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06 Trigonometric Identities and Equations

A progressive guide to verifying trigonometric identities, applying angle formulas, and solving trigonometric equations with exact values and complete solution sets.

Verify identities with fundamental relationships

A is true for every angle in the common domain of its expressions. This differs from a , which usually holds only for selected values of a variable.

To verify an identity, begin with the more complicated side and transform it into the other side using valid algebraic steps. Keep track of restrictions: an original denominator cannot equal zero. For example,

1−cos⁡2θsin⁡θ=sin⁡2θsin⁡θ=sin⁡θ,\frac{1-\cos^2\theta}{\sin\theta} =\frac{\sin^2\theta}{\sin\theta} =\sin\theta,

where sin⁡θ≠0\sin\theta\ne0, as required by the original expression.

Core relationships

The provide the main tools for rewriting expressions:

  • Reciprocal identities:

    csc⁡θ=1sin⁡θ,sec⁡θ=1cos⁡θ,cot⁡θ=1tan⁡θ.\csc\theta=\frac{1}{\sin\theta},\qquad \sec\theta=\frac{1}{\cos\theta},\qquad \cot\theta=\frac{1}{\tan\theta}.

  • Quotient identities:

    tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ.\tan\theta=\frac{\sin\theta}{\cos\theta},\qquad \cot\theta=\frac{\cos\theta}{\sin\theta}.

  • Pythagorean identities:

    sin⁡2θ+cos⁡2θ=1,\sin^2\theta+\cos^2\theta=1,
    1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=csc⁡2θ.1+\tan^2\theta=\sec^2\theta, \qquad 1+\cot^2\theta=\csc^2\theta.

The even–odd relationships describe behavior under a negative angle:

sin⁡(−θ)=−sin⁡θ,tan⁡(−θ)=−tan⁡θ,\sin(-\theta)=-\sin\theta,\qquad \tan(-\theta)=-\tan\theta,
cos⁡(−θ)=cos⁡θ,sec⁡(−θ)=sec⁡θ.\cos(-\theta)=\cos\theta,\qquad \sec(-\theta)=\sec\theta.

Takeaway: Rewrite unfamiliar functions using reciprocal or quotient identities, then use a Pythagorean identity to complete the simplification. Always preserve the original domain restrictions.

Expand combined angles

The expand a function whose angle is the sum or difference of two angles. They are especially useful for exact values and for converting complicated equations into simpler ones.

Sine and cosine

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta
sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta
cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta
cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta

A frequent error is carrying the inner sign unchanged through the cosine formula. For cosine, a sum produces subtraction between the products, while a difference produces addition.

Tangent

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
tan⁡(α−β)=tan⁡α−tan⁡β1+tan⁡αtan⁡β\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}

These formulas apply only where all expressions are defined.

Exact-value example

To find cos⁡π12\cos\frac{\pi}{12}, write

π12=π3−π4.\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}.

Then

cos⁡π12=cos⁡(π3−π4)=cos⁡π3cos⁡π4+sin⁡π3sin⁡π4=(12)(22)+(32)(22)=2+64.\begin{aligned} \cos\frac{\pi}{12} &=\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\\ &=\cos\frac{\pi}{3}\cos\frac{\pi}{4}+\sin\frac{\pi}{3}\sin\frac{\pi}{4}\\ &=\left(\frac{1}{2}\right)\left(\frac{\sqrt{2}}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right)\\ &=\frac{\sqrt{2}+\sqrt{6}}{4}. \end{aligned}

Takeaway: Decompose an angle into familiar special angles, select the correct sum or difference formula, and substitute exact unit-circle values.

Use double-angle relationships

The follow from the sum formulas by setting both angles equal to θ\theta. They are useful for changing between expressions involving θ\theta and expressions involving 2θ2\theta.

sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta)=2\sin\theta\cos\theta
cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos(2\theta)=\cos^2\theta-\sin^2\theta

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, the cosine formula also becomes

cos⁡(2θ)=1−2sin⁡2θ\cos(2\theta)=1-2\sin^2\theta

or

cos⁡(2θ)=2cos⁡2θ−1.\cos(2\theta)=2\cos^2\theta-1.

Choose the version that matches the function already present. The tangent form is

tan⁡(2θ)=2tan⁡θ1−tan⁡2θ,\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta},

provided the denominator is nonzero and both sides are defined.

Simplification example

Factor a difference of squares:

sin⁡4θ−cos⁡4θ=(sin⁡2θ−cos⁡2θ)(sin⁡2θ+cos⁡2θ)=sin⁡2θ−cos⁡2θ=−cos⁡(2θ).\begin{aligned} \sin^4\theta-\cos^4\theta &=(\sin^2\theta-\cos^2\theta)(\sin^2\theta+\cos^2\theta)\\ &=\sin^2\theta-\cos^2\theta\\ &=-\cos(2\theta). \end{aligned}

Takeaway: Factor first when possible, use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, and select the double-angle form that minimizes the number of different trigonometric functions.

Evaluate half-angle expressions

The are obtained from the by replacing the angle with α2\frac{\alpha}{2}. Their square-root signs depend on the quadrant of the half-angle.

sin⁡α2=±1−cos⁡α2,cos⁡α2=±1+cos⁡α2.\sin\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{2}}, \qquad \cos\frac{\alpha}{2}=\pm\sqrt{\frac{1+\cos\alpha}{2}}.

For tangent,

tan⁡α2=±1−cos⁡α1+cos⁡α\tan\frac{\alpha}{2}=\pm\sqrt{\frac{1-\cos\alpha}{1+\cos\alpha}}

and, when the denominators are nonzero,

tan⁡α2=sin⁡α1+cos⁡α=1−cos⁡αsin⁡α.\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha} =\frac{1-\cos\alpha}{\sin\alpha}.

Exact-value example

To evaluate sin⁡π8\sin\frac{\pi}{8}, observe that π8\frac{\pi}{8} lies in Quadrant I, so the positive square root is selected:

sin⁡π8=1−cos⁡(π4)2=1−222=2−22.\begin{aligned} \sin\frac{\pi}{8} &=\sqrt{\frac{1-\cos\left(\frac{\pi}{4}\right)}{2}}\\ &=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\\ &=\frac{\sqrt{2-\sqrt{2}}}{2}. \end{aligned}

Takeaway: Find the quadrant of the half-angle before choosing the sign; the quadrant of α\alpha alone is not sufficient.

Solve trigonometric equations

A is solved by reducing it to a basic sine, cosine, or tangent equation and then listing every permitted value. is essential: sine and cosine repeat every 2π2\pi, while tangent repeats every π\pi.

Basic equations

For ∣a∣≤1\lvert a\rvert\le1,

sin⁡x=a⟹x=arcsin⁡(a)+2πk\sin x=a \quad\Longrightarrow\quad x=\arcsin(a)+2\pi k

or

x=π−arcsin⁡(a)+2πk,k∈Z.x=\pi-\arcsin(a)+2\pi k, \qquad k\in\mathbb Z.

For ∣a∣≤1\lvert a\rvert\le1,

cos⁡x=a⟹x=±arccos⁡(a)+2πk,k∈Z.\cos x=a \quad\Longrightarrow\quad x=\pm\arccos(a)+2\pi k, \qquad k\in\mathbb Z.

For any real number aa,

tan⁡x=a⟹x=arctan⁡(a)+πk,k∈Z.\tan x=a \quad\Longrightarrow\quad x=\arctan(a)+\pi k, \qquad k\in\mathbb Z.

On [0,2π)[0,2\pi), use the unit circle to retain only solutions in that interval.

Quadratic trigonometric equations

For

2sin⁡2x−sin⁡x−1=0,2\sin^2x-\sin x-1=0,

let u=sin⁡xu=\sin x. Then

2u2−u−1=0=(2u+1)(u−1),2u^2-u-1=0=(2u+1)(u-1),

so sin⁡x=−12\sin x=-\frac{1}{2} or sin⁡x=1\sin x=1. On [0,2π)[0,2\pi), the solutions are

x=7π6,11π6,π2.x=\frac{7\pi}{6},\quad \frac{11\pi}{6},\quad \frac{\pi}{2}.

Equations requiring an identity

Use the cosine difference formula to simplify

cos⁡xcos⁡(2x)+sin⁡xsin⁡(2x)=32.\cos x\cos(2x)+\sin x\sin(2x)=\frac{\sqrt{3}}{2}.

The left side is

cos⁡(x−2x)=cos⁡(−x)=cos⁡x,\cos(x-2x)=\cos(-x)=\cos x,

because cosine is even. Therefore,

cos⁡x=32,\cos x=\frac{\sqrt{3}}{2},

and on [0,2π)[0,2\pi),

x=π6,11π6.x=\frac{\pi}{6},\quad \frac{11\pi}{6}.

Reliable procedure

  1. Identify the interval and whether exact or decimal answers are required.

  2. Simplify algebraically by collecting terms, factoring, or isolating a function.

  3. Apply an identity that reduces the number of functions or simplifies the angles.

  4. Solve the resulting basic .

  5. List every solution in the requested interval, or use k∈Zk\in\mathbb Z for a general solution.

  6. Check candidates in the original equation and reject values that make an original denominator zero.

Takeaway: Do not stop after finding one angle. Use the interval, quadrant information, and to produce the complete solution set.