Trigonometric Identities

Study Sheet

Trigonometric Identities

Fundamental identities, simplifying, and verifying

The Fundamental Identities

These are the building blocks. Memorize them: every simplification and every proof in this topic comes from combining a few of these.

Concept
Reciprocal Identities
csc⁡θ=1sin⁡θsec⁡θ=1cos⁡θcot⁡θ=1tan⁡θ\csc\theta=\frac{1}{\sin\theta}\qquad \sec\theta=\frac{1}{\cos\theta}\qquad \cot\theta=\frac{1}{\tan\theta}
sin⁡θ=1csc⁡θcos⁡θ=1sec⁡θtan⁡θ=1cot⁡θ\sin\theta=\frac{1}{\csc\theta}\qquad \cos\theta=\frac{1}{\sec\theta}\qquad \tan\theta=\frac{1}{\cot\theta}
Concept
Quotient Identities
tan⁡θ=sin⁡θcos⁡θcot⁡θ=cos⁡θsin⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}\qquad\qquad \cot\theta=\frac{\cos\theta}{\sin\theta}
Concept
Pythagorean Identities
sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
1+tan⁡2θ=sec⁡2θ1+cot⁡2θ=csc⁡2θ1+\tan^2\theta=\sec^2\theta\qquad\qquad 1+\cot^2\theta=\csc^2\theta

Each has useful rearrangements, e.g. sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta and tan⁡2θ=sec⁡2θ−1\tan^2\theta=\sec^2\theta-1.

Concept
Cofunction Identities (angles complementary)
sin⁡ ⁣(90∘−θ)=cos⁡θtan⁡ ⁣(90∘−θ)=cot⁡θsec⁡ ⁣(90∘−θ)=csc⁡θ\sin\!\left(90^\circ-\theta\right)=\cos\theta\qquad \tan\!\left(90^\circ-\theta\right)=\cot\theta\qquad \sec\!\left(90^\circ-\theta\right)=\csc\theta
cos⁡ ⁣(90∘−θ)=sin⁡θcot⁡ ⁣(90∘−θ)=tan⁡θcsc⁡ ⁣(90∘−θ)=sec⁡θ\cos\!\left(90^\circ-\theta\right)=\sin\theta\qquad \cot\!\left(90^\circ-\theta\right)=\tan\theta\qquad \csc\!\left(90^\circ-\theta\right)=\sec\theta

In radians replace 90∘90^\circ with π2\dfrac{\pi}{2}.

Concept
Even/Odd (Negative-Angle) Identities

Even (unchanged by a sign flip):

cos⁡(−θ)=cos⁡θsec⁡(−θ)=sec⁡θ\cos(-\theta)=\cos\theta\qquad\qquad \sec(-\theta)=\sec\theta

Odd (pick up a negative sign):

sin⁡(−θ)=−sin⁡θtan⁡(−θ)=−tan⁡θcsc⁡(−θ)=−csc⁡θcot⁡(−θ)=−cot⁡θ\sin(-\theta)=-\sin\theta\quad \tan(-\theta)=-\tan\theta\quad \csc(-\theta)=-\csc\theta\quad \cot(-\theta)=-\cot\theta
Example
Applying one identity

Write sec⁡θcot⁡θ\sec\theta\cot\theta as a single function.

sec⁡θcot⁡θ=1cos⁡θ⋅cos⁡θsin⁡θreciprocal and quotient=1sin⁡θ=csc⁡θ.\begin{aligned} \sec\theta\cot\theta &=\frac{1}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta} &&\text{reciprocal and quotient}\\ &=\frac{1}{\sin\theta}=\csc\theta. \end{aligned}
Tip

Tip. When you are stuck, rewrite everything in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta. Almost every identity untangles once only sines and cosines remain.

Simplifying Trigonometric Expressions

To simplify means to reduce an expression to its shortest equivalent form. Strategy: convert to sines and cosines, combine fractions, then use a Pythagorean identity to collapse the result.

Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Example
A full simplification

Simplify 1−sin⁡2θcos⁡θ\dfrac{1-\sin^2\theta}{\cos\theta}.

1−sin⁡2θcos⁡θ=cos⁡2θcos⁡θsince 1−sin⁡2θ=cos⁡2θ=cos⁡θ.\begin{aligned} \frac{1-\sin^2\theta}{\cos\theta} &=\frac{\cos^2\theta}{\cos\theta} &&\text{since }1-\sin^2\theta=\cos^2\theta\\ &=\cos\theta. \end{aligned}
Example
Combine, then use sin⁡2+cos⁡2=1\sin^2+\cos^2=1

Simplify cos⁡θ (tan⁡θ+cot⁡θ)\cos\theta\,(\tan\theta+\cot\theta).

cos⁡θ(tan⁡θ+cot⁡θ)=cos⁡θ(sin⁡θcos⁡θ+cos⁡θsin⁡θ)=sin⁡θ+cos⁡2θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θ=1sin⁡θ=csc⁡θ.\begin{aligned} \cos\theta\left(\tan\theta+\cot\theta\right) &=\cos\theta\left(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\right)\\ &=\sin\theta+\frac{\cos^2\theta}{\sin\theta} =\frac{\sin^2\theta+\cos^2\theta}{\sin\theta}\\ &=\frac{1}{\sin\theta}=\csc\theta. \end{aligned}
Tip

Tip. Watch for the Pythagorean patterns hiding in an expression: 1−sin⁡2θ1-\sin^2\theta, sec⁡2θ−1\sec^2\theta-1, and csc⁡2θ−1\csc^2\theta-1 each collapse to a single squared function.

Rewriting in Terms of Sine and Cosine (or as One Function)

Any of the six functions can be written using only sin⁡θ\sin\theta and cos⁡θ\cos\theta. This is the single most useful move for both simplifying and verifying.

Opposite, adjacent and hypotenuse are named from the angle.

Concept
The universal rewrite
tan⁡θ=sin⁡θcos⁡θcot⁡θ=cos⁡θsin⁡θsec⁡θ=1cos⁡θcsc⁡θ=1sin⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta}\quad \cot\theta=\frac{\cos\theta}{\sin\theta}\quad \sec\theta=\frac{1}{\cos\theta}\quad \csc\theta=\frac{1}{\sin\theta}
Example
Rewrite, then combine into one function

Express tan⁡θsec⁡θ\dfrac{\tan\theta}{\sec\theta} as a single function.

tan⁡θsec⁡θ= sin⁡θcos⁡θ  1cos⁡θ =sin⁡θcos⁡θ⋅cos⁡θ1=sin⁡θ.\begin{aligned} \frac{\tan\theta}{\sec\theta} &=\frac{\ \dfrac{\sin\theta}{\cos\theta}\ }{\ \dfrac{1}{\cos\theta}\ } =\frac{\sin\theta}{\cos\theta}\cdot\frac{\cos\theta}{1}\\ &=\sin\theta. \end{aligned}
Tip

Tip. A complex fraction is divided by multiplying by the reciprocal. Rewriting in sin⁡\sin/cos⁡\cos usually turns a messy quotient into a clean cancellation.

Verifying Trigonometric Identities

To verify (prove) an identity, transform one side until it matches the other. You may not move terms across the == sign as in solving an equation---each side is worked independently.

Concept
Strategies
  • [leftmargin=*,itemsep=2pt]
  • Start with the more complicated side; there is more to simplify.
  • Convert everything to sine and cosine.
  • Get a common denominator to combine fractions.
  • Multiply by a conjugate to create a Pythagorean form (e.g. multiply by 1+sin⁡θ1+\sin\theta to get 1−sin⁡2θ1-\sin^2\theta).
  • Factor and look for a Pythagorean identity to substitute.
  • Work both sides to a common third expression if neither side alone reaches the other.
Example
Verification by converting to sin/cos

Verify sec⁡θ−sin⁡θtan⁡θ=cos⁡θ\sec\theta-\sin\theta\tan\theta=\cos\theta.

sec⁡θ−sin⁡θtan⁡θ=1cos⁡θ−sin⁡θ⋅sin⁡θcos⁡θ=1cos⁡θ−sin⁡2θcos⁡θ=1−sin⁡2θcos⁡θ=cos⁡2θcos⁡θ=cos⁡θ.✓\begin{aligned} \sec\theta-\sin\theta\tan\theta &=\frac{1}{\cos\theta}-\sin\theta\cdot\frac{\sin\theta}{\cos\theta}\\ &=\frac{1}{\cos\theta}-\frac{\sin^2\theta}{\cos\theta} =\frac{1-\sin^2\theta}{\cos\theta}\\ &=\frac{\cos^2\theta}{\cos\theta}=\cos\theta.\qquad\checkmark \end{aligned}
Example
Verification using a conjugate

Verify cos⁡θ1−sin⁡θ=1+sin⁡θcos⁡θ\dfrac{\cos\theta}{1-\sin\theta}=\dfrac{1+\sin\theta}{\cos\theta}.

cos⁡θ1−sin⁡θ=cos⁡θ1−sin⁡θ⋅1+sin⁡θ1+sin⁡θmultiply by conjugate=cos⁡θ(1+sin⁡θ)1−sin⁡2θ=cos⁡θ(1+sin⁡θ)cos⁡2θ=1+sin⁡θcos⁡θ.✓\begin{aligned} \frac{\cos\theta}{1-\sin\theta} &=\frac{\cos\theta}{1-\sin\theta}\cdot\frac{1+\sin\theta}{1+\sin\theta} &&\text{multiply by conjugate}\\ &=\frac{\cos\theta(1+\sin\theta)}{1-\sin^2\theta}\\ &=\frac{\cos\theta(1+\sin\theta)}{\cos^2\theta}\\ &=\frac{1+\sin\theta}{\cos\theta}.\qquad\checkmark \end{aligned}
Tip

Tip. Never divide by an expression that could be zero, and never “cross-multiply.” A verification is a chain of equalities down one side, ending exactly at the other side.

Finding the Other Five Values from One

Given one function value and the quadrant of θ\theta, you can find all six. Use a Pythagorean identity to get a partner value, then the reciprocal and quotient identities for the rest. The quadrant fixes each sign.

Concept
Signs by quadrant (ASTC)

“All Students Take Calculus”: the function that is positive in QI, QII, QIII, QIV.

Example
Full worked value problem

Given sin⁡θ=35\sin\theta=\dfrac{3}{5} with θ\theta in Quadrant II, find the other five.

cos⁡2θ=1−sin⁡2θ=1−925=1625cos⁡θ=−45(QII: cosine negative)tan⁡θ=sin⁡θcos⁡θ=3/5−4/5=−34csc⁡θ=1sin⁡θ=53,sec⁡θ=1cos⁡θ=−54cot⁡θ=1tan⁡θ=−43.\begin{aligned} \cos^2\theta&=1-\sin^2\theta=1-\tfrac{9}{25}=\tfrac{16}{25}\\ \cos\theta&=-\tfrac{4}{5}&&\text{(QII: cosine negative)}\\ \tan\theta&=\frac{\sin\theta}{\cos\theta}=\frac{3/5}{-4/5}=-\tfrac{3}{4}\\ \csc\theta&=\frac{1}{\sin\theta}=\tfrac{5}{3},\quad \sec\theta=\frac{1}{\cos\theta}=-\tfrac{5}{4}\\ \cot\theta&=\frac{1}{\tan\theta}=-\tfrac{4}{3}. \end{aligned}
Tip

Tip. Solve for the partner value (cos⁡\cos if you were given sin⁡\sin) with a Pythagorean identity first, choosing its sign from the quadrant. Everything else follows from reciprocal and quotient identities---no square roots needed.

Going Deeper: Advanced Identities

The identities above are enough to simplify and verify. The next level combines them into multi-step proofs and into conditional problems, where a single fact about sin⁡x\sin x and cos⁡x\cos x unlocks a whole family of expressions. The recurring trick: treat s=sin⁡xs=\sin x and c=cos⁡xc=\cos x as two unknowns tied by s2+c2=1s^2+c^2=1, and rewrite every target as a symmetric function of them.

Concept
Symmetric-Function Toolkit

Write s=sin⁡xs=\sin x, c=cos⁡xc=\cos x. Every symmetric expression in ss and cc is built from the sum s+cs+c and the product scsc. The Pythagorean identity locks them together:

s2+c2=1,(s+c)2=1+2sc,(s−c)2=1−2sc.s^2+c^2=1,\qquad (s+c)^2=1+2sc,\qquad (s-c)^2=1-2sc.

So knowing either s+cs+c or scsc immediately gives the other. Higher powers reduce the same way:

s3+c3=(s+c) ⁣(1−sc),s3−c3=(s−c) ⁣(1+sc).s^3+c^3=(s+c)\!\left(1-sc\right),\qquad s^3-c^3=(s-c)\!\left(1+sc\right).
Concept
Power-Sum Reductions (memorize the pattern)

Since s2+c2=1s^2+c^2=1, every even power sum collapses to a polynomial in p=scp=sc:

sin⁡4x+cos⁡4x=1−2p2,\sin^4 x+\cos^4 x=1-2p^2,
sin⁡6x+cos⁡6x=1−3p2,\sin^6 x+\cos^6 x=1-3p^2,
sin⁡4x+cos⁡4x=1−2sin⁡2xcos⁡2x,sin⁡6x+cos⁡6x=1−3sin⁡2xcos⁡2x.\sin^4 x+\cos^4 x = 1-2\sin^2 x\cos^2 x,\quad \sin^6 x+\cos^6 x = 1-3\sin^2 x\cos^2 x.

These come from (s2+c2)2=s4+c4+2s2c2(s^2+c^2)^2=s^4+c^4+2s^2c^2 and (s2+c2)3=s6+c6+3s2c2(s2+c2)(s^2+c^2)^3=s^6+c^6+3s^2c^2(s^2+c^2).

Example
Conditional identity: given sin⁡x+cos⁡x\sin x+\cos x

Given sin⁡x+cos⁡x=12\sin x+\cos x=\dfrac{1}{2}, find sin⁡xcos⁡x\sin x\cos x and sin⁡3x+cos⁡3x\sin^3 x+\cos^3 x.

(sin⁡x+cos⁡x)2=sin⁡2x+cos⁡2x+2sin⁡xcos⁡x14=1+2sin⁡xcos⁡xsquare the givensin⁡xcos⁡x=14−12=−38.\begin{aligned} (\sin x+\cos x)^2&=\sin^2 x+\cos^2 x+2\sin x\cos x\\ \tfrac{1}{4}&=1+2\sin x\cos x &&\text{square the given}\\ \sin x\cos x&=\frac{\tfrac14-1}{2}=-\frac{3}{8}. \end{aligned}

Now use the sum-of-cubes factoring with s2+c2=1s^2+c^2=1:

sin⁡3x+cos⁡3x=(sin⁡x+cos⁡x) ⁣(sin⁡2x−sin⁡xcos⁡x+cos⁡2x)=(sin⁡x+cos⁡x) ⁣(1−sin⁡xcos⁡x)=12(1−(−38))=12⋅118=1116.\begin{aligned} \sin^3 x+\cos^3 x &=(\sin x+\cos x)\!\left(\sin^2 x-\sin x\cos x+\cos^2 x\right)\\ &=(\sin x+\cos x)\!\left(1-\sin x\cos x\right)\\ &=\frac{1}{2}\left(1-\left(-\frac{3}{8}\right)\right) =\frac{1}{2}\cdot\frac{11}{8}=\frac{11}{16}. \end{aligned}
Example
“If sin⁡4x+cos⁡4x=k\sin^4 x+\cos^4 x=k, find sin⁡xcos⁡x\sin x\cos x”

Reduce the power sum, then solve for the product p=sin⁡xcos⁡xp=\sin x\cos x.

sin⁡4x+cos⁡4x=(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x=1−2p2since sin⁡2x+cos⁡2x=1k=1−2p2p2=1−k2sin⁡xcos⁡x=±1−k2.\begin{aligned} \sin^4 x+\cos^4 x &=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\\ &=1-2p^2 &&\text{since }\sin^2 x+\cos^2 x=1\\ k&=1-2p^2\\ p^2&=\frac{1-k}{2}\\ \sin x\cos x&=\pm\sqrt{\frac{1-k}{2}}. \end{aligned}

The sign depends on the quadrant of xx. As a check, k=1k=1 gives p=0p=0 (an axis angle), and k=12k=\tfrac12 gives p=±12p=\pm\tfrac12, i.e. sin⁡2x=±1\sin 2x=\pm1.

Tip

Tip. For any “given one symmetric fact, find another” problem: square the given (or use s2+c2=1s^2+c^2=1) to extract scsc, then express the target through s+cs+c and scsc only. You almost never need the individual values of sin⁡x\sin x and cos⁡x\cos x.

Concept
Conditional identities from tan⁡x\tan x

If you are given tan⁡x=t\tan x=t, divide numerator and denominator by cos⁡2x\cos^2 x to turn symmetric expressions into rational functions of tt:

sin⁡xcos⁡x=tan⁡x1+tan⁡2x=t1+t2,\sin x\cos x=\frac{\tan x}{1+\tan^2 x}=\frac{t}{1+t^2},
sin⁡2x−cos⁡2x=tan⁡2x−1tan⁡2x+1=t2−1t2+1.\sin^2 x-\cos^2 x=\frac{\tan^2 x-1}{\tan^2 x+1}=\frac{t^2-1}{t^2+1}.

Both follow from dividing by sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1 written as cos⁡2x(tan⁡2x+1)\cos^2 x(\tan^2 x+1).

Example
Multi-step proof: common denominator, then a conjugate collapse

Verify 1+sin⁡xcos⁡x+cos⁡x1+sin⁡x=2sec⁡x\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=2\sec x.

1+sin⁡xcos⁡x+cos⁡x1+sin⁡x=(1+sin⁡x)2+cos⁡2xcos⁡x (1+sin⁡x)common denominator=1+2sin⁡x+sin⁡2x+cos⁡2xcos⁡x (1+sin⁡x)=2+2sin⁡xcos⁡x (1+sin⁡x)use sin⁡2x+cos⁡2x=1=2(1+sin⁡x)cos⁡x (1+sin⁡x)=2cos⁡x=2sec⁡x.✓\begin{aligned} \frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} &=\frac{(1+\sin x)^2+\cos^2 x}{\cos x\,(1+\sin x)} &&\text{common denominator}\\ &=\frac{1+2\sin x+\sin^2 x+\cos^2 x}{\cos x\,(1+\sin x)}\\ &=\frac{2+2\sin x}{\cos x\,(1+\sin x)} &&\text{use }\sin^2 x+\cos^2 x=1\\ &=\frac{2(1+\sin x)}{\cos x\,(1+\sin x)} =\frac{2}{\cos x}=2\sec x.\qquad\checkmark \end{aligned}
Example
Multi-step proof: factor out, then substitute Pythagorean

Verify tan⁡2x−sin⁡2x=tan⁡2x sin⁡2x\tan^2 x-\sin^2 x=\tan^2 x\,\sin^2 x.

tan⁡2x−sin⁡2x=sin⁡2xcos⁡2x−sin⁡2xconvert to sin/cos=sin⁡2x ⁣(1cos⁡2x−1)factor out sin⁡2x=sin⁡2x⋅1−cos⁡2xcos⁡2x=sin⁡2x⋅sin⁡2xcos⁡2x=sin⁡2x tan⁡2x.✓\begin{aligned} \tan^2 x-\sin^2 x &=\frac{\sin^2 x}{\cos^2 x}-\sin^2 x &&\text{convert to sin/cos}\\ &=\sin^2 x\!\left(\frac{1}{\cos^2 x}-1\right) &&\text{factor out }\sin^2 x\\ &=\sin^2 x\cdot\frac{1-\cos^2 x}{\cos^2 x}\\ &=\sin^2 x\cdot\frac{\sin^2 x}{\cos^2 x} =\sin^2 x\,\tan^2 x.\qquad\checkmark \end{aligned}
Example
A long expression that is secretly a constant

Show that sin⁡6x+cos⁡6x+3sin⁡2xcos⁡2x=1\sin^6 x+\cos^6 x+3\sin^2 x\cos^2 x=1 for every xx.

sin⁡6x+cos⁡6x=(sin⁡2x+cos⁡2x)3−3sin⁡2xcos⁡2x(sin⁡2x+cos⁡2x)=1−3sin⁡2xcos⁡2xsince sin⁡2x+cos⁡2x=1sin⁡6x+cos⁡6x+3sin⁡2xcos⁡2x=(1−3sin⁡2xcos⁡2x)+3sin⁡2xcos⁡2x=1.✓\begin{aligned} \sin^6 x+\cos^6 x &=(\sin^2 x+\cos^2 x)^3-3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x)\\ &=1-3\sin^2 x\cos^2 x &&\text{since }\sin^2 x+\cos^2 x=1\\ \sin^6 x+\cos^6 x+3\sin^2 x\cos^2 x &=\left(1-3\sin^2 x\cos^2 x\right)+3\sin^2 x\cos^2 x\\ &=1.\qquad\checkmark \end{aligned}

The sin⁡2xcos⁡2x\sin^2 x\cos^2 x terms cancel exactly---the whole expression is constant.

Tip

Tip. A long trig expression that reduces to a constant almost always hides a power-sum reduction: rewrite sin⁡4\sin^4/sin⁡6\sin^6 (and their cosine partners) using sin⁡2x+cos⁡2x=1\sin^2 x+\cos^2 x=1, and watch the leftover product terms cancel. If they do not cancel, recheck a sign before doubting the identity.

Formulas, Proofs & Tips

Tip
The Pythagorean identity
sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1

What it means. Knowing one of sin⁡\sin or cos⁡\cos (plus the quadrant) determines the other.

Example. If sin⁡θ=35\sin\theta=\tfrac35, then cos⁡θ=1−925=45\cos\theta=\sqrt{1-\tfrac{9}{25}}=\tfrac45.

Why it works. On the unit circle the point at angle θ\theta is (cos⁡θ, sin⁡θ)(\cos\theta,\ \sin\theta) and lies at distance 11 from the origin. The distance formula gives cos⁡2θ+sin⁡2θ=12\cos^2\theta+\sin^2\theta=1^2 — it is Pythagoras on a radius.

Tip. Dividing through by cos⁡2θ\cos^2\theta gives 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta; by sin⁡2θ\sin^2\theta gives 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta. Use the quadrant to pick the sign when you take the square root.