos\theta = -\frac{\sqrt{5}}{3}, and \displaystyle \cos 2\theta = 1 - 2\cdot\frac{4}{9}.
\displaystyle \cos 2\theta = -\frac{41}{50} and \displaystyle \sin 2\theta = \pm\frac{3\sqrt{91}}{50}. \displaystyle 2\cdot\frac{9}{100} - 1 = -\frac{82}{100}. \displaystyle \sin\theta = \pm\frac{\sqrt{91}}{10}, positive in Quadrant I and negative in Quadrant IV, the two quadrants where cosine is positive.
\displaystyle \sin 2\theta = -\frac{4\sqrt{2}}{9}, \displaystyle \cos 2\theta = -\frac{7}{9}. \displaystyle \cos\theta = -\frac{1}{3} and \displaystyle \sin\theta = \frac{2\sqrt{2}}{3}; then \displaystyle 2\cdot\frac{2\sqrt{2}}{3}\left(-\frac{1}{3}\right) and \displaystyle 2\cdot\frac{1}{9} - 1.
\displaystyle \frac{1}{2}. It is \displaystyle \sin\frac{5\pi}{6}.
\displaystyle -\frac{\sqrt{3}}{2}. It is \cos 210^\circ.
\displaystyle \frac{\sqrt{3}}{2}. It is \displaystyle \cos\frac{\pi}{6}.
(a) \cos 10x, the form 2\cos^2 x - 1 with 5x for x. (b) 3\sin 4x, since 6\sin 2x\cos 2x = 3(2\sin 2x\cos 2x).
4\sin x\cos x = 2\sin 2x, so the left side is 2\sin 2x\cos 2x, which is \sin 4x by the sine formula with 2x for x.
\displaystyle \tan 2\theta = \frac{4}{3}, \displaystyle \sin 2\theta = \frac{4}{5}, \displaystyle \cos 2\theta = \frac{3}{5}. \displaystyle \frac{1}{1 - \frac{1}{4}} = \frac{4}{3}. The point (2, 1) is on the terminal side, so \displaystyle \sin\theta = \frac{1}{\sqrt{5}} and \displaystyle \cos\theta = \frac{2}{\sqrt{5}}.
\displaystyle \sin 2\theta = \frac{2184}{7225}, \displaystyle \cos 2\theta = -\frac{6887}{7225}, \displaystyle \tan 2\theta = -\frac{2184}{6887}. \displaystyle \sin\theta = -\frac{84}{85}, so \displaystyle 2\cdot\frac{84\cdot 13}{7225} and \displaystyle \frac{169 - 7056}{7225}; the tangent is their quotient, and the formula with \displaystyle \tan\theta = \frac{84}{13} agrees.
\displaystyle \tan 2\theta = \frac{\sqrt{3}}{3}, so \theta = 15^\circ. \displaystyle (2 - \sqrt{3})^2 = 7 - 4\sqrt{3}, so the denominator 1 - \tan^2\theta is \displaystyle 4\sqrt{3} - 6, which factors as \displaystyle 2\sqrt{3}(2 - \sqrt{3}). The numerator is \displaystyle 2(2 - \sqrt{3}), so the quotient is \displaystyle \frac{1}{\sqrt{3}}. As 2\theta is between 0^\circ and 180^\circ, 2\theta = 30^\circ.
\tan 6x, wherever \tan 3x exists and \tan 3x \ne \pm 1, that is, wherever both \tan 3x and \tan 6x exist.
By the cosine sum formula, \cos 3x = \cos 2x\cos x - \sin 2x\sin x. The double angle formulas make this (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x. With \sin^2 x = 1 - \cos^2 x it becomes 2\cos^3 x - \cos x - 2\cos x + 2\cos^3 x, which is 4\cos^3 x - 3\cos x.
\cos 4x = 8\cos^4 x - 8\cos^2 x + 1. First \cos 4x = 2\cos^2 2x - 1, and \cos 2x = 2\cos^2 x - 1, so \cos 4x = 2(2\cos^2 x - 1)^2 - 1. The square is 4\cos^4 x - 4\cos^2 x + 1.
\displaystyle \frac{56\sqrt{2}}{81}. \displaystyle \cos x = \frac{2\sqrt{2}}{3}, so \displaystyle \sin 2x = \frac{4\sqrt{2}}{9} and \displaystyle \cos 2x = 1 - \frac{2}{9} = \frac{7}{9}; then \displaystyle \sin 4x = 2\cdot\frac{4\sqrt{2}}{9}\cdot\frac{7}{9}.
\displaystyle 2\sin x\cos x = 2\cdot\frac{\sin x}{\cos x}\cdot\cos^2 x = \frac{2\tan x}{\sec^2 x}, and \sec^2 x = 1 + \tan^2 x (divide the Pythagorean identity by \cos^2 x).
About 19.88 meters, and 60^\circ. \displaystyle R = \frac{225\sin 60^\circ}{9.8}; the complement of 30^\circ gives the same \sin 2\theta.
About 22.14 meters per second. At 45^\circ, \displaystyle R = \frac{v^2}{9.8} = 50, so \displaystyle v = \sqrt{490}.
True at x = \pi: both sides are 0. False at \displaystyle x = \frac{\pi}{2}: \sin\pi = 0 but \displaystyle 2\sin\frac{\pi}{2} = 2. An identity must hold at every value, so one failure rules it out; the true rule has the extra factor \cos x.
The form 1 - 2\sin^2\theta uses only \sin^2\theta, and a square does not depend on any sign. The sine formula needs \cos\theta itself, and \sin\theta fixes \cos\theta only up to sign, which depends on the quadrant.
The formula is built from \tan x, and \displaystyle \tan\frac{\pi}{2} does not exist, so nothing can be substituted; \displaystyle \tan\pi = \frac{\sin\pi}{\cos\pi} = 0 is found from sine and cosine instead. The denominator 1 - \tan^2 x is 0 when \tan x = \pm 1: \displaystyle x = \frac{\pi}{4} + k\cdot\frac{\pi}{2}, k any integer.
\displaystyle \frac{1}{2}, at \displaystyle x = \frac{\pi}{4} + k\pi, k any integer. \displaystyle \sin x\cos x = \frac{1}{2}\sin 2x, and \sin 2x is at most 1, reaching it when \displaystyle 2x = \frac{\pi}{2} + 2k\pi.
\sin 2\theta = 2\sin\theta\cos\theta, and 0 < \cos\theta < 1 while \sin\theta > 0, so multiplying 2\sin\theta by \cos\theta makes it smaller. In the figure, the height BF = \sin 2\theta is shorter than the side AB = 2\sin\theta, as it must be: BF is the perpendicular from B to line OA, the shortest path from B to that line.