Precalculus

Power reducing and half angle formulas

A squared sine or cosine is awkward to graph and to add up. Can the square be traded for a plain cosine of a doubled angle, and does the same trade give the sine and cosine of half an angle?

You need

Double angle formulasPrecalculus Signs of the trig functions by quadrantPrecalculus

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

A square such as \sin^2 x is harder to work with than a first power. Its graph is not one of the familiar waves at sight, and a sum of squares and fourth powers resists simplifying. The double angle formulas already link squares to a doubled angle: \cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1. Solved for the square, they turn every \sin^2 x or \cos^2 x into a first power of \cos 2x. Read the other way, with x standing for half an angle, the same equations give the sine and cosine of half an angle from the cosine of the whole one. That is how exact values such as \cos 112.5^\circ are found, and the only new difficulty is choosing a sign.

Definition

The power reducing formulas write a square as a first power of a cosine of the doubled angle:

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

The first two hold for every real x; the third holds wherever \tan x exists, that is, wherever \cos x \ne 0.

The half angle formulas are the same facts with \displaystyle \frac{x}{2} in place of x:

\begin{aligned} \sin\frac{x}{2} &= \pm\sqrt{\frac{1 - \cos x}{2}} \\ \cos\frac{x}{2} &= \pm\sqrt{\frac{1 + \cos x}{2}} \end{aligned}

Each \pm means one sign, not both: it is the sign that \displaystyle \sin\frac{x}{2} or \displaystyle \cos\frac{x}{2} actually has, read from the quadrant where the angle \displaystyle \frac{x}{2} ends, by the rules of Signs of the trig functions by quadrant. The quadrant of x itself does not decide it. The tangent of half an angle needs no sign choice at all:

\tan\frac{x}{2} = \frac{1 - \cos x}{\sin x} = \frac{\sin x}{1 + \cos x}

The first form holds where \sin x \ne 0, the second where \cos x \ne -1.

Intuition

Figure 1\displaystyle \sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x and \displaystyle \cos^2 x = \frac{1}{2} + \frac{1}{2}\cos 2x: two waves of period \pi, centered on \displaystyle y = \frac{1}{2}, mirror images of each other across that line.

A square is never negative, and \sin^2 x runs between 0 and 1. Squaring also erases the difference between \sin x and -\sin x, so the graph repeats every half turn instead of every full turn: its period is \pi, not 2\pi. A curve that swings between 0 and 1 with period \pi has midline \displaystyle y = \frac{1}{2}, amplitude \displaystyle \frac{1}{2}, and the shape of \cos 2x, which also has period \pi. At x = 0 the square is at its lowest, so the wave is flipped: \displaystyle \sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x. The cosine square is the same wave right side up, and the two curves add to 1 at every x, which is the Pythagorean identity seen on a graph.

Results

Power reducing. For all real x, \sin^2 x = \dfrac{1 - \cos 2x}{2} and \cos^2 x = \dfrac{1 + \cos 2x}{2}. Where \cos x \ne 0, \tan^2 x = \dfrac{1 - \cos 2x}{1 + \cos 2x}.

Half angle. For all real x, \displaystyle \left|\sin\frac{x}{2}\right| = \sqrt{\frac{1 - \cos x}{2}} and \displaystyle \left|\cos\frac{x}{2}\right| = \sqrt{\frac{1 + \cos x}{2}}; the sign is the sign of the function at the angle \displaystyle \frac{x}{2}.

Tangent of a half angle. \displaystyle \tan\frac{x}{2} = \dfrac{1 - \cos x}{\sin x} where \sin x \ne 0, and \displaystyle \tan\frac{x}{2} = \dfrac{\sin x}{1 + \cos x} where \cos x \ne -1.

Why it is true

Power reducing. The double angle formula \cos 2x = 1 - 2\sin^2 x holds for every real x. Add 2\sin^2 x to both sides, subtract \cos 2x, and divide by 2: \displaystyle \sin^2 x = \frac{1 - \cos 2x}{2}. The form \cos 2x = 2\cos^2 x - 1 gives \displaystyle \cos^2 x = \frac{1 + \cos 2x}{2} the same way. Where \cos x \ne 0, tangent is sine over cosine (Reciprocal and quotient identities), so \displaystyle \tan^2 x = \frac{\sin^2 x}{\cos^2 x}; the twos cancel and the tangent formula follows. Its denominator 1 + \cos 2x = 2\cos^2 x is 0 exactly where \tan x does not exist.

Half angle. The power reducing formulas hold for every real number, so they hold for \displaystyle \frac{x}{2}, whose double is x:

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

Two numbers with this square are \displaystyle \pm\sqrt{\frac{1 - \cos x}{2}}, and \displaystyle \sin\frac{x}{2} is one of them. The equation cannot say which, because squaring threw the sign away. The cosine formula follows the same way.

Tangent of a half angle. Write \displaystyle u = \frac{x}{2}. Where \sin u \ne 0 and \cos u \ne 0, multiply the top and bottom of \displaystyle \frac{\sin u}{\cos u} by 2\sin u:

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

using 2\sin^2 u = 1 - \cos 2u and 2\sin u\cos u = \sin 2u. Multiplying instead by 2\cos u gives \displaystyle \frac{2\sin u\cos u}{2\cos^2 u} = \frac{\sin x}{1 + \cos x}. No square root was taken, so no sign was lost: 1 - \cos x \ge 0 and 1 + \cos x \ge 0, and the sign of \displaystyle \tan\frac{x}{2} is simply the sign of \sin x.

Worked examples

Example 1

A product of squares

Write \sin^2 x\cos^2 x with no power of a trig function above the first.

Solution

The product is a square of a product, and \displaystyle \sin x\cos x = \frac{1}{2}\sin 2x by the double angle formula read backward:

\sin^2 x\cos^2 x = (\sin x\cos x)^2 = \tfrac{1}{4}\sin^2 2x

The power reducing formula for sine, with 2x in place of x, gives \displaystyle \sin^2 2x = \frac{1 - \cos 4x}{2}. So

\sin^2 x\cos^2 x = \frac{1 - \cos 4x}{8}

Check at \displaystyle x = \frac{\pi}{4}: the left side is \displaystyle \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4}, and the right side is \displaystyle \frac{1 - \cos\pi}{8} = \frac{2}{8} = \frac{1}{4}.

Example 2

A fourth power

Write \sin^4 x with no power above the first.

Solution

A fourth power is the square of a square, so reduce the inner square first:

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

The square of a binomial left a new square, \cos^2 2x. Reduce it too, with 2x in place of x: \displaystyle \cos^2 2x = \frac{1 + \cos 4x}{2}. Then multiply the top and bottom by 2 to clear the inner fraction:

\begin{aligned} \sin^4 x &= \frac{2 - 4\cos 2x + 1 + \cos 4x}{8} \\ &= \frac{3 - 4\cos 2x + \cos 4x}{8} \end{aligned}

Check at \displaystyle x = \frac{\pi}{2}: \displaystyle \sin^4\frac{\pi}{2} = 1, and \displaystyle \frac{3 + 4 + 1}{8} = 1.

Example 3

An exact value from a half angle

Bhavya wants exact values of \cos 112.5^\circ and \sin 112.5^\circ.

Solution

The angle 112.5^\circ is half of 225^\circ, whose cosine is known: \displaystyle \cos 225^\circ = -\frac{\sqrt{2}}{2}. The half angle formula gives the size of the cosine:

\sqrt{\frac{1 + \cos 225^\circ}{2}} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2}

The inside was simplified by multiplying the top and bottom of \displaystyle \frac{1 - \frac{\sqrt{2}}{2}}{2} by 2. Now the sign: 112.5^\circ ends in Quadrant II, where cosine is negative. So

\cos 112.5^\circ = -\frac{\sqrt{2 - \sqrt{2}}}{2} \approx -0.3827

Sine is positive in Quadrant II, so \displaystyle \sin 112.5^\circ = \sqrt{\frac{1 - \cos 225^\circ}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \approx 0.9239. The quadrant of 225^\circ (Quadrant III) played no part.

Check. \displaystyle \frac{2 - \sqrt{2}}{4} + \frac{2 + \sqrt{2}}{4} = 1, as the Pythagorean identity requires.

Example 4

The sign comes from the half angle

Given \displaystyle \cos x = -\frac{7}{25} with \displaystyle \pi < x < \frac{3\pi}{2}, find \displaystyle \sin\frac{x}{2}, \displaystyle \cos\frac{x}{2} and \displaystyle \tan\frac{x}{2}. Then redo the signs for \displaystyle -\pi < x < -\frac{\pi}{2}.

Solution

Halve the interval: \displaystyle \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}, so \displaystyle \frac{x}{2} ends in Quadrant II, where sine is positive and cosine negative.

Figure 2Example 4. The angle x ends in Quadrant III, but \displaystyle \frac{x}{2} ends in Quadrant II, so \displaystyle \sin\frac{x}{2} > 0 and \displaystyle \cos\frac{x}{2} < 0. The signs come from where \displaystyle \frac{x}{2} lands.

\begin{aligned} \sin\frac{x}{2} &= +\sqrt{\frac{1 + \frac{7}{25}}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \\ \cos\frac{x}{2} &= -\sqrt{\frac{1 - \frac{7}{25}}{2}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \end{aligned}

and \displaystyle \tan\frac{x}{2} = \frac{4/5}{-3/5} = -\frac{4}{3}. As a check, the tangent formula needs \sin x, which is negative in Quadrant III: \displaystyle \sin x = -\frac{24}{25}. Then \displaystyle \frac{1 - \cos x}{\sin x} = \frac{32/25}{-24/25} = -\frac{4}{3}.

For \displaystyle -\pi < x < -\frac{\pi}{2}, the angle x still ends in Quadrant III with the same sine and cosine. But now \displaystyle -\frac{\pi}{2} < \frac{x}{2} < -\frac{\pi}{4}, in Quadrant IV, so \displaystyle \sin\frac{x}{2} = -\frac{4}{5} and \displaystyle \cos\frac{x}{2} = \frac{3}{5}. The tangent is \displaystyle -\frac{4}{3} again. Knowing the quadrant of x was not enough; the interval for x was needed.

Example 5

The tangent of a half angle, with no sign to choose

Find \displaystyle \tan\frac{3\pi}{8} exactly. Then Aarush knows that \displaystyle \sin x = \frac{5}{13} and \displaystyle \cos x = -\frac{12}{13}, and nothing else; find \displaystyle \tan\frac{x}{2}.

Solution

The angle \displaystyle \frac{3\pi}{8} is half of \displaystyle \frac{3\pi}{4}, where \displaystyle \cos\frac{3\pi}{4} = -\frac{\sqrt{2}}{2} and \displaystyle \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}. The first tangent form gives

\tan\frac{3\pi}{8} = \frac{1 + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 + \sqrt{2}}{\sqrt{2}} = \sqrt{2} + 1

which is about 2.4142. Since \displaystyle \frac{3\pi}{8} and \displaystyle \frac{\pi}{8} are complementary, this should be the reciprocal of \displaystyle \tan\frac{\pi}{8} = \sqrt{2} - 1 (Cofunction identities), and indeed \displaystyle (\sqrt{2} + 1)(\sqrt{2} - 1) = 1.

For Aarush, the second form is quickest:

\tan\frac{x}{2} = \frac{\frac{5}{13}}{1 - \frac{12}{13}} = \frac{5/13}{1/13} = 5

He did not need x, its turn, or a quadrant for \displaystyle \frac{x}{2}. Indeed x ends in Quadrant II, so \displaystyle \frac{x}{2} ends in Quadrant I or III, and tangent is positive in both.

Example 6

Reading off a range and a period

Santino needs the largest and smallest values and the period of f(x) = 3\sin^2 x + 7\cos^2 x.

Solution

Two squares with different coefficients are hard to read. Reduce both:

\begin{aligned} f(x) &= \frac{3 - 3\cos 2x}{2} + \frac{7 + 7\cos 2x}{2} \\ &= \frac{10 + 4\cos 2x}{2} = 5 + 2\cos 2x \end{aligned}

This is a single cosine wave with midline 5 and amplitude 2. Its largest value is 7, at x = k\pi, and its smallest is 3, at \displaystyle x = \frac{\pi}{2} + k\pi, for k any integer. Its period is \displaystyle \frac{2\pi}{2} = \pi. Check at x = 0: 3\cdot 0 + 7\cdot 1 = 7.

Example 7

Halving twice

Find \displaystyle \cos\frac{\pi}{16} exactly.

Solution

Halve \displaystyle \frac{\pi}{4} first. The angle \displaystyle \frac{\pi}{8} is acute, so its cosine is positive:

\cos\frac{\pi}{8} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 + \sqrt{2}}}{2}

Halve again, and \displaystyle \frac{\pi}{16} is acute too:

\begin{aligned} \cos\frac{\pi}{16} &= \sqrt{\frac{1 + \frac{\sqrt{2 + \sqrt{2}}}{2}}{2}} \\ &= \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \approx 0.9808 \end{aligned}

Each halving adds one more \displaystyle \sqrt{2 + \ldots} under the root, so the cosine of any acute angle reached by halving a known angle can be written exactly with square roots.

Boundaries

The power reducing formulas for sine and cosine hold for every real number. The one for tangent fails where \cos x = 0, which is exactly where \tan x does not exist.
A half angle formula gives a value only up to sign. Without knowing where \displaystyle \frac{x}{2} ends, both signs remain possible: knowing that x is in Quadrant II puts \displaystyle \frac{x}{2} in Quadrant I or III, depending on the turn (Exercise 15).
The form \displaystyle \frac{1 - \cos x}{\sin x} cannot be used where \sin x = 0, even at x = 0, where \tan 0 = 0 exists; the form \displaystyle \frac{\sin x}{1 + \cos x} works there. Where \cos x = -1 both forms fail, and \displaystyle \tan\frac{x}{2} does not exist.
The formulas reduce a square to a first power of the doubled angle. A product of two different functions of different angles, such as \sin 3x\cos x, needs the Product to sum and sum to product formulas instead.

Where this goes next

The Product to sum and sum to product formulas do for \sin a\cos b what Example 1 did for \sin x\cos x. Example 6 is a first case of Combining a sine and a cosine into one wave, where a sine and a cosine combine into one wave. In Solving trigonometric equations and Solving trig equations by factoring, quadratics and identities, an equation in \sin^2 x and \cos 2x becomes an equation in one function once a square is reduced. In calculus, these formulas are how even powers of sine and cosine are handled in Trigonometric integrals, and the substitution \displaystyle t = \tan\frac{x}{2} of Exercise 18 turns any rational expression in \sin x and \cos x into one in t.

Exercises

Reducing powers

Write \cos^2 4x with no squares.
Write \displaystyle \sin^2\frac{x}{3} with no squares.
Write \tan^2 3x in terms of \cos 6x.
Write \cos^4 x with no power above the first.
Write \sin^4 x + \cos^4 x with no power above the first.
Find the largest and smallest values and the period of \displaystyle g(x) = 6\sin^2\frac{x}{2}.

Exact values

Find \cos 157.5^\circ.
Find \displaystyle \cos\frac{7\pi}{12} with a half angle formula.
Find \displaystyle \sin\left(-\frac{\pi}{8}\right).
Find \tan 165^\circ.
Find \displaystyle \tan\left(-\frac{5\pi}{12}\right).

From a value and an interval

\displaystyle \cos x = \frac{1}{9} with \displaystyle \frac{3\pi}{2} < x < 2\pi. Find \displaystyle \sin\frac{x}{2}, \displaystyle \cos\frac{x}{2} and \displaystyle \tan\frac{x}{2}.
\displaystyle \sin x = -\frac{3}{5} with \displaystyle \pi < x < \frac{3\pi}{2}. Find \displaystyle \sin\frac{x}{2}, \displaystyle \cos\frac{x}{2} and \displaystyle \tan\frac{x}{2}.
\displaystyle \tan x = \frac{12}{5} with \displaystyle -\pi < x < -\frac{\pi}{2}. Find \displaystyle \sin\frac{x}{2}, \displaystyle \cos\frac{x}{2} and \displaystyle \tan\frac{x}{2}.
Bhavya knows that \displaystyle \cos x = -\frac{1}{8} and that x ends in Quadrant II, but not which turn. Find every possible value of \displaystyle \cos\frac{x}{2}.

The tangent of a half angle, and halving twice

Show that \displaystyle \tan\frac{x}{2} = \csc x - \cot x wherever \sin x \ne 0.
\displaystyle \sin x = \frac{8}{17} and \displaystyle \cos x = -\frac{15}{17}. Find \displaystyle \tan\frac{x}{2} both ways.
With \displaystyle t = \tan\frac{x}{2}, show that \cos x = \dfrac{1 - t^2}{1 + t^2} wherever t exists.
Find \displaystyle \cos\frac{\pi}{24} exactly by halving \displaystyle \frac{\pi}{12}, starting from \displaystyle \cos\frac{\pi}{6}.

Reasoning

Aarush finds \sin 150^\circ with the half angle formula and x = 300^\circ. Because 300^\circ is in Quadrant IV, where sine is negative, he takes the minus sign. Find the correct value and explain which sign is right and why.
Explain why \displaystyle \tan\frac{x}{2} = \frac{1 - \cos x}{\sin x} needs no \pm although \displaystyle \sin\frac{x}{2} and \displaystyle \cos\frac{x}{2} do.
Use the power reducing formulas to explain the first figure: why the two curves have period \pi, why they are mirror images across \displaystyle y = \frac{1}{2}, and why they add to 1.
Explain why the formula for \tan^2 x cannot be used at \displaystyle x = \frac{\pi}{2}, and what its denominator is there.
Explain how \displaystyle \cos\frac{x}{2} = \pm\sqrt{\frac{1 + \cos x}{2}} comes from the power reducing formula for cosine, and where the \pm comes from.

Answers

\displaystyle \frac{1 + \cos 8x}{2}. The cosine formula with 4x for x.
\displaystyle \frac{1 - \cos\frac{2x}{3}}{2}. The sine formula with \displaystyle \frac{x}{3} for x, whose double is \displaystyle \frac{2x}{3}.
\displaystyle \frac{1 - \cos 6x}{1 + \cos 6x}, wherever \cos 3x \ne 0. The tangent formula with 3x for x.
\displaystyle \frac{3 + 4\cos 2x + \cos 4x}{8}. \displaystyle \cos^4 x = \frac{1 + 2\cos 2x + \cos^2 2x}{4}, and \displaystyle \cos^2 2x = \frac{1 + \cos 4x}{2}.
\displaystyle \frac{3 + \cos 4x}{4}. (\sin^2 x + \cos^2 x)^2 = 1, so the sum is \displaystyle 1 - 2\sin^2 x\cos^2 x = 1 - \frac{1 - \cos 4x}{4}, using Example 1.
Largest 6, smallest 0, period 2\pi. g(x) = 3 - 3\cos x, a wave with midline 3 and amplitude 3.
\displaystyle -\frac{\sqrt{2 + \sqrt{2}}}{2} \approx -0.9239. Half of 315^\circ, where \displaystyle \cos 315^\circ = \frac{\sqrt{2}}{2}; 157.5^\circ is in Quadrant II, so cosine is negative.
\displaystyle -\frac{\sqrt{2 - \sqrt{3}}}{2} \approx -0.2588. Half of \displaystyle \frac{7\pi}{6}, where cosine is \displaystyle -\frac{\sqrt{3}}{2}, so \displaystyle \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}; \displaystyle \frac{7\pi}{12} is in Quadrant II, where cosine is negative.
\displaystyle -\frac{\sqrt{2 - \sqrt{2}}}{2} \approx -0.3827. Half of \displaystyle -\frac{\pi}{4}, where cosine is \displaystyle \frac{\sqrt{2}}{2}; \displaystyle -\frac{\pi}{8} is in Quadrant IV, where sine is negative.
\displaystyle \sqrt{3} - 2 \approx -0.2679. Half of 330^\circ: \displaystyle \frac{1 - \frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -(2 - \sqrt{3}).
\displaystyle -2 - \sqrt{3} \approx -3.7321. Half of \displaystyle -\frac{5\pi}{6}, where cosine is \displaystyle -\frac{\sqrt{3}}{2} and sine is \displaystyle -\frac{1}{2}: \displaystyle \frac{1 + \frac{\sqrt{3}}{2}}{-\frac{1}{2}} = -(2 + \sqrt{3}). The negative sine carries the sign.
\displaystyle \sin\frac{x}{2} = \frac{2}{3}, \displaystyle \cos\frac{x}{2} = -\frac{\sqrt{5}}{3}, \displaystyle \tan\frac{x}{2} = -\frac{2\sqrt{5}}{5}. Here \displaystyle \frac{3\pi}{4} < \frac{x}{2} < \pi, Quadrant II; \displaystyle \frac{1 - 1/9}{2} = \frac{4}{9} and \displaystyle \frac{1 + 1/9}{2} = \frac{5}{9}.
\displaystyle \sin\frac{x}{2} = \frac{3\sqrt{10}}{10}, \displaystyle \cos\frac{x}{2} = -\frac{\sqrt{10}}{10}, \displaystyle \tan\frac{x}{2} = -3. In Quadrant III, \displaystyle \cos x = -\frac{4}{5}; \displaystyle \frac{x}{2} is in Quadrant II; \displaystyle \frac{1 + 4/5}{2} = \frac{9}{10} and \displaystyle \frac{1 - 4/5}{2} = \frac{1}{10}. Check: \displaystyle \frac{1 - \cos x}{\sin x} = \frac{9/5}{-3/5} = -3.
\displaystyle \sin\frac{x}{2} = -\frac{3\sqrt{13}}{13}, \displaystyle \cos\frac{x}{2} = \frac{2\sqrt{13}}{13}, \displaystyle \tan\frac{x}{2} = -\frac{3}{2}. In Quadrant III, \displaystyle \cos x = -\frac{5}{13} and \displaystyle \sin x = -\frac{12}{13}; \displaystyle -\frac{\pi}{2} < \frac{x}{2} < -\frac{\pi}{4} is Quadrant IV; \displaystyle \frac{1 + 5/13}{2} = \frac{9}{13} and \displaystyle \frac{1 - 5/13}{2} = \frac{4}{13}.
\displaystyle \cos\frac{x}{2} = \pm\frac{\sqrt{7}}{4}, both possible. \displaystyle \frac{1 - 1/8}{2} = \frac{7}{16}. If \displaystyle \frac{\pi}{2} + 2k\pi < x < \pi + 2k\pi, then \displaystyle \frac{\pi}{4} + k\pi < \frac{x}{2} < \frac{\pi}{2} + k\pi: Quadrant I when the integer k is even, Quadrant III when it is odd.
\displaystyle \frac{1 - \cos x}{\sin x} = \frac{1}{\sin x} - \frac{\cos x}{\sin x}, which is \csc x - \cot x.
\displaystyle \tan\frac{x}{2} = 4. \displaystyle \frac{1 + 15/17}{8/17} = \frac{32}{8} and \displaystyle \frac{8/17}{1 - 15/17} = \frac{8}{2}.
\displaystyle t^2 = \frac{1 - \cos x}{1 + \cos x} (the tangent formula with \displaystyle \frac{x}{2} for x), so t^2(1 + \cos x) = 1 - \cos x, \cos x\,(1 + t^2) = 1 - t^2, and 1 + t^2 > 0.
\displaystyle \cos\frac{\pi}{24} = \frac{\sqrt{2 + \sqrt{2 + \sqrt{3}}}}{2} \approx 0.9914. First \displaystyle \cos\frac{\pi}{12} = \frac{\sqrt{2 + \sqrt{3}}}{2}, then halve again; both angles are acute.
\displaystyle \sin 150^\circ = \frac{1}{2}. \displaystyle \sqrt{\frac{1 - \cos 300^\circ}{2}} = \sqrt{\frac{1}{4}}. The sign belongs to the angle being found, 150^\circ, which is in Quadrant II, where sine is positive. The quadrant of 300^\circ says nothing about it.
The tangent forms come from multiplying \displaystyle \frac{\sin u}{\cos u} by \displaystyle \frac{2\sin u}{2\sin u}; no square root is taken, so no sign is lost. The sign is carried by \sin x, since 1 - \cos x \ge 0.
\displaystyle \sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x and \displaystyle \cos^2 x = \frac{1}{2} + \frac{1}{2}\cos 2x. Each is \displaystyle \frac{1}{2} plus or minus the same wave \displaystyle \frac{1}{2}\cos 2x, which has period \pi; so the curves are reflections of each other across \displaystyle y = \frac{1}{2} and their sum is 1.
At \displaystyle x = \frac{\pi}{2}, \cos 2x = \cos\pi = -1, so the denominator 1 + \cos 2x is 0. This matches \displaystyle \tan\frac{\pi}{2} not existing.
Put \displaystyle \frac{x}{2} in place of x in \displaystyle \cos^2 x = \frac{1 + \cos 2x}{2}, giving \displaystyle \cos^2\frac{x}{2} = \frac{1 + \cos x}{2}. Taking a square root leaves two candidates, because a number and its opposite have the same square; the formula alone cannot tell which, so the quadrant of \displaystyle \frac{x}{2} must.