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.

An identity is an equation true for every angle. The Pythagorean identities and the reciprocal and quotient relations are the toolkit; verifying an identity means transforming one side into the other with those tools — usually by rewriting everything in sines and cosines, combining fractions, and factoring. Never move terms across the equals sign when verifying.

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
θ1tan θsec θ1sin θcos θ
Small triangle: cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1. Large triangle (legs 11 and tan⁡θ\tan\theta): 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta.
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.

All three are the Pythagorean theorem. The unit-circle triangle has legs cos⁡θ\cos\theta, sin⁡θ\sin\theta and hypotenuse 11. Scale it by 1cos⁡θ\dfrac{1}{\cos\theta} and you get the larger similar triangle in the figure, with legs 11 and tan⁡θ\tan\theta and hypotenuse sec⁡θ\sec\theta; algebraically that is dividing sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1 by cos⁡2θ\cos^2\theta. Dividing by sin⁡2θ\sin^2\theta instead gives tan⁡\tan's partner: 1+cot⁡2θ=csc⁡2θ1+\cot^2\theta=\csc^2\theta.

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}
Side note
Where it goes wrong

Verifying by working on both sides at once and meeting in the middle. That assumes what you are trying to prove. Transform one side only.

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.

Try it
Quick check

If tan⁡θ=2\tan\theta=2, find sec⁡2θ\sec^2\theta and cos⁡2θ\cos^2\theta without finding θ\theta.

Show the answer

Answer: sec⁡2θ=5\sec^2\theta=5 and cos⁡2θ=15\cos^2\theta=\tfrac15. The identity 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta gives 1+4=51+4=5, and cos⁡2θ\cos^2\theta is the reciprocal of sec⁡2θ\sec^2\theta. (The sign of cos⁡θ\cos\theta itself is unknown until the quadrant is given; its square is not.)

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.

Try it
Quick check

Simplify sec⁡2θ−1tan⁡θ\dfrac{\sec^2\theta-1}{\tan\theta}.

Show the answer

Answer: tan⁡θ\tan\theta. The numerator is a Pythagorean pattern: sec⁡2θ−1=tan⁡2θ\sec^2\theta-1=\tan^2\theta. So the expression is tan⁡2θtan⁡θ=tan⁡θ\dfrac{\tan^2\theta}{\tan\theta}=\tan\theta (wherever tan⁡θ≠0\tan\theta\neq0 and is defined).

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.

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.

Try it
Quick check

Write csc⁡θcot⁡θ\dfrac{\csc\theta}{\cot\theta} as a single trigonometric function.

Show the answer

Answer: sec⁡θ\sec\theta. In sines and cosines, 1/sin⁡θcos⁡θ/sin⁡θ=1sin⁡θ⋅sin⁡θcos⁡θ=1cos⁡θ=sec⁡θ\dfrac{1/\sin\theta}{\cos\theta/\sin\theta}=\dfrac{1}{\sin\theta}\cdot\dfrac{\sin\theta}{\cos\theta}=\dfrac{1}{\cos\theta}=\sec\theta.

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
  • 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.

A verification must work in one direction only: a chain LHS=⋯=RHS\text{LHS}=\cdots=\text{RHS} in which every step is a known identity or algebra that is valid for every allowed θ\theta. Treating the identity like an equation (doing the same thing to both sides) assumes the very statement you are proving, and some of those operations, like squaring, can turn a false statement into a true one.

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}
Side note
Where it goes wrong

Operating on both sides. Start from the false claim sin⁡θ=−sin⁡θ\sin\theta=-\sin\theta and square both sides: sin⁡2θ=sin⁡2θ\sin^2\theta=\sin^2\theta, which is true for every θ\theta. Nothing false was ever ruled out; squaring hid the minus sign. That is why a proof must transform one side into the other, never both at once.

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.

Try it
Quick check

Verify (1+tan⁡2θ)cos⁡2θ=1(1+\tan^2\theta)\cos^2\theta=1.

Show the answer

Proof. Start with the left side. By the Pythagorean identity, 1+tan⁡2θ=sec⁡2θ=1cos⁡2θ1+\tan^2\theta=\sec^2\theta=\dfrac{1}{\cos^2\theta}, so (1+tan⁡2θ)cos⁡2θ=1cos⁡2θ⋅cos⁡2θ=1(1+\tan^2\theta)\cos^2\theta=\dfrac{1}{\cos^2\theta}\cdot\cos^2\theta=1, which is the right 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)
Allall six > 0Sinesin, csc > 0Tangenttan, cot > 0Cosinecos, sec > 0
The positive functions in each quadrant; every other function is negative there.

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

The signs come from the coordinates of the unit-circle point. Sine is yy, positive above the xx-axis (Quadrants I, II); cosine is xx, positive right of the yy-axis (I, IV); tangent is yx\tfrac{y}{x}, positive where the two agree in sign (I, III). A reciprocal never changes sign, so cosecant, secant and cotangent follow their partners.

Example
Full worked value problem
(−4, 3)5−43
Quadrant II: sin⁡θ=35\sin\theta=\tfrac35 puts the point at (−4,3)(-4,3) on a circle of radius 55.

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.

Try it
Quick check

Given cos⁡θ=−513\cos\theta=-\dfrac{5}{13} with θ\theta in Quadrant III, find sin⁡θ\sin\theta, tan⁡θ\tan\theta, and csc⁡θ\csc\theta.

Show the answer

Answer: sin⁡θ=−1213\sin\theta=-\dfrac{12}{13}, tan⁡θ=125\tan\theta=\dfrac{12}{5}, csc⁡θ=−1312\csc\theta=-\dfrac{13}{12}. sin⁡2θ=1−25169=144169\sin^2\theta=1-\dfrac{25}{169}=\dfrac{144}{169}, and sine is negative in Quadrant III, so sin⁡θ=−1213\sin\theta=-\tfrac{12}{13}. Then tan⁡θ=−12/13−5/13=125\tan\theta=\dfrac{-12/13}{-5/13}=\dfrac{12}{5}, positive as ASTC predicts for Quadrant III.

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.

Why It Works

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is the Pythagorean theorem for the unit circle: the point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) sits at distance 11 from the origin. Divide by cos⁡2θ\cos^2\theta to get tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta, or by sin⁡2θ\sin^2\theta for 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta. Every other identity in the course is built by algebra from this one fact plus the definitions of the six functions.

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

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.