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Trigonometry Formula List: Ratios, Table and Identities

12 formulasCalculatorUpdated 22 Oct 2026

The first trigonometry formula you learn is sin⁡θ=perpendicularhypotenuse\sin\theta = \dfrac{\text{perpendicular}}{\text{hypotenuse}}, and almost everything else is built from it together with sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. From these two you get the double angle results such as cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x, the triple angle results, and the sum to product rules that class 11 tests lean on.

This page is the trigonometry all formula list a school student needs, grouped the way NCERT teaches them: ratios and the value table for class 10, then allied angles, compound angles, 2x and 3x formulas and sum to product for class 11. Each group has a short worked example. The trig calculator just below the key box gives sin, cos, tan and their reciprocals for any angle, so you can check every answer here yourself.

Trigonometry formulas

Key formulas

  • sin cos tan formula

    sin⁡θ=PH, cos⁡θ=BH, tan⁡θ=PB\sin\theta=\dfrac{P}{H},\ \cos\theta=\dfrac{B}{H},\ \tan\theta=\dfrac{P}{B}

    P perpendicular, B base, H hypotenuse

  • Pythagorean identity

    sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
  • sin(A + B) formula

    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B
  • cos(A + B) formula

    cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B-\sin A\sin B
  • sin 2x formula

    sin⁡2x=2sin⁡xcos⁡x=2tan⁡x1+tan⁡2x\sin 2x=2\sin x\cos x=\dfrac{2\tan x}{1+\tan^2 x}
  • cos 2x formula

    cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x=\cos^2 x-\sin^2 x=2\cos^2 x-1=1-2\sin^2 x

    same as the cos 2 theta formula with x written as θ

  • tan 2x formula

    tan⁡2x=2tan⁡x1−tan⁡2x\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}
  • sin 3x formula

    sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x=3\sin x-4\sin^3 x
  • cos 3x formula

    cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x=4\cos^3 x-3\cos x
  • tan 3x formula

    tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x\tan 3x=\dfrac{3\tan x-\tan^3 x}{1-3\tan^2 x}
  • sin A sin B formula

    2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B=\cos(A-B)-\cos(A+B)
  • Sum to product (sines)

    sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}

Sin cos tan formula: the six trigonometric ratios

Take a right triangle and fix one acute angle θ\theta. The side opposite θ\theta is the perpendicular (P), the side next to it is the base (B), and the longest side, facing the right angle, is the hypotenuse (H). The six ratios are:

sin⁡θ=PH,cos⁡θ=BH,tan⁡θ=PB\sin\theta=\dfrac{P}{H},\quad \cos\theta=\dfrac{B}{H},\quad \tan\theta=\dfrac{P}{B}
cosec⁡θ=HP,sec⁡θ=HB,cot⁡θ=BP\operatorname{cosec}\theta=\dfrac{H}{P},\quad \sec\theta=\dfrac{H}{B},\quad \cot\theta=\dfrac{B}{P}

So the sin theta formula is perpendicular over hypotenuse, the cos theta formula is base over hypotenuse (cos is base upon hypotenuse), and the tan theta formula is perpendicular over base. A memory line many teachers use is "Pandit Badri Prasad, Har Har Bole": the first letters pair up as P/H, B/H, P/B. The other three ratios are just the flips, so cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta = \dfrac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta=\dfrac{1}{\cos\theta} and cot⁡θ=1tan⁡θ\cot\theta=\dfrac{1}{\tan\theta}. You will also see the useful link tan⁡θ=sin⁡θcos⁡θ\tan\theta=\dfrac{\sin\theta}{\cos\theta}.

How do you find the hypotenuse in trigonometry when only two legs are given? Use Pythagoras, H2=P2+B2H^2=P^2+B^2. If you know one side and one angle instead, rearrange a ratio: H=Psin⁡θH = \dfrac{P}{\sin\theta}. The same right triangle idea also gives you the distance formula between two points.

Example 1: all six ratios from one value

Given sin⁡θ=817\sin\theta=\dfrac{8}{17} for an acute angle. Find the other five ratios.

  1. Take P=8P=8 and H=17H=17. Then B=172−82=289−64=225=15B=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15.
  2. cos⁡θ=1517\cos\theta=\dfrac{15}{17}, tan⁡θ=815\tan\theta=\dfrac{8}{15}.
  3. Flip each: cosec⁡θ=178\operatorname{cosec}\theta=\dfrac{17}{8}, sec⁡θ=1715\sec\theta=\dfrac{17}{15}, cot⁡θ=158\cot\theta=\dfrac{15}{8}.

Answer: cos⁡θ=1517, tan⁡θ=815, cosec⁡θ=178, sec⁡θ=1715, cot⁡θ=158\cos\theta=\tfrac{15}{17},\ \tan\theta=\tfrac{8}{15},\ \operatorname{cosec}\theta=\tfrac{17}{8},\ \sec\theta=\tfrac{17}{15},\ \cot\theta=\tfrac{15}{8}.

Trigonometry table formula for 0, 30, 45, 60 and 90 degrees

Board papers expect you to know these values without a calculator. Here is the trigonometry table for the five standard angles.

Values of sin, cos and tan
Anglesincostan
0°001100
30°12\tfrac{1}{2}32\tfrac{\sqrt3}{2}13\tfrac{1}{\sqrt3}
45°12\tfrac{1}{\sqrt2}12\tfrac{1}{\sqrt2}11
60°32\tfrac{\sqrt3}{2}12\tfrac{1}{2}3\sqrt3
90°1100not defined
Values of cosec, sec and cot
Anglecosecseccot
0°not defined11not defined
30°2223\tfrac{2}{\sqrt3}3\sqrt3
45°2\sqrt22\sqrt211
60°23\tfrac{2}{\sqrt3}2213\tfrac{1}{\sqrt3}
90°11not defined00

A quick way to rebuild the sin row in the exam hall: write 0, 1, 2, 3, 4, divide each by 4 and take the square root. You get 0,12,12,32,10, \tfrac12, \tfrac{1}{\sqrt2}, \tfrac{\sqrt3}{2}, 1. Write the same row backwards for cos. Then divide sin by cos for tan, and flip the rows for cosec, sec and cot.

tan⁡90∘\tan 90^\circ is not defined because cos⁡90∘=0\cos 90^\circ = 0 and you cannot divide by zero. Do not write it as infinity in a board answer.

Trigonometric identities

An identity is true for every angle where both sides make sense. Class 10 uses three, all coming from Pythagoras divided through by H2H^2, B2B^2 or P2P^2:

sin⁡2θ+cos⁡2θ=1,1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=cosec⁡2θ\sin^2\theta+\cos^2\theta=1,\qquad 1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\operatorname{cosec}^2\theta

Rewrite them as you need: sec⁡2θ−tan⁡2θ=1\sec^2\theta-\tan^2\theta=1 is often handy because it factors as (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1. The sec 1 formula students ask about is this one, sec⁡2θ=1+tan⁡2θ\sec^2\theta = 1+\tan^2\theta.

Example 2: prove an identity

Show that (1−cos⁡2A)cosec⁡2A=1(1-\cos^2 A)\operatorname{cosec}^2 A = 1.

  1. 1−cos⁡2A=sin⁡2A1-\cos^2 A = \sin^2 A from the first identity.
  2. So the left side is sin⁡2A⋅1sin⁡2A=1\sin^2 A\cdot\dfrac{1}{\sin^2 A}=1.

Answer: the left side reduces to 1, which equals the right side.

Allied angle formulas and quadrant signs

Class 11 extends the ratios to any angle using the unit circle. Angles such as 90∘±θ90^\circ\pm\theta, 180∘±θ180^\circ\pm\theta, 270∘±θ270^\circ\pm\theta and 360∘−θ360^\circ-\theta are called allied angles. Two rules cover all of them:

  • Name rule: with 90° or 270°, sin becomes cos, tan becomes cot, sec becomes cosec (and the reverse). With 180° or 360°, the name stays.
  • Sign rule: the sign is the sign of the original ratio in the quadrant where the angle lands. Only sin and cosec are positive in quadrant II, only tan and cot in quadrant III, only cos and sec in quadrant IV, and all are positive in quadrant I. Remember it as "All Silver Tea Cups".
Common allied angles
Anglesincostan
−θ-\theta−sin⁡θ-\sin\thetacos⁡θ\cos\theta−tan⁡θ-\tan\theta
90∘−θ90^\circ-\thetacos⁡θ\cos\thetasin⁡θ\sin\thetacot⁡θ\cot\theta
90∘+θ90^\circ+\thetacos⁡θ\cos\theta−sin⁡θ-\sin\theta−cot⁡θ-\cot\theta
180∘−θ180^\circ-\thetasin⁡θ\sin\theta−cos⁡θ-\cos\theta−tan⁡θ-\tan\theta
180∘+θ180^\circ+\theta−sin⁡θ-\sin\theta−cos⁡θ-\cos\thetatan⁡θ\tan\theta
360∘−θ360^\circ-\theta−sin⁡θ-\sin\thetacos⁡θ\cos\theta−tan⁡θ-\tan\theta

For example, cos⁡150∘=cos⁡(180∘−30∘)=−cos⁡30∘=−32\cos 150^\circ=\cos(180^\circ-30^\circ)=-\cos 30^\circ=-\tfrac{\sqrt3}{2}. The 90∘−θ90^\circ-\theta row is the one you are most likely to have met before class 11; it is why sin⁡30∘=cos⁡60∘\sin 30^\circ=\cos 60^\circ. In radians, 180∘=π180^\circ=\pi, so these rows also read sin⁡(π−θ)=sin⁡θ\sin(\pi-\theta)=\sin\theta and so on.

Compound angle formulas: sin(A+B), cos(A+B), tan(A+B)

These give the ratio of a sum or difference of two angles. They are the root of every class 11 formula that follows.

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\begin{aligned}\sin(A\pm B)&=\sin A\cos B\pm\cos A\sin B\\ \cos(A\pm B)&=\cos A\cos B\mp\sin A\sin B\\ \tan(A\pm B)&=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\end{aligned}

Watch the cosine line: the sign in the middle flips. The cos(a+b) formula has a minus, while cos⁡(A−B)\cos(A-B) has a plus. The sin a+b formula keeps the same sign on both sides.

Example 3: exact value of sin 75° and tan 15°

  1. sin⁡75∘=sin⁡(45∘+30∘)=12⋅32+12⋅12=3+122\sin 75^\circ=\sin(45^\circ+30^\circ)=\tfrac{1}{\sqrt2}\cdot\tfrac{\sqrt3}{2}+\tfrac{1}{\sqrt2}\cdot\tfrac12=\dfrac{\sqrt3+1}{2\sqrt2}.
  2. tan⁡15∘=tan⁡(45∘−30∘)=1−131+13=3−13+1\tan 15^\circ=\tan(45^\circ-30^\circ)=\dfrac{1-\tfrac{1}{\sqrt3}}{1+\tfrac{1}{\sqrt3}}=\dfrac{\sqrt3-1}{\sqrt3+1}.
  3. Multiply top and bottom by 3−1\sqrt3-1: (3−1)23−1=4−232=2−3\dfrac{(\sqrt3-1)^2}{3-1}=\dfrac{4-2\sqrt3}{2}=2-\sqrt3.

Answer: sin⁡75∘=3+122≈0.9659\sin 75^\circ=\dfrac{\sqrt3+1}{2\sqrt2}\approx 0.9659 and tan⁡15∘=2−3≈0.2679\tan 15^\circ=2-\sqrt3\approx 0.2679.

Double angle trigonometry formula: sin 2x, cos 2x and tan 2x

Put B=AB=A in the compound angle results and you get the double angle set. Searches for the sin2x formula, cos2x formula, sin2a formula, cos 2 theta formula and tan 2 theta formula all point here; only the letter changes.

sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x
cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x
tan⁡2x=2tan⁡x1−tan⁡2x\tan 2x=\dfrac{2\tan x}{1-\tan^2x}

The three forms of cos 2x come from swapping sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x or the reverse. Pick the form that leaves only the ratio you know. In terms of tan alone:

sin⁡2x=2tan⁡x1+tan⁡2x,cos⁡2x=1−tan⁡2x1+tan⁡2x\sin 2x=\dfrac{2\tan x}{1+\tan^2x},\qquad \cos 2x=\dfrac{1-\tan^2x}{1+\tan^2x}

1 - cos 2x and 1 + cos 2x

Rearranging the last two forms of cos 2x gives the power reducing pair, used all the time in integration:

1−cos⁡2x=2sin⁡2x,1+cos⁡2x=2cos⁡2x1-\cos 2x=2\sin^2x,\qquad 1+\cos 2x=2\cos^2x

Replace 2x2x by AA and you get the half angle version, cos⁡A=2cos⁡2A2−1\cos A = 2\cos^2\tfrac{A}{2}-1, which answers the cos a/2 search.

1 + sin x formula

Since sin⁡2x2+cos⁡2x2=1\sin^2\tfrac{x}{2}+\cos^2\tfrac{x}{2}=1 and sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin\tfrac{x}{2}\cos\tfrac{x}{2}, you can write 1+sin⁡x1+\sin x as a perfect square:

1+sin⁡x=(sin⁡x2+cos⁡x2)2,1−sin⁡x=(sin⁡x2−cos⁡x2)21+\sin x=\left(\sin\tfrac{x}{2}+\cos\tfrac{x}{2}\right)^2,\qquad 1-\sin x=\left(\sin\tfrac{x}{2}-\cos\tfrac{x}{2}\right)^2

Example 4: double angle values from sin x = 3/5

x is acute and sin⁡x=35\sin x=\tfrac35. Find sin⁡2x\sin 2x, cos⁡2x\cos 2x and tan⁡2x\tan 2x.

  1. cos⁡x=1−925=45\cos x=\sqrt{1-\tfrac{9}{25}}=\tfrac45, so tan⁡x=34\tan x=\tfrac34.
  2. sin⁡2x=2⋅35⋅45=2425\sin 2x=2\cdot\tfrac35\cdot\tfrac45=\tfrac{24}{25}.
  3. cos⁡2x=1−2⋅925=725\cos 2x=1-2\cdot\tfrac{9}{25}=\tfrac{7}{25}.
  4. tan⁡2x=2⋅341−916=32716=247\tan 2x=\dfrac{2\cdot\tfrac34}{1-\tfrac{9}{16}}=\dfrac{\tfrac32}{\tfrac{7}{16}}=\tfrac{24}{7}. Check: 24/257/25=247\tfrac{24/25}{7/25}=\tfrac{24}{7}.

Answer: sin⁡2x=2425\sin 2x=\tfrac{24}{25}, cos⁡2x=725\cos 2x=\tfrac{7}{25}, tan⁡2x=247\tan 2x=\tfrac{24}{7}.

Triple angle formulas: sin 3x, cos 3x and tan 3x

Write 3x=2x+x3x=2x+x, expand with the compound angle rule, and replace the 2x terms. The results are:

sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x=3\sin x-4\sin^3x
cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x=4\cos^3x-3\cos x
tan⁡3x=3tan⁡x−tan⁡3x1−3tan⁡2x\tan 3x=\dfrac{3\tan x-\tan^3x}{1-3\tan^2x}

Notice the pattern: the sin3x formula starts with the 3 term and subtracts the cube, while the cos3x formula starts with the cube and subtracts the 3 term. A fast self check is x=30∘x=30^\circ: 3⋅12−4⋅18=1=sin⁡90∘3\cdot\tfrac12-4\cdot\tfrac18=1=\sin 90^\circ.

Example 5: use the triple angle result

With sin⁡x=35\sin x=\tfrac35 and cos⁡x=45\cos x=\tfrac45 as in example 4, find sin⁡3x\sin 3x and cos⁡3x\cos 3x.

  1. sin⁡3x=3⋅35−4⋅27125=225125−108125=117125\sin 3x=3\cdot\tfrac35-4\cdot\tfrac{27}{125}=\tfrac{225}{125}-\tfrac{108}{125}=\tfrac{117}{125}.
  2. cos⁡3x=4⋅64125−3⋅45=256125−300125=−44125\cos 3x=4\cdot\tfrac{64}{125}-3\cdot\tfrac45=\tfrac{256}{125}-\tfrac{300}{125}=-\tfrac{44}{125}.

Answer: sin⁡3x=117125\sin 3x=\tfrac{117}{125} and cos⁡3x=−44125\cos 3x=-\tfrac{44}{125}. The negative cosine tells you 3x has crossed into quadrant II.

Product to sum and sum to product formulas

Adding and subtracting the compound angle lines turns a product of two ratios into a sum:

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)\begin{aligned}2\sin A\cos B&=\sin(A+B)+\sin(A-B)\\ 2\cos A\sin B&=\sin(A+B)-\sin(A-B)\\ 2\cos A\cos B&=\cos(A+B)+\cos(A-B)\\ 2\sin A\sin B&=\cos(A-B)-\cos(A+B)\end{aligned}

The last line is the sin a sin b formula. Note the order: cos⁡(A−B)\cos(A-B) comes first. Now put A+B=CA+B=C and A−B=DA-B=D to go the other way:

sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\begin{aligned}\sin C+\sin D&=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2}\\ \sin C-\sin D&=2\cos\tfrac{C+D}{2}\sin\tfrac{C-D}{2}\\ \cos C+\cos D&=2\cos\tfrac{C+D}{2}\cos\tfrac{C-D}{2}\\ \cos C-\cos D&=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2}\end{aligned}

Example 6: sin 75° + sin 15°

  1. Here C=75∘C=75^\circ, D=15∘D=15^\circ, so C+D2=45∘\tfrac{C+D}{2}=45^\circ and C−D2=30∘\tfrac{C-D}{2}=30^\circ.
  2. sin⁡75∘+sin⁡15∘=2sin⁡45∘cos⁡30∘=2⋅12⋅32=32\sin 75^\circ+\sin 15^\circ=2\sin 45^\circ\cos 30^\circ=2\cdot\tfrac{1}{\sqrt2}\cdot\tfrac{\sqrt3}{2}=\tfrac{\sqrt3}{\sqrt2}.

Answer: 62≈1.2247\dfrac{\sqrt6}{2}\approx 1.2247.

Inverse trigonometry formula basics for class 12

Class 12 chapter 2, Inverse Trigonometric Functions, asks the reverse question: which angle has a given sine? To make the answer unique, each inverse function has a principal value range, for example sin⁡−1x∈[−π2,π2]\sin^{-1}x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right], cos⁡−1x∈[0,π]\cos^{-1}x\in[0,\pi] and tan⁡−1x∈(−π2,π2)\tan^{-1}x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right). The results students use most are:

sin⁡−1x+cos⁡−1x=π2 (−1≤x≤1),tan⁡−1x+cot⁡−1x=π2,sin⁡−1(−x)=−sin⁡−1x\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}\ (-1\le x\le 1),\qquad \tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2},\qquad \sin^{-1}(-x)=-\sin^{-1}x

Keep in mind that sin⁡−1x\sin^{-1}x is not 1sin⁡x\dfrac{1}{\sin x}; that reciprocal is cosec⁡x\operatorname{cosec} x. The derivatives of these functions are listed on the differentiation formulas page, and they turn up again as answers on the integration formulas page.

Trigonometry formula class 10 vs class 11: what is in your syllabus

If you only need the class 10 trigonometry formula list, you can stop after the identities section. Here is how the topic is split across NCERT books.

ClassChapterFormulas you need
Class 10Introduction to Trigonometry, Some Applications of TrigonometrySix ratios, value table for 0° to 90°, three identities, heights and distances
Class 11Chapter 3, Trigonometric FunctionsRadian measure, signs in quadrants, allied angles, compound, double and triple angle, sum to product
Class 12Inverse Trigonometric FunctionsPrincipal values and the inverse function results

Karnataka 1st PUC students can use the class 11 list. BSc first year maths and engineering entrance papers assume all of it, plus the radian form of every result. In class 10 the heights and distances chapter mostly uses tan, since you usually know a horizontal distance and want a height.

One radian conversion worth keeping handy: π rad=180∘\pi\text{ rad}=180^\circ, so 1∘=π1801^\circ=\tfrac{\pi}{180} rad. Angles in class 11 problems are often given as π6\tfrac{\pi}{6}, π4\tfrac{\pi}{4}, π3\tfrac{\pi}{3}, which are simply 30°, 45° and 60°.

Common mistakes with trig formulas

  • Writing sin⁡(A+B)=sin⁡A+sin⁡B\sin(A+B)=\sin A+\sin B. Try A = B = 45° and you get sin⁡90∘=1\sin 90^\circ=1 on the left and 2\sqrt2 on the right, so it is wrong.
  • Mixing up sin⁡2θ\sin^2\theta and sin⁡θ2\sin\theta^2. The first squares the value of sine; the second squares the angle.
  • Picking the wrong form of cos 2x. If a question gives cos x, use 2cos⁡2x−12\cos^2x-1; if it gives sin x, use 1−2sin⁡2x1-2\sin^2x.
  • Forgetting the sign in quadrants II to IV for allied angles. Draw a quick quadrant sketch before you write the answer.
  • Using degrees in a formula that needs radians, such as arc length l=rθl=r\theta.
  • Writing tan⁡90∘\tan 90^\circ as a number. It has no value.

For the algebra steps that keep coming up inside trig proofs, like a2−b2=(a−b)(a+b)a^2-b^2=(a-b)(a+b), see algebra identities. The complete list across topics sits on the maths formulas page, and the formula sheet PDF of this page's key box is available to download below.

Download the PDF

Questions students ask

What is the formula of cos 2x?

There are three equivalent forms: cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x. In terms of tan it is 1−tan⁡2x1+tan⁡2x\dfrac{1-\tan^2x}{1+\tan^2x}. Choose the form that uses the ratio given in the question.

What is the formula of sin 2x?
sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x. If only tan x is known, use sin⁡2x=2tan⁡x1+tan⁡2x\sin 2x=\dfrac{2\tan x}{1+\tan^2x}. The sin 2a and sin 2 theta formulas are the same result with a different letter.
What is the formula of sin 3x and cos 3x?
sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x=3\sin x-4\sin^3x and cos⁡3x=4cos⁡3x−3cos⁡x\cos 3x=4\cos^3x-3\cos x. Both come from writing 3x as 2x + x and applying the compound angle results.
Which trigonometry formulas are in class 10?

Class 10 covers the six ratios, the values at 0°, 30°, 45°, 60° and 90°, the three identities such as sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1, and their use in heights and distances. Compound and double angle formulas start in class 11.

Is cos the base over hypotenuse?

Yes. cos⁡θ=basehypotenuse\cos\theta=\dfrac{\text{base}}{\text{hypotenuse}}, where the base is the side next to the angle (not the hypotenuse). Sine is perpendicular over hypotenuse and tan is perpendicular over base.

What is 1 minus cos 2x equal to?
1−cos⁡2x=2sin⁡2x1-\cos 2x=2\sin^2x, and similarly 1+cos⁡2x=2cos⁡2x1+\cos 2x=2\cos^2x. Both follow straight from the forms of cos 2x and are used to integrate sin⁡2x\sin^2x and cos⁡2x\cos^2x.
Is it formula or formulas?

Both are correct plurals of formula: "formulas" is common in everyday and Indian school English, while "formulae" is the older form you see in some textbooks. Use either, but stay consistent in one answer.