Trig identities

Worked solutions

Problem 1

1. Take any $x$ with $\cos x \neq 0$.$\tan x$ and $\sec x$ are defined only where $\cos x \neq 0$, so the identity is a claim about those angles alone.
2. $\dfrac{\sin^2x}{\cos^2x} + \dfrac{\cos^2x}{\cos^2x} = \dfrac{1}{\cos^2x}$.Both terms of the Pythagorean identity have a factor $\cos^2x$ to divide by, which is nonzero where $\tan x$ is defined, and dividing a true equation by a nonzero number gives a true equation. This starts from a known fact, not from the identity to be proved.
3. $\tan^2x + 1 = \sec^2x$$\sin^2x/\cos^2x = (\sin x/\cos x)^2 = \tan^2x$, $\cos^2x/\cos^2x = 1$ and $1/\cos^2x = \sec^2x$; valid wherever $\cos x \neq 0$.

Problem 2

1. $\sin 2x = \sin(x + x) = \sin x\cos x + \cos x\sin x$.The sine sum formula holds for every pair of angles, so it holds with $y = x$, and $x + x = 2x$.
2. $\sin 2x = 2\sin x\cos x$.The two terms are the same product written in a different order.
3. $\cos 2x = \cos(x+x) = \cos x\cos x - \sin x\sin x = \cos^2x - \sin^2x$.The cosine sum formula with $y = x$.
4. $\cos^2x - \sin^2x = \cos^2x - (1 - \cos^2x) = 2\cos^2x - 1$.The Pythagorean identity gives $\sin^2x = 1 - \cos^2x$; substituting it leaves a form with cosines only.
5. $\cos^2x - \sin^2x = (1 - \sin^2x) - \sin^2x = 1 - 2\sin^2x$.Likewise $\cos^2x = 1 - \sin^2x$ leaves a form with sines only.
6. $\sin 2x = 2\sin x\cos x$; $\cos 2x = \cos^2x - \sin^2x = 2\cos^2x - 1 = 1 - 2\sin^2x$Steps 2 to 5. Which form of $\cos 2x$ to use depends on which function the target keeps: Problems 3, 5 and 9 pick the form that suits their target.

Problem 3

1. $\cos 2x = 2\cos^2x - 1$.Problem 2's second form is the one whose only square is $\cos^2x$, so solving it for $\cos^2x$ gives the first identity. It is already proved, so rearranging it is a deduction from a true equation, not an assumption.
2. $2\cos^2x = 1 + \cos 2x$, so $\cos^2x = \tfrac{1+\cos 2x}{2}$.Add 1 to both sides, then divide both by 2; each step keeps a true equation true.
3. $\cos 2x = 1 - 2\sin^2x$.Problem 2's third form is the one whose only square is $\sin^2x$.
4. $2\sin^2x = 1 - \cos 2x$, so $\sin^2x = \tfrac{1-\cos 2x}{2}$.Add $2\sin^2x - \cos 2x$ to both sides, then divide both by 2.
5. $\cos^2x = \tfrac{1+\cos2x}{2}$, $\sin^2x = \tfrac{1-\cos 2x}{2}$Steps 2 and 4. Each trades a square for a first power of a cosine at twice the angle, which is how $\int\cos^2x\,dx$ and $\int\sin^2x\,dx$ are done.

Problem 4

1. $\sin(x+y) = \sin x\cos y + \cos x\sin y$ and $\sin(x-y) = \sin x\cos y - \cos x\sin y$.The sine sum formula with the upper and with the lower sign.
2. $\sin(x+y)\sin(x-y) = \sin^2x\cos^2y - \cos^2x\sin^2y$.The two factors are the sum and the difference of the same two products, $\sin x\cos y$ and $\cos x\sin y$, and a sum times a difference is the difference of the squares.
3. $= \sin^2x\,(1 - \sin^2y) - (1 - \sin^2x)\sin^2y$.The target contains only sines, so replace each $\cos^2$ by $1 - \sin^2$ (Pythagorean identity).
4. $= \sin^2x - \sin^2x\sin^2y - \sin^2y + \sin^2x\sin^2y$.Multiply out both products.
5. $\sin(x+y)\sin(x-y) = \sin^2x - \sin^2y$The two $\sin^2x\sin^2y$ terms cancel. The left side has been transformed into the right side.

Problem 5

1. $x = 2\cdot\tfrac x2$.The right side is a function of the angle $\tfrac x2$, so write $x$ as twice that angle and use Problem 2, whose identities hold for every angle, at $\tfrac x2$.
2. $1 - \cos x = 1 - \big(1 - 2\sin^2\tfrac x2\big) = 2\sin^2\tfrac x2$.Problem 2's form $\cos 2x = 1 - 2\sin^2x$, at the angle $\tfrac x2$, gives $\cos x = 1 - 2\sin^2\tfrac x2$; this form is chosen because its 1 cancels the 1 in the numerator.
3. $\sin x = 2\sin\tfrac x2\cos\tfrac x2$.Problem 2's $\sin 2x = 2\sin x\cos x$ at the angle $\tfrac x2$.
4. $\dfrac{1-\cos x}{\sin x} = \dfrac{2\sin^2\tfrac x2}{2\sin\tfrac x2\cos\tfrac x2}$.Steps 2 and 3 in the numerator and denominator. The denominator is $\sin x$, which is nonzero by hypothesis, so neither $\sin\tfrac x2$ nor $\cos\tfrac x2$ is 0.
5. $\tan\tfrac x2$Cancel $2\sin\tfrac x2$, nonzero by step 4, leaving $\sin\tfrac x2/\cos\tfrac x2$. Since $\cos\tfrac x2 \neq 0$, $\tan\tfrac x2$ is defined, so both sides exist for every $x$ with $\sin x \neq 0$.

Problem 6

1. Let $u = \tfrac{x+y}{2}$ and $v = \tfrac{x-y}{2}$, the half-sum and half-difference of $x$ and $y$.The right side is written in these two angles, so naming them lets the left side be expanded with the sum formulas.
2. $u + v = x$ and $u - v = y$.Adding the two definitions gives $\tfrac{2x}{2} = x$; subtracting them gives $\tfrac{2y}{2} = y$. So every pair $x$, $y$ is a sum and a difference of $u$ and $v$.
3. $\sin x = \sin(u+v) = \sin u\cos v + \cos u\sin v$.Step 2, then the sine sum formula with the upper sign.
4. $\sin y = \sin(u-v) = \sin u\cos v - \cos u\sin v$.Step 2, then the sine sum formula with the lower sign.
5. $\sin x + \sin y = 2\sin u\cos v$.Add steps 3 and 4: the $\cos u\sin v$ terms cancel and the $\sin u\cos v$ terms double.
6. $2\sin\tfrac{x+y}2\cos\tfrac{x-y}2$Put back $u = \tfrac{x+y}2$ and $v = \tfrac{x-y}2$ from step 1.

Problem 7

1. The identity needs $\sin x \neq 0$ and $1 + \cos x \neq 0$; the first implies the second.Both are denominators. If $\cos x = -1$ then $\sin^2x = 1 - \cos^2x = 0$ by the Pythagorean identity, so excluding $\sin x = 0$ also excludes $\cos x = -1$.
2. $\dfrac{\sin x}{1+\cos x} + \dfrac{1+\cos x}{\sin x} = \dfrac{\sin^2x + (1+\cos x)^2}{(1+\cos x)\sin x}$.Common denominator: multiply the top and bottom of each fraction by the other's denominator, which is nonzero by step 1.
3. $\sin^2x + (1+\cos x)^2 = \sin^2x + 1 + 2\cos x + \cos^2x = 2 + 2\cos x$.Expand the square; $\sin^2x + \cos^2x = 1$ collapses two of the terms to 1.
4. $\dfrac{2 + 2\cos x}{(1+\cos x)\sin x} = \dfrac{2(1+\cos x)}{(1+\cos x)\sin x}$.Factoring the 2 out of the numerator exposes the factor $1 + \cos x$ it shares with the denominator.
5. $\dfrac{2}{\sin x}$, valid when $\sin x\neq0$ (which also makes $1+\cos x \neq 0$)Cancel $1 + \cos x$, nonzero by step 1. So the identity holds for every $x$ except $x = k\pi$, $k$ an integer, where $\sin x = 0$.

Problem 8

1. $\tan(x+y) = \dfrac{\sin(x+y)}{\cos(x+y)} = \dfrac{\sin x\cos y + \cos x\sin y}{\cos x\cos y - \sin x\sin y}$.The definition of $\tan$, valid where $\cos(x+y) \neq 0$, then the two sum formulas with the upper sign.
2. Suppose $\cos x \neq 0$ and $\cos y \neq 0$.These are the conditions for $\tan x$ and $\tan y$ on the right side to be defined, and they make $\cos x\cos y$ a nonzero number to divide by.
3. $\tan(x+y) = \dfrac{\tan x + \tan y}{1 - \tan x\tan y}$.Dividing the top and bottom of a fraction by the same nonzero number, $\cos x\cos y$, leaves it unchanged: $\tfrac{\sin x\cos y}{\cos x\cos y} = \tan x$, $\tfrac{\cos x\sin y}{\cos x\cos y} = \tan y$, $\tfrac{\cos x\cos y}{\cos x\cos y} = 1$ and $\tfrac{\sin x\sin y}{\cos x\cos y} = \tan x\tan y$.
4. $1 - \tan x\tan y = \dfrac{\cos(x+y)}{\cos x\cos y}$.Step 3 divided the denominator of step 1, which is $\cos(x+y)$, by $\cos x\cos y$. So $\tan x\tan y \neq 1$ is the same condition as $\cos(x+y) \neq 0$, the one step 1 needed.
5. $\tan(x+y) = \dfrac{\tan x+\tan y}{1-\tan x\tan y}$, for $\cos x, \cos y \neq 0$ and $\tan x\tan y\neq1$Steps 1 to 4: the first two conditions make the right side's tangents exist, and the third makes both sides' denominators nonzero.

Problem 9

1. $\cos 3x = \cos(2x + x) = \cos 2x\cos x - \sin 2x\sin x$.$3x = 2x + x$, and the cosine sum formula with the angles $2x$ and $x$ produces $\cos 2x$ and $\sin 2x$, which Problem 2 already expresses in $\sin x$ and $\cos x$.
2. $= (2\cos^2x - 1)\cos x - (2\sin x\cos x)\sin x$.Problem 2. Of the three forms of $\cos 2x$, take $2\cos^2x - 1$, because the target contains only cosines.
3. $= 2\cos^3x - \cos x - 2\sin^2x\cos x$.Multiply out.
4. $= 2\cos^3x - \cos x - 2(1 - \cos^2x)\cos x = 2\cos^3x - \cos x - 2\cos x + 2\cos^3x$.The Pythagorean identity $\sin^2x = 1 - \cos^2x$ removes the last sine; then multiply out.
5. $\cos 3x = 4\cos^3x - 3\cos x$Collect the $\cos^3x$ terms and the $\cos x$ terms. A check at $x = 0$: $\cos 0 = 1$ and $4 - 3 = 1$.

Problem 10

1. $R\sin(x+\varphi) = R\sin x\cos\varphi + R\cos x\sin\varphi = (R\cos\varphi)\sin x + (R\sin\varphi)\cos x$.The sine sum formula with the angles $x$ and $\varphi$; $R$ and $\varphi$ are constants, so $R\cos\varphi$ and $R\sin\varphi$ are fixed coefficients.
2. $R\cos\varphi = 3$ and $R\sin\varphi = 4$.Matching the coefficients of $\sin x$ and $\cos x$ with $3\sin x + 4\cos x$ makes the two expressions equal for every $x$, and the match is forced: $x = 0$ gives $R\sin\varphi = 4$ and $x = \tfrac\pi2$ gives $R\cos\varphi = 3$.
3. $R^2 = R^2\cos^2\varphi + R^2\sin^2\varphi = 3^2 + 4^2 = 25$, so $R = 5$.Square the two equations of step 2 and add; the Pythagorean identity removes $\varphi$. $R > 0$ picks the positive root.
4. $\cos\varphi = \tfrac35$, $\sin\varphi = \tfrac45$, so $\tan\varphi = \tfrac43$.Divide both equations of step 2 by $R = 5$, then divide the second by the first.
5. $R = 5$, $\varphi = \arctan\tfrac43$ (first quadrant, since $\cos\varphi = \tfrac35 > 0$ and $\sin\varphi = \tfrac45 > 0$)$\tan\varphi = \tfrac43$ alone has two solutions in a full turn: $\arctan\tfrac43$ in the first quadrant and $\arctan\tfrac43 + \pi$ in the third, where sine and cosine are both negative. Both are positive here, so $\varphi$ is the first-quadrant one, which is the one $\arctan$ returns (about $0.927$ rad; adding any multiple of $2\pi$ gives the same sine). So $3\sin x + 4\cos x = 5\sin\big(x + \arctan\tfrac43\big)$, a sine wave of amplitude 5.