Trigonometry

Trigonometric identities (sec, cosec, tan, cot)

Trigonometric identities involving sec, cosec, tan and cot are equations that hold for every admissible angle, meaning every angle for which both sides are defined. They all grow from the reciprocal relations \sec\theta=\frac{1}{\cos\theta}, \operatorname{cosec}\theta=\frac{1}{\sin\theta}, \tan\theta=\frac{\sin\theta}{\cos\theta} and \cot\theta=\frac{\cos\theta}{\sin\theta}, which give \tan\theta\cot\theta=1. In most problems the first step is to rewrite every function in terms of \sin\theta and \cos\theta and then simplify.

The Pythagorean identity \sin^2\theta+\cos^2\theta=1 gives two more forms. Dividing by \cos^2\theta yields 1+\tan^2\theta=\sec^2\theta, and dividing by \sin^2\theta yields 1+\cot^2\theta=\operatorname{cosec}^2\theta. Subtracting gives \sec^2\theta-\tan^2\theta=1 and \operatorname{cosec}^2\theta-\cot^2\theta=1. Because these are differences of squares, (\operatorname{cosec}\theta+\cot\theta)(\operatorname{cosec}\theta-\cot\theta)=1, so the two factors are reciprocals. If \operatorname{cosec}\theta-\cot\theta=m, then \operatorname{cosec}\theta+\cot\theta=\frac{1}{m}, and adding and subtracting these gives \operatorname{cosec}\theta=\frac{m}{2}+\frac{1}{2m} and \cot\theta=\frac{1}{2m}-\frac{m}{2}.

A pair of the form x=m\sec A+n\tan A and y=m\tan A+n\sec A shows why difference-of-squares factorisation matters. Writing x^2-y^2=(x+y)(x-y) and recognising the factors (\sec A+\tan A) and (\sec A-\tan A) whose product is 1 reduces the expression to a constant combination of m and n. Always look for this pair before expanding.

Two conversions turn mixed expressions into sines and cosines. Since \sec\theta-\tan\theta=\frac{1-\sin\theta}{\cos\theta} and \operatorname{cosec}\theta+\cot\theta=\frac{1+\cos\theta}{\sin\theta}, the combination p+q(p-1) with these two quantities as p and q simplifies to the constant -1 for every admissible \theta. Products of the form (1+\cot\theta-\operatorname{cosec}\theta)(1+\tan\theta+\sec\theta) are handled the same way. Over the common denominator \sin\theta\cos\theta the numerator becomes (\sin\theta+\cos\theta)^2-1=2\sin\theta\cos\theta, so the product is the constant 2.

Ratios such as \frac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1} are verified by multiplying numerator and denominator by \cos\theta and cross-multiplying; once \sin^2\theta+\cos^2\theta=1 is used, both sides agree. This particular ratio equals \frac{1+\sin\theta}{\cos\theta}, which is \sec\theta+\tan\theta. When the denominator is \sin\theta+\cos\theta-1, multiply by \sin\theta+\cos\theta+1, because the denominator then becomes (\sin\theta+\cos\theta)^2-1=2\sin\theta\cos\theta. The ratio \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} therefore reduces to \sec\theta+\tan\theta, and the expression can be written as p\sec\theta+q\tan\theta with p=q=1.

When two relations are linear in \sec\theta and \tan\theta, squaring and adding removes the cross terms. The identity (a+bt)^2+(b-at)^2=(a^2+b^2)(1+t^2) holds for every t, so with t=\tan\theta the right side becomes (a^2+b^2)\sec^2\theta. Comparing this with a second relation fixes the unknown constant. The same pattern works for (1+ab)^2+(a-b)^2=(1+a^2)(1+b^2). Substituting a=\tan\alpha and b=\tan\beta turns a sum of squares into a product of secant squares, and expanding with \sec^2=1+\tan^2 gives a constant quickly.

If \sec\theta+\tan\theta=x, then \sec\theta-\tan\theta=\frac{1}{x}, so \sec\theta=\frac{x+x^{-1}}{2} and \tan\theta=\frac{x-x^{-1}}{2}. Any linear expression in sec and tan can then be evaluated. For example, if \sec\theta+\tan\theta=5, then \sec\theta=\frac{13}{5} and \tan\theta=\frac{12}{5}, so 2\sec\theta+\tan\theta=\frac{38}{5}. The same halving rule explains the pair \sec\theta and \tan\theta hidden in an expression such as \frac{1+\sin\theta}{\cos\theta}=p+\sqrt{p^2+1}. Because this expression is \sec\theta+\tan\theta, its reciprocal is \sqrt{p^2+1}-p, so \sec\theta=\sqrt{p^2+1} and \tan\theta=p. The options \frac{1}{\sqrt{p^2+1}} and \frac{p}{\sqrt{p^2+1}} are the cosine and sine under the same substitution, so they are wrong for \sec\theta and \tan\theta.

An equation mixing \tan^2\theta and \sec\theta is solved by substituting \tan^2\theta=\sec^2\theta-1, which produces a quadratic in \sec\theta. For example, \tan^2\theta=3\sec\theta-3 becomes \sec^2\theta-3\sec\theta+2=0, with roots 1 and 2. Because 0<\theta<90^\circ forces \sec\theta>1, the root is \sec\theta=2, so \theta=60^\circ. Once the angle is known, the requested combination of \cot\theta and \operatorname{cosec}\theta is evaluated, and the result is often written as a perfect square such as (\sqrt3+1)^2.

An equation mixing \tan\theta and \sec^2\theta becomes a quadratic in t=\tan\theta after the substitution \sec^2\theta=1+t^2. A quadratic such as t^2-2t-1=0 has roots 1\pm\sqrt2, and the quadrant decides which root is admissible. In the first quadrant only the positive root is valid, and \cot\theta is its reciprocal, \frac{1}{1+\sqrt2}=\sqrt2-1. Rationalising such reciprocals is a routine step.

If \tan\theta=\frac{\sqrt{b^2-a^2}}{a} with 0<\theta<90^\circ, a right triangle with base a and hypotenuse b has perpendicular \sqrt{b^2-a^2}. Then \cos\theta=\frac{a}{b} and \sec\theta=\frac{b}{a}, and \sec\theta+\cos\theta+2=\frac{a^2+2ab+b^2}{ab}=\frac{(a+b)^2}{ab}.

If \sin\theta+\cos\theta=p, squaring gives \sin\theta\cos\theta=\frac{p^2-1}{2}. Hence \tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}=\frac{2}{p^2-1}. Writing q=\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} then gives q=\frac{2p}{p^2-1}, that is, 2p=q(p^2-1).

Since \tan\theta+\cot\theta=\frac{1}{\sin\theta\cos\theta}, a quantity such as \sqrt{k(\tan\theta+\cot\theta)} depends only on \sin\theta\cos\theta. That value can be fixed with the bound a\sin\theta+b\cos\theta\le\sqrt{a^2+b^2}, whose equality holds when (\sin\theta,\cos\theta) is proportional to (a,b). For example, if 3\sin\theta+4\cos\theta=5, then \sin\theta=\frac{3}{5} and \cos\theta=\frac{4}{5}, so \tan\theta+\cot\theta=\frac{25}{12}.

A statement with an equals sign is an identity only if it holds for all admissible angles, so each candidate is tested by converting both sides to sines and cosines. After factoring \frac{\sin\theta}{\cos\theta} out of numerator and denominator, \frac{\tan\theta+\sin\theta}{\tan\theta-\sin\theta} becomes \frac{1+\cos\theta}{1-\cos\theta}, which is the same as \frac{\sec\theta+1}{\sec\theta-1}, so this equality is an identity. To disprove a candidate, substitute a convenient angle such as 30^\circ or 45^\circ and compare the values. A mismatch is enough to reject the equality. For instance, the square root \sqrt{\tan^2\theta+\cot^2\theta+4} cannot equal \sec\theta+\operatorname{cosec}\theta in general, because (\sec\theta+\operatorname{cosec}\theta)^2=\tan^2\theta+\cot^2\theta+2+2\sec\theta\operatorname{cosec}\theta. By contrast, \sqrt{\sec^2\theta+\operatorname{cosec}^2\theta}=\tan\theta+\cot\theta holds in the first quadrant, since both sides square to \frac{1}{\sin^2\theta\cos^2\theta} and are positive there.

Equations such as \cos^4\theta-\sin^4\theta=\tan2\theta fail because \cos^4\theta-\sin^4\theta factors into \cos2\theta, which is not \tan2\theta in general. In contrast, \frac{1-\tan^2\theta}{1+\tan^2\theta}=\cos^2\theta-\sin^2\theta and \operatorname{cosec}\theta+\cot\theta=\frac{1}{\operatorname{cosec}\theta-\cot\theta} are both genuine identities.

A root such as \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} is simplified by writing the fraction as \frac{(1-\sin\theta)^2}{\cos^2\theta}. In the first quadrant \cos\theta>0, so the root equals \frac{1-\sin\theta}{\cos\theta}, which is exactly \sec\theta-\tan\theta; their difference is therefore 0. Sign and quadrant must always be checked before removing a square root.

If \cot\theta is given, then \operatorname{cosec}^2\theta=1+\cot^2\theta and \tan^2\theta=\frac{1}{\cot^2\theta}, which gives \sec^2\theta=1+\tan^2\theta. For example, if \cot\theta=2, then \operatorname{cosec}^2\theta=5 and \sec^2\theta=\frac{5}{4}, so the ratio \frac{\operatorname{cosec}^2\theta-\sec^2\theta}{\operatorname{cosec}^2\theta+\sec^2\theta} equals \frac{3}{5} without finding \theta.

When the squared sine and cosine of an angle are known as fractions, \tan^2 is their quotient, \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}. For example, if \sin^2B=\frac{2}{7}, then \cos^2B=\frac{5}{7} and \tan^2B=\frac{2}{5}. Such values are fixed by the Pythagorean identity once each square is found from the relations given.

An expression \frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta} is the tangent of a shifted angle. Dividing by \cos\theta gives \frac{1+\tan\theta}{\tan\theta-1}=-\tan\left(\frac{\pi}{4}+\theta\right)=\tan\left(\frac{3\pi}{4}-\theta\right). Since \sin\left(\frac{3\pi}{4}-\theta\right)=\frac{\sin\theta+\cos\theta}{\sqrt2}, the quantity \sqrt2\sin x returns \sin\theta+\cos\theta. The angle x must still be checked to lie in the correct interval for the given range of \theta.

If \tan A+\cot A=2, multiplying by \tan A gives (\tan A-1)^2=0, so \tan A=1. The sum of a positive number and its reciprocal equals 2 only at equality. Then every power of \tan A equals 1, and a sum of consecutive powers from \tan^2A to \tan^nA is just the number of terms, namely n-1.

• admissible angle — an angle for which every function in the expression is defined and no denominator vanishes, for example \cos\theta\neq0 for \sec\theta.
• principal value — the single angle in the stated interval that a trigonometric equation selects, which must be checked after solving.

Trigonometric identities (sin²+cos²=1)

An identity is an equation that holds for every admissible value of the variable, whereas a conditional equation holds only for some values. The fundamental identity \sin^2\theta+\cos^2\theta=1 follows from the Pythagorean theorem applied to a point on the unit circle, and most other identities in this chapter are derived from it. Dividing by \cos^2\theta gives 1+\tan^2\theta=\sec^2\theta, and dividing by \sin^2\theta gives 1+\cot^2\theta=\csc^2\theta.

In problems, the identity is used in four main ways. First, replace \cos^2\theta by 1-\sin^2\theta (or the reverse) so that the whole expression is in a single function. Second, expand higher powers by taking a=\sin^2\theta and b=\cos^2\theta with a+b=1. This gives the standard forms:

• \sin^4\theta+\cos^4\theta — equals 1-2\sin^2\theta\cos^2\theta. The cross term has a minus sign, since (a+b)^2=a^2+2ab+b^2.
• \sin^6\theta+\cos^6\theta — equals 1-3\sin^2\theta\cos^2\theta, from a^3+b^3=(a+b)^3-3ab(a+b).
• \sin^4\theta-\sin^2\theta — equals -\sin^2\theta\cos^2\theta, and \cos^4\theta-\cos^2\theta equals the same expression.

Third, a ratio that combines a squared term with a linear term can be simplified by writing everything in terms of one ratio. Fourth, a relation such as a\sin^2\theta+b\cos^2\theta=c becomes linear in \sin^2\theta after \cos^2\theta is eliminated, so it can be solved directly.

To check whether two expressions are identically equal, reduce one side until it matches the other, or cross-multiply and reduce the result to \sin^2\theta+\cos^2\theta=1. A single numerical check can disprove an identity but can never prove it. The sign of a square root must also be decided by the quadrant, since \sin\theta=\sqrt{1-\cos^2\theta} holds only when \sin\theta\ge 0.

In the first quadrant, 0\le\sin\theta,\cos\theta\le 1, so a sum of two such quantities reaches its upper bound only when each term is at its own bound.

A worked pattern is \sec\theta+\cos\theta=k. Multiplying by \cos\theta gives a quadratic in \cos\theta, and roots outside [-1,1] are rejected before \sin^2\theta=1-\cos^2\theta is applied.

A product of consecutive integers equals zero only when one of the factors is zero, so the sequence must contain 0. When the extreme values of a symmetric set are -k and k, their arithmetic mean is 0.

For an expression such as \tan\theta+\cot\theta, the identity gives \tan\theta+\cot\theta=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}, so the sum is fixed once \sin\theta\cos\theta is known, and the angle itself is found from the given relation.

A statement is an identity only if both sides agree for all admissible \theta. Expressions such as \sin^4\theta-\sin^2\theta and \cos^4\theta-\cos^2\theta both reduce to -\sin^2\theta\cos^2\theta, while a sign error in the cross term of \sin^4\theta+\cos^4\theta makes an equation false.

For \tan^4\theta+\tan^2\theta=\tan^2\theta\sec^2\theta, and \sec^4\theta-\sec^2\theta=\sec^2\theta\tan^2\theta, so the two sides match.

When a result is built from squares of several terms, collect the squared terms, factor out the common factor, and apply \sin^2+\cos^2=1 to the remaining bracket.

To verify a proportion between trigonometric fractions, cross-multiply and expand both sides. If the difference reduces to \sin^2\theta+\cos^2\theta-1, the proportion is an identity.

When fractions with \csc\theta\pm 1 are added, the product (\csc\theta+1)(\csc\theta-1)=\csc^2\theta-1=\cot^2\theta simplifies the denominator. The result then reduces to a single ratio such as \tan\theta, which fixes the angle in the given quadrant.

In a right triangle with the right angle at C, the acute angles are complementary, so \cos B=\sin A and \cos^2A+\cos^2B=1. Since \cos C=0, the sum of squared cosines of the three angles is 1.

The same complementary relation gives \sin^2A+\sin^2B=1, and \sin^2C=1 for the right angle, so \sin^2A+\sin^2B+\sin^2C=2.

If p=\sin^2\theta\cos^2\theta, then the combinations \sin^6\theta+\cos^6\theta=1-3p and \sin^4\theta+\cos^4\theta=1-2p make the p-terms cancel in many expressions, leaving a constant.

Using 1+\cot^2\theta=\csc^2\theta and 1-\cos^2\theta=\sin^2\theta, the product (1+\cot^2\theta)(1-\cos^2\theta) equals \csc^2\theta\sin^2\theta=1. The parallel product with \tan and \sin also equals 1.

If an equation in \cos^2\theta and \sin\theta is rewritten with \cos^2\theta=1-\sin^2\theta, it becomes a quadratic in \sin\theta. Each factor is solved separately, and the roots outside the given range of \theta are rejected.

For a linear form a\sin\theta+b\cos\theta and its companion b\sin\theta-a\cos\theta, the sum of their squares is (a^2+b^2)(\sin^2\theta+\cos^2\theta)=a^2+b^2. Once one of them is fixed, the square of the other is known, but its sign must be checked against the condition.

When an expression in \sin x and \cos x contains \sin x\cos x and (\sin x\pm\cos x)^2, substitute t=\sin x\cos x and use (\sin x\pm\cos x)^2=1\pm 2t. The t-terms then cancel, and the expression is a constant.

A product of two quadratic factors in \sin\theta is zero only when a factor is zero, and each root must be tested against the admissible interval. If the interval is open, a root at its endpoint is excluded.

If x\sin\theta=y\cos\theta, then (x,y) is proportional to (\cos\theta,\sin\theta), so we write x=t\cos\theta and y=t\sin\theta. Then x^2+y^2=t^2(\sin^2\theta+\cos^2\theta)=t^2.

The sum of cubes factors as \sin^3\theta+\cos^3\theta=(\sin\theta+\cos\theta)(1-\sin\theta\cos\theta), and 1-\sin^6\theta=\cos^2\theta(1+\sin^2\theta+\sin^4\theta). Both forms are checked with the identity, and both expressions are identities.

An equation of the form 7s^2+9(1-s)^2+42s=16 in s=\sin^2\theta simplifies to a perfect square (4s+3)^2=16. Since 4s+3 is positive, the square root is taken as 4, and \sin^2\theta follows directly.

Since \operatorname{cosec}\theta=\dfrac{1}{\sin\theta}, an expression such as \operatorname{cosec}\theta(1-\cos\theta) equals \dfrac{1-\cos\theta}{\sin\theta}. For an angle in the first quadrant, \sin\theta=\sqrt{1-\cos^2\theta} is positive, and the result is rationalized to the form \sqrt{\dfrac{1-t}{1+t}}.