Precalculus · F.TF.C.9

Sine and tangent of a sum or difference

Cosine already has a rule for the sum of two angles. What are the matching rules for sine and tangent, and what do they say about the angle at which two lines cross?

You need

Cosine of a sum or differencePrecalculus Exact values on the unit circlePrecalculus Cofunction identitiesPrecalculus

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Combining a sine and a cosine into one wavePrecalculus Double angle formulasPrecalculus

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Sine and tangent of a sum or difference

Combining a sine and a cosine into one wave Double angle formulas ▾ Show all 13 (1 more)

Why this exists

The Cosine of a sum or difference turns the cosine of a + b into the sines and cosines of a and b. That gives exact cosines at angles such as 15^\circ and 75^\circ, but a sine is just as often the thing wanted: the height of a point on a wheel, the y-coordinate of a terminal point, the sine in the law of sines. A rule for \sin(a + b) is the missing half.

The rule does not need a new figure or a new proof from scratch. The Cofunction identities say a sine is the cosine of the complementary angle, and the complement of a + b is a difference, which the cosine formula already handles. Dividing the sine rule by the cosine rule then gives a rule for tangent. Tangent is the slope of a line through the origin, so that last rule answers a question from coordinate geometry: at what angle do two lines with known slopes cross?

Definition

For all real numbers a and b (or angles in degrees), the sum and difference formulas for sine are the identities

\begin{aligned} \sin(a + b) &= \sin a\cos b + \cos a\sin b \\ \sin(a - b) &= \sin a\cos b - \cos a\sin b \end{aligned}

Read in words: “sine cosine plus cosine sine”. The sign between the two products matches the sign between the angles, unlike cosine, where it flips.

The sum and difference formulas for tangent are

\begin{aligned} \tan(a + b) &= \frac{\tan a + \tan b}{1 - \tan a\tan b} \\[4pt] \tan(a - b) &= \frac{\tan a - \tan b}{1 + \tan a\tan b} \end{aligned}

The sine formulas hold for every pair a, b. The tangent sum formula holds whenever \tan a, \tan b and \tan(a + b) all exist, that is, when none of a, b, a + b is an odd multiple of \displaystyle \frac{\pi}{2}; the difference formula needs the same of a, b and a - b. They are identities in the sense of What an identity is: true at every value where both sides are defined.

Intuition

Figure 1The height of P is \sin(a + b), and the right edge shows it as \sin a\cos b + \cos a\sin b. The width of the rectangle, read along the top and along the bottom, gives the cosine sum formula as well.

Stand a segment of length 1 at angle a + b above the ground. Its height is \sin(a + b). Reach that height in two stages: first go out along a ray at angle a for a distance \cos b, to the point Q, then turn a right angle and go \sin b more, to P. These are the two legs of a right triangle with hypotenuse 1 and angle b at O.

The first stage climbs \cos b times \sin a, since it is a length \cos b tilted at angle a. The second stage is tilted the other way, at angle a from vertical (the marked angle at Q), so it climbs \sin b times \cos a. The two climbs add, and that is the sum formula. The same picture read sideways gives the cosine formula: the width \cos a\cos b is \cos(a + b) plus the overhang \sin a\sin b. The picture only shows acute angles with an acute sum; the proof below covers every real a and b.

Results

Sine of a sum and a difference. For all real a and b,

\begin{aligned} \sin(a + b) &= \sin a\cos b + \cos a\sin b \\ \sin(a - b) &= \sin a\cos b - \cos a\sin b \end{aligned}

Tangent of a sum and a difference. When \tan a, \tan b and \tan(a \pm b) exist,

\tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a\tan b}

and in that case the denominator is never 0.

The angle between two lines. Two lines with slopes m_1 \ne m_2 (so neither line is vertical, and they are not parallel), where 1 + m_1m_2 \ne 0, cross at an acute angle \varphi with

\tan\varphi = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right|

Two lines that both have slopes are perpendicular exactly when 1 + m_1m_2 = 0, that is, m_1m_2 = -1.

Why it is true

Sine of a sum. Start from the cofunction identity \displaystyle \sin t = \cos\left(\frac{\pi}{2} - t\right) with t = a + b, and regroup the angle:

\begin{aligned} \sin(a + b) &= \cos\left(\tfrac{\pi}{2} - a - b\right) \\ &= \cos\left(\left(\tfrac{\pi}{2} - a\right) - b\right) \end{aligned}

Now the cosine difference formula applies with first angle \displaystyle \frac{\pi}{2} - a and second angle b:

\begin{aligned} &\cos\left(\tfrac{\pi}{2} - a\right)\cos b + \sin\left(\tfrac{\pi}{2} - a\right)\sin b \\ &= \sin a\cos b + \cos a\sin b \end{aligned}

The last step is the cofunction identities again, once for each factor. Every step holds for all real a and b, so the formula does too.

Sine of a difference. Write a - b = a + (-b) and use the sum formula with the even and odd identities \cos(-b) = \cos b and \sin(-b) = -\sin b from Unit circle symmetry and even-odd identities:

\begin{aligned} \sin(a - b) &= \sin a\cos(-b) + \cos a\sin(-b) \\ &= \sin a\cos b - \cos a\sin b \end{aligned}

Tangent of a sum. Tangent is sine over cosine (Reciprocal and quotient identities), so where \cos(a + b) \ne 0,

\tan(a + b) = \frac{\sin a\cos b + \cos a\sin b}{\cos a\cos b - \sin a\sin b}

When \tan a and \tan b exist, \cos a \ne 0 and \cos b \ne 0, so the numerator and the denominator may both be divided by \cos a\cos b without changing the fraction. Term by term, \displaystyle \frac{\sin a\cos b}{\cos a\cos b} = \tan a, \displaystyle \frac{\cos a\sin b}{\cos a\cos b} = \tan b, \displaystyle \frac{\cos a\cos b}{\cos a\cos b} = 1 and \displaystyle \frac{\sin a\sin b}{\cos a\cos b} = \tan a\tan b, which gives the formula. The same division shows

1 - \tan a\tan b = \frac{\cos(a + b)}{\cos a\cos b}

so the denominator is 0 exactly when \cos(a + b) = 0, the case already excluded. The difference formula follows by putting -b for b, since \tan(-b) = -\tan b.

The angle between two lines. A line with slope m rises m for each 1 across, so it makes an angle \theta with the positive x-direction, measured counterclockwise and taken in [0^\circ, 180^\circ), with \tan\theta = m.

Figure 2Each line makes an angle with the x-axis whose tangent is its slope. The angle where the lines cross is \varphi = \theta_2 - \theta_1, the exterior angle of the triangle at the steeper line’s foot minus the interior angle at the other foot.

In the figure the triangle’s exterior angle at the steeper line’s foot is \theta_2, and it equals the sum of the two interior angles not next to it: \theta_1, and the angle at the crossing, which is vertical to the marked \varphi and so equal to it. So \varphi = \theta_2 - \theta_1, and the difference formula gives

\tan(\theta_2 - \theta_1) = \frac{m_2 - m_1}{1 + m_1m_2}

Two crossing lines make two angles, \varphi and 180^\circ - \varphi, whose tangents are opposites, so the absolute value picks the acute one. For the right-angle case, the cosine difference formula with \cos\theta_1\cos\theta_2 factored out gives \cos(\theta_2 - \theta_1) = \cos\theta_1\cos\theta_2\,(1 + m_1m_2). Neither line is vertical, so \cos\theta_1 \ne 0 and \cos\theta_2 \ne 0, and the product is 0 exactly when 1 + m_1m_2 = 0. So the lines meet at a right angle if and only if m_1m_2 = -1. This is the rule from Slopes of perpendicular lines, now proved in both directions. A vertical line is left out because it has no slope; a vertical line and a horizontal line are perpendicular, but no product of slopes can say so.

Worked examples

Examples 1 to 4 belong to Day 1, the sine of a sum or difference. Examples 5 to 7 belong to Day 2, the tangent of a sum or difference and the angle between two lines.

Example 1

An exact sine in degrees

Find the exact value of \sin 255^\circ.

Solution

Write 255^\circ as a sum of angles with known values: 255^\circ = 210^\circ + 45^\circ. On the unit circle, \displaystyle \sin 210^\circ = -\frac{1}{2} and \displaystyle \cos 210^\circ = -\frac{\sqrt{3}}{2}. The sum formula gives

\begin{aligned} \sin 255^\circ &= \sin 210^\circ\cos 45^\circ + \cos 210^\circ\sin 45^\circ \\ &= \left(-\tfrac{1}{2}\right)\tfrac{\sqrt{2}}{2} + \left(-\tfrac{\sqrt{3}}{2}\right)\tfrac{\sqrt{2}}{2} \\ &= -\frac{\sqrt{2} + \sqrt{6}}{4} \end{aligned}

Check. This is about -0.9659. The angle 255^\circ is in Quadrant III, where sine is negative, and it sits 75^\circ past 180^\circ, so its sine should be close to -1.

Example 2

An exact sine in radians, from a difference

Find the exact value of \displaystyle \sin\frac{11\pi}{12}.

Solution

Over the denominator 12, the angles with known values are multiples of \displaystyle \frac{\pi}{6} = \frac{2\pi}{12} and \displaystyle \frac{\pi}{4} = \frac{3\pi}{12}. One difference that works is \displaystyle \frac{11\pi}{12} = \frac{15\pi}{12} - \frac{4\pi}{12} = \frac{5\pi}{4} - \frac{\pi}{3}. With \displaystyle \sin\frac{5\pi}{4} = \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2},

\begin{aligned} \sin\tfrac{11\pi}{12} &= \sin\tfrac{5\pi}{4}\cos\tfrac{\pi}{3} - \cos\tfrac{5\pi}{4}\sin\tfrac{\pi}{3} \\ &= \left(-\tfrac{\sqrt{2}}{2}\right)\tfrac{1}{2} - \left(-\tfrac{\sqrt{2}}{2}\right)\tfrac{\sqrt{3}}{2} \\ &= \frac{\sqrt{6} - \sqrt{2}}{4} \end{aligned}

Check. The value is about 0.2588. The angle \displaystyle \frac{11\pi}{12} is \displaystyle \frac{\pi}{12} short of \pi, in Quadrant II, so its sine is small and positive. The split \displaystyle \frac{3\pi}{4} + \frac{\pi}{6} gives the same number, as an identity must.

Example 3

Shiori’s two angles

Shiori knows that \displaystyle \cos a = \frac{28}{53} with a in Quadrant IV, and \displaystyle \sin b = \frac{11}{61} with b in Quadrant II. Find \sin(a + b) and \sin(a - b), and the quadrant of a + b.

Solution

Each formula needs all four values. The Pythagorean identity (The Pythagorean identities) gives each missing one up to sign, and the quadrant settles the sign. In Quadrant IV sine is negative, and in Quadrant II cosine is negative:

\begin{aligned} \sin a &= -\sqrt{1 - \tfrac{784}{2809}} = -\frac{45}{53} \\ \cos b &= -\sqrt{1 - \tfrac{121}{3721}} = -\frac{60}{61} \end{aligned}

The common denominator is 53 \cdot 61 = 3233. The two products are \displaystyle \sin a\cos b = \frac{2700}{3233} and \displaystyle \cos a\sin b = \frac{308}{3233}, so

\begin{aligned} \sin(a + b) &= \tfrac{2700}{3233} + \tfrac{308}{3233} = \frac{3008}{3233} \\ \sin(a - b) &= \tfrac{2700}{3233} - \tfrac{308}{3233} = \frac{2392}{3233} \end{aligned}

The sine alone cannot place a + b, since sine is positive in Quadrants I and II. The cosine sum formula settles it:

\cos(a + b) = \tfrac{28}{53}\left(-\tfrac{60}{61}\right) - \left(-\tfrac{45}{53}\right)\tfrac{11}{61} = -\frac{1185}{3233}

Sine positive and cosine negative put the terminal point of a + b in Quadrant II. Check: 3008^2 + 1185^2 = 10{,}452{,}289 = 3233^2, as the Pythagorean identity requires.

Example 4

Takeru reads the formula backward

Takeru simplifies three expressions: (a) \sin 70^\circ\cos 25^\circ - \cos 70^\circ\sin 25^\circ; (b) \sin 5x\cos 2x + \cos 5x\sin 2x; (c) \displaystyle \sin\left(x - \frac{3\pi}{2}\right).

Solution

(a) “Sine cosine minus cosine sine”, with the sine of the first angle in front, is the right side of the difference formula with a = 70^\circ and b = 25^\circ. So the expression equals \displaystyle \sin(70^\circ - 25^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}.

(b) The same pattern with a plus sign is the sum formula with a = 5x, b = 2x, so the expression is \sin(5x + 2x) = \sin 7x for every real x.

(c) Here the formula runs forward. The sine and cosine of \displaystyle \frac{3\pi}{2} are -1 and 0 (Exact values on the unit circle), so

\begin{aligned} \sin\left(x - \tfrac{3\pi}{2}\right) &= \sin x\cos\tfrac{3\pi}{2} - \cos x\sin\tfrac{3\pi}{2} \\ &= \sin x\cdot 0 - \cos x\cdot(-1) = \cos x \end{aligned}

A shift by a multiple of \displaystyle \frac{\pi}{2} always collapses this way, because one of the two products has a factor 0.

Example 5

An exact tangent

Find the exact value of \tan 195^\circ.

Solution

Write 195^\circ = 225^\circ - 30^\circ. Here \tan 225^\circ = 1 and \displaystyle \tan 30^\circ = \frac{\sqrt{3}}{3}, and none of 225^\circ, 30^\circ, 195^\circ is an odd multiple of 90^\circ, so the difference formula applies:

\tan 195^\circ = \frac{1 - \frac{\sqrt{3}}{3}}{1 + \frac{\sqrt{3}}{3}} = \frac{3 - \sqrt{3}}{3 + \sqrt{3}}

The second form multiplies top and bottom by 3. To clear the root from the denominator (Rationalizing a denominator), multiply top and bottom by \displaystyle 3 - \sqrt{3}; the denominator becomes 9 - 3 = 6:

\frac{(3 - \sqrt{3})^2}{6} = \frac{12 - 6\sqrt{3}}{6} = 2 - \sqrt{3}

Check. \displaystyle 2 - \sqrt{3} \approx 0.2679, positive, as a tangent in Quadrant III must be. It also equals \tan 15^\circ, since tangent repeats every 180^\circ.

Example 6

Yuzuki adds two angles she cannot name

Angles a and b are acute, with \tan a = 2 and \tan b = 3. Find a + b exactly. Then do the same for a and an acute angle c with \displaystyle \tan c = \frac{1}{2}.

Solution

Neither a nor b is a special angle, but their sum may be. The sum formula gives

\tan(a + b) = \frac{2 + 3}{1 - 2\cdot 3} = \frac{5}{-5} = -1

A tangent of -1 happens at 135^\circ, 315^\circ, and every 180^\circ from them, so the size of a + b is needed. Both tangents exceed 1 = \tan 45^\circ, and tangent increases on (0^\circ, 90^\circ), so a and b are each between 45^\circ and 90^\circ. Then a + b is between 90^\circ and 180^\circ, and there the only angle with tangent -1 is 135^\circ. So \displaystyle a + b = 135^\circ = \frac{3\pi}{4}.

For a and c the denominator is \displaystyle 1 - 2\cdot\frac{1}{2} = 0, and the formula breaks. That is the formula’s way of saying \tan(a + c) does not exist: a + c is an odd multiple of 90^\circ. Since both are acute, a + c = 90^\circ. The cofunction identities agree: \displaystyle \tan c = \frac{1}{2} = \cot a, and the cotangent of a is the tangent of its complement.

Example 7

The angle between two lines

Find the acute angle between the lines \displaystyle y = \frac{1}{3}x - 4 and y = -2x + 5, to two decimal places.

Solution

Take \displaystyle m_1 = \frac{1}{3} and m_2 = -2. Then \displaystyle 1 + m_1m_2 = 1 - \frac{2}{3} = \frac{1}{3}, not 0, so the lines are not perpendicular and the formula applies:

\frac{m_2 - m_1}{1 + m_1m_2} = \frac{-2 - \frac{1}{3}}{\frac{1}{3}} = \frac{-\frac{7}{3}}{\frac{1}{3}} = -7

The acute angle has \tan\varphi = |-7| = 7, so \varphi = \tan^{-1} 7 \approx 81.87^\circ, using the inverse tangent, which returns the angle in (-90^\circ, 90^\circ) with the given tangent. The other angle at the crossing is 180^\circ - 81.87^\circ = 98.13^\circ.

Check with the angles of the lines themselves: \displaystyle \tan^{-1}\frac{1}{3} \approx 18.43^\circ, and the line of slope -2 makes 180^\circ - \tan^{-1} 2 \approx 116.57^\circ with the positive x-direction. The difference is 98.13^\circ, one of the two angles found.

Boundaries

The sine formulas hold for every pair of real numbers, in radians or degrees, as long as a, b and a \pm b are in the same unit.
Sine does not distribute: \sin(a + b) = \sin a + \sin b is false in general, though it holds at some pairs, such as any pair with a = 0 (Exercise 23).
The tangent formulas need \tan a and \tan b to exist. For \displaystyle \tan\left(\frac{3\pi}{2} - x\right) the formula cannot even be written, since \displaystyle \tan\frac{3\pi}{2} does not exist; go back to sine over cosine (Exercise 18). They also need \tan(a \pm b) to exist, which is exactly when the denominator is not 0.
An exact answer needs angles whose exact values are known. \sin 52^\circ = \sin(30^\circ + 22^\circ) is true but gives nothing exact.
The angle formula for lines needs both slopes. A vertical line has none; use its angle, 90^\circ, directly (Exercise 25). When 1 + m_1m_2 = 0 the formula has no value and the lines are perpendicular. When m_1 = m_2 it gives 0: the lines are parallel or the same line, and do not cross.

Where this goes next

Setting b = a in the sine sum formula gives \sin 2a = 2\sin a\cos a, the first of the double angle formulas, and the half angle formulas grow from those (Power reducing and half angle formulas). Adding the sum and difference formulas, as in Exercise 12, gives the Product to sum and sum to product formulas. Expanding A\sin(t + c) writes a shifted sine wave as a mix of \sin t and \cos t; run backward, that is Combining a sine and a cosine into one wave. Later, the sine and cosine sum formulas together are why angles add when complex numbers in polar form are multiplied (Multiplying, powers and De Moivre's theorem).

Exercises

Day 1: exact values

Find the exact value of \sin 15^\circ.
Find the exact value of \sin 285^\circ.
Find the exact value of \displaystyle \sin\frac{7\pi}{12}.
Find the exact value of \displaystyle \sin\left(-\frac{\pi}{12}\right), using a difference.
Find the exact values of \displaystyle \sin\frac{17\pi}{12} and \displaystyle \cos\frac{17\pi}{12}, and check that the squares of the two add to 1.

Day 1: from a value and a quadrant

\displaystyle \sin a = \frac{5}{13} with a in Quadrant II, and \displaystyle \cos b = \frac{3}{5} with b in Quadrant IV. Find \sin(a + b) and \sin(a - b).
Takeru knows \displaystyle \cos a = -\frac{2}{3} with a in Quadrant III, and \displaystyle \sin b = \frac{1}{3} with b acute. Find \sin(a + b) exactly.

Day 1: reading backward and identities

Find the exact value of \sin 50^\circ\cos 10^\circ + \cos 50^\circ\sin 10^\circ.
Find the exact value of \displaystyle \sin\frac{5\pi}{8}\cos\frac{\pi}{8} - \cos\frac{5\pi}{8}\sin\frac{\pi}{8}.
Simplify \sin(\pi + x).
Expand and simplify \displaystyle \sin\left(x + \frac{\pi}{4}\right).
Simplify \sin(a + b) + \sin(a - b).
Show that \sin(a + b)\sin(a - b) = \sin^2 a - \sin^2 b.

Day 2: tangent of a sum or difference

Find the exact value of \tan 75^\circ.
Find the exact value of \displaystyle \tan\frac{11\pi}{12}.
Angles a and b are acute, with \displaystyle \tan a = \frac{1}{4} and \displaystyle \tan b = \frac{3}{5}. Find a + b exactly.
\displaystyle \cos a = -\frac{7}{25} with a in Quadrant II, and \displaystyle \tan b = \frac{3}{4}. Find \tan(a - b).
Simplify \displaystyle \tan\left(\frac{3\pi}{2} - x\right), and say why the tangent difference formula cannot do it.
Write \dfrac{\tan 3x - \tan x}{1 + \tan 3x\tan x} as a single tangent.

Day 2: angles between lines

Find the acute angle between y = 4x + 3 and y = -x - 2, to two decimal places.
Yuzuki wants every line through the origin that crosses y = 2x at an angle of 45^\circ. Find their equations.

Reasoning

Show that \cot(a + b) = \dfrac{\cot a\cot b - 1}{\cot a + \cot b} wherever all the cotangents exist.
Shiori claims \sin(a + b) = \sin a + \sin b. Give a pair a, b where this is true and a pair where it is false, and explain why that does not make it an identity.
Suppose \tan a and \tan b exist. Explain why 1 - \tan a\tan b = 0 exactly when a + b is an odd multiple of \displaystyle \frac{\pi}{2}, and what this says about two lines whose slopes multiply to -1.
The vertical line x = 1 has no slope. Find the acute angle between it and y = 2x, to two decimal places, and explain how \tan^{-1} of a reciprocal slope gives the same answer.

Answers

\displaystyle \frac{\sqrt{6} - \sqrt{2}}{4}. 15^\circ = 45^\circ - 30^\circ: \displaystyle \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2}.
\displaystyle -\frac{\sqrt{6} + \sqrt{2}}{4}. 285^\circ = 240^\circ + 45^\circ: \displaystyle \left(-\frac{\sqrt{3}}{2}\right)\frac{\sqrt{2}}{2} + \left(-\frac{1}{2}\right)\frac{\sqrt{2}}{2}.
\displaystyle \frac{\sqrt{6} + \sqrt{2}}{4}. \displaystyle \frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}: \displaystyle \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2}.
\displaystyle \frac{\sqrt{2} - \sqrt{6}}{4}. \displaystyle -\frac{\pi}{12} = \frac{\pi}{6} - \frac{\pi}{4}: \displaystyle \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}.
\displaystyle \sin\frac{17\pi}{12} = -\frac{\sqrt{6} + \sqrt{2}}{4} and \displaystyle \cos\frac{17\pi}{12} = \frac{\sqrt{2} - \sqrt{6}}{4}. With \displaystyle \frac{17\pi}{12} = \frac{5\pi}{4} + \frac{\pi}{6}, both formulas use \displaystyle \sin\frac{5\pi}{4} = \cos\frac{5\pi}{4} = -\frac{\sqrt{2}}{2}. The squares are \displaystyle \frac{8 + 4\sqrt{3}}{16} and \displaystyle \frac{8 - 4\sqrt{3}}{16}, which add to 1.
\displaystyle \sin(a + b) = \frac{63}{65} and \displaystyle \sin(a - b) = -\frac{33}{65}. Here \displaystyle \cos a = -\frac{12}{13} and \displaystyle \sin b = -\frac{4}{5}, so the products are \displaystyle \frac{15}{65} and \displaystyle \frac{48}{65}.
\displaystyle -\frac{2\sqrt{10} + 2}{9}. \displaystyle \sin a = -\frac{\sqrt{5}}{3} (negative in Quadrant III) and \displaystyle \cos b = \frac{2\sqrt{2}}{3}; then \displaystyle \left(-\frac{\sqrt{5}}{3}\right)\frac{2\sqrt{2}}{3} + \left(-\frac{2}{3}\right)\frac{1}{3}.
\displaystyle \frac{\sqrt{3}}{2}. It is \sin(50^\circ + 10^\circ) = \sin 60^\circ.
1. It is \displaystyle \sin\left(\frac{5\pi}{8} - \frac{\pi}{8}\right) = \sin\frac{\pi}{2}.
-\sin x. \sin\pi\cos x + \cos\pi\sin x = 0 - \sin x.
\displaystyle \frac{\sqrt{2}}{2}(\sin x + \cos x). \displaystyle \sin x\cos\frac{\pi}{4} + \cos x\sin\frac{\pi}{4}, and both values are \displaystyle \frac{\sqrt{2}}{2}.
2\sin a\cos b. The \cos a\sin b terms cancel.
The product is (\sin a\cos b)^2 - (\cos a\sin b)^2, a difference of squares. Replace \cos^2 b by 1 - \sin^2 b and \cos^2 a by 1 - \sin^2 a: \sin^2 a - \sin^2 a\sin^2 b - \sin^2 b + \sin^2 a\sin^2 b, which is \sin^2 a - \sin^2 b.
\displaystyle 2 + \sqrt{3}. \displaystyle \frac{1 + \frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}}; multiply top and bottom by \displaystyle 3 + \sqrt{3} to get \displaystyle \frac{12 + 6\sqrt{3}}{6}.
\displaystyle \sqrt{3} - 2. \displaystyle \frac{11\pi}{12} = \frac{3\pi}{4} + \frac{\pi}{6}: \displaystyle \frac{-1 + \frac{\sqrt{3}}{3}}{1 + \frac{\sqrt{3}}{3}} = \frac{\sqrt{3} - 3}{3 + \sqrt{3}} = \frac{-12 + 6\sqrt{3}}{6}.
\displaystyle a + b = \frac{\pi}{4}, or 45^\circ. \displaystyle \tan(a + b) = \frac{\frac{17}{20}}{1 - \frac{3}{20}} = 1, and a + b is between 0^\circ and 180^\circ, where only 45^\circ has tangent 1.
\displaystyle \frac{117}{44}. \displaystyle \sin a = \frac{24}{25}, so \displaystyle \tan a = -\frac{24}{7}; then \displaystyle \frac{-\frac{24}{7} - \frac{3}{4}}{1 + \left(-\frac{24}{7}\right)\frac{3}{4}} = \frac{-\frac{117}{28}}{-\frac{11}{7}}. The quadrant of b is not needed: the formula uses only \tan b.
\cot x, for x not a multiple of \pi. The formula needs \displaystyle \tan\frac{3\pi}{2}, which does not exist. Instead, the sine and cosine difference formulas give \displaystyle \sin\left(\frac{3\pi}{2} - x\right) = -\cos x and \displaystyle \cos\left(\frac{3\pi}{2} - x\right) = -\sin x, so the quotient is \displaystyle \frac{-\cos x}{-\sin x} = \cot x.
\tan 2x, the difference formula with a = 3x, b = x, wherever \tan 3x, \tan x and \tan 2x exist.
About 59.04^\circ. \displaystyle \frac{-1 - 4}{1 + 4(-1)} = \frac{-5}{-3} = \frac{5}{3}, and \displaystyle \tan^{-1}\frac{5}{3} \approx 59.04^\circ.
y = -3x and \displaystyle y = \frac{1}{3}x. A slope m needs \displaystyle \left|\frac{m - 2}{1 + 2m}\right| = 1. Then m - 2 = 1 + 2m, so m = -3, or m - 2 = -(1 + 2m), so \displaystyle m = \frac{1}{3}.
\displaystyle \cot(a + b) = \frac{\cos a\cos b - \sin a\sin b}{\sin a\cos b + \cos a\sin b}. Divide top and bottom by \sin a\sin b, which is not 0 where \cot a and \cot b exist: the top becomes \cot a\cot b - 1 and the bottom \cot b + \cot a.
True at a = 0, \displaystyle b = \frac{\pi}{3}: both sides are \displaystyle \frac{\sqrt{3}}{2}. False at \displaystyle a = b = \frac{\pi}{2}: \sin\pi = 0 but \displaystyle \sin\frac{\pi}{2} + \sin\frac{\pi}{2} = 2. An identity must hold at every value where both sides are defined, so one failure is enough to rule it out.
Dividing the cosine sum formula by \cos a\cos b (not 0, since the tangents exist) gives \displaystyle 1 - \tan a\tan b = \frac{\cos(a + b)}{\cos a\cos b}, which is 0 exactly when \cos(a + b) = 0, that is, when a + b is an odd multiple of \displaystyle \frac{\pi}{2}. With b replaced by -b: 1 + \tan a\tan b = 0 exactly when a - b is an odd multiple of \displaystyle \frac{\pi}{2}. For lines at angles \theta_1, \theta_2 with m_1m_2 = -1, the angles differ by 90^\circ, so the lines are perpendicular.
About 26.57^\circ. The line y = 2x makes \tan^{-1} 2 \approx 63.43^\circ with the x-axis, and the vertical line makes 90^\circ, so they cross at 90^\circ - 63.43^\circ = 26.57^\circ. By the cofunction identities, the tangent of the complement of an angle is the cotangent of the angle, \displaystyle \frac{1}{2}, so the angle is \displaystyle \tan^{-1}\frac{1}{2}, the same number.