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Trigonometry · TOPIC LESSON

Trigonometric Identities

Transform with purpose, solve with complete periods: identities and equations with domain control

JEE Main · JEE Advanced100 min study guide13 indexed sections

BY THE END, YOU CAN

  • Derive and select Pythagorean, addition, double-angle and product-sum identities.
  • Track denominator restrictions and quadrant signs during simplification.
  • Solve trigonometric equations completely over ℝ or a specified interval.
  • Use amplitude forms, half-angle substitutions and interval reasoning without losing or inventing solutions.

BEFORE YOU START

  • Unit-circle sine and cosine, radians and exact special-angle values.
  • Factoring, quadratic equations and interval notation.
  • Principal inverse trigonometric ranges.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. An identity is an equality on its common domain

A trigonometric identity holds for every input where both displayed sides are defined. An equation to be solved holds only for selected inputs. Proving an identity and solving an equation therefore have different goals, even when they use the same algebraic transformations.

The unit circle gives sin²x+cos²x=1. Dividing by cos²x gives 1+tan²x=sec²x where cos x≠0. Dividing by sin²x gives 1+cot²x=csc²x where sin x≠0. The derived identities inherit these denominator restrictions.

Reciprocal and quotient definitions are tan x=sin x/cos x, cot x=cos x/sin x, sec x=1/cos x and csc x=1/sin x on their respective domains. Rewriting in sine and cosine often reveals a common denominator, but it does not make excluded inputs valid.

Unless a degree symbol is shown explicitly, angles here are in radians. A full sine or cosine period is 2π, and a tangent or cotangent period is π. Mixing degrees and radians in one calculation changes the actual angle rather than merely its notation.

02 / UNDERSTAND & APPLY

2. Reduce angles using geometry, not sign mnemonics alone

Sine is odd and cosine is even: sin(−x)=−sin x and cos(−x)=cos x. Tangent is odd on its domain. Unit-circle reflection gives these signs directly, and a half-turn gives sin(x+π)=−sin x and cos(x+π)=−cos x.

Cofunction identities include sin(π/2−x)=cos x and cos(π/2−x)=sin x. Combining them with periodicity and symmetry reduces angles to familiar intervals while preserving signs. For example, sin(π+x)=−sin x and cos(π−x)=−cos x.

A period T>0 satisfies f(x+T)=f(x) with the domain preserved. A common period of two functions is a period of their sum or product, but it need not be the least positive period after simplification. The expression sin²x+cos²x is constant and has no least positive period.

For a nonconstant sine or cosine with argument kx+b, k≠0, the fundamental period is 2π/|k|. Squared sine or cosine has period π/|k|. Tangent has period π/|k|, with its excluded poles shifting accordingly. For an equation, use a period of the entire equation rather than one term alone.

ORIGINAL WORKED EXAMPLE

Reduce a large angle and identify a squared-function period

Evaluate sin(17π/6), and find the fundamental period of sin²(3x−π/4).

  1. Subtract one full period: 17π/6−2π=5π/6. Its sine is 1/2.
  2. For the squared function, sin²u=(1−cos2u)/2, so sin²(3x−π/4)=(1−cos(6x−π/2))/2.
  3. The nonconstant cosine has angular coefficient 6, giving fundamental period 2π/6=π/3.
  4. The horizontal phase shift changes the locations of peaks but not the period.

Answer sin(17π/6)=1/2; fundamental period π/3.

03 / UNDERSTAND & APPLY

3. Addition formulas describe rotations

A rotation through an angle u sends a unit vector at angle v to one at angle u+v. Multiplying the rotation matrix [[cos u,−sin u],[sin u,cos u]] by (cos v,sin v) gives cos(u+v)=cos u cos v−sin u sin v and sin(u+v)=sin u cos v+cos u sin v.

Replacing v by −v and using parity produces the subtraction formulas. The sine formula keeps the sign of the angle operation, while cosine changes it in the sine-product term. Deriving the pair from rotation is safer than memorizing an isolated sign pattern.

The tangent addition formula is tan(u+v)=(tan u+tan v)/(1−tan u tan v) when all displayed tangents and the denominator are defined. It follows by dividing the sine and cosine addition formulas. Cases with a zero denominator may describe a vertical direction, not an ordinary finite tangent value.

Addition formulas can evaluate exact nonstandard angles, combine oscillations or expose factors in equations. They do not imply sin(u+v)=sin u+sin v. The products involving both angles are essential.

ORIGINAL WORKED EXAMPLE

An exact angle from two familiar angles

Find sin(π/12) and cos(π/12).

  1. Use π/12=π/4−π/6.
  2. The sine subtraction formula gives sin(π/12)=(√2/2)(√3/2)−(√2/2)(1/2)=(√6−√2)/4.
  3. The cosine subtraction formula gives cos(π/12)=(√2/2)(√3/2)+(√2/2)(1/2)=(√6+√2)/4.
  4. Both are positive because π/12 is in the first quadrant, and their squares add to 1.

Answer sin(π/12)=(√6−√2)/4; cos(π/12)=(√6+√2)/4.

04 / UNDERSTAND & APPLY

4. Match the identity to the desired variable

Setting the two angles equal gives sin2x=2sin x cos x and cos2x=cos²x−sin²x=1−2sin²x=2cos²x−1. The three cosine forms are interchangeable, but one may reduce an equation to a single trigonometric function more efficiently.

Power reduction gives sin²x=(1−cos2x)/2 and cos²x=(1+cos2x)/2. Half-angle square roots therefore involve signs: sin(x/2)=±√[(1−cos x)/2] and cos(x/2)=±√[(1+cos x)/2], with the sign determined by the quadrant of x/2.

The identity tan(x/2)=sin x/(1+cos x) holds where 1+cos x≠0, exactly the excluded odd multiples of π for the half-angle tangent. The alternative (1−cos x)/sin x needs sin x≠0 and thus has additional exclusions where the first form may still work.

Higher powers often simplify through the Pythagorean identity before any angle expansion. For instance, sin⁴x+cos⁴x=(sin²x+cos²x)²−2sin²x cos²x=1−(1/2)sin²2x, immediately showing its range [1/2,1].

ORIGINAL WORKED EXAMPLE

A half-angle magnitude still needs a quadrant sign

Given cos x=−3/5 and π<x<3π/2, find sin(x/2), cos(x/2) and sin2x.

  1. The angle x is in quadrant III, so sin x=−4/5. The half-angle lies in (π/2,3π/4), quadrant II.
  2. Thus sin(x/2)=+√[(1+3/5)/2]=2/√5.
  3. Cos(x/2) is negative: −√[(1−3/5)/2]=−1/√5.
  4. The double-angle value is sin2x=2(−4/5)(−3/5)=24/25. The signs come from the specified interval, not from taking every square root positive.

Answer sin(x/2)=2/√5; cos(x/2)=−1/√5; sin2x=24/25.

05 / UNDERSTAND & APPLY

5. Convert sums to factors or to one shifted wave

Adding the sine addition and subtraction formulas gives sin u+sin v=2sin((u+v)/2)cos((u−v)/2). Subtracting gives sin u−sin v=2cos((u+v)/2)sin((u−v)/2). Analogous formulas convert cosine sums and differences into products.

Products can also be converted to sums: 2sin u cos v=sin(u+v)+sin(u−v), 2cos u cos v=cos(u+v)+cos(u−v), and 2sin u sin v=cos(u−v)−cos(u+v). These forms are especially useful when frequencies differ.

For real A,B not both zero, set R=√(A²+B²). Choose φ with cos φ=A/R and sin φ=B/R. Then A sin x+B cos x=R sin(x+φ), so its exact range is [−R,R]. Using only tan φ=B/A without quadrant information can choose the wrong phase.

The amplitude form turns a linear combination into one standard equation and provides an immediate feasibility test. If A=B=0, the expression is identically zero and must be treated separately; dividing by R would then be invalid.

ORIGINAL WORKED EXAMPLE

An amplitude rewrite gives both solutions and a range bound

Solve sin x+√3 cos x=1 on 0≤x<2π.

  1. The amplitude is 2. Since cos(π/3)=1/2 and sin(π/3)=√3/2, the left side is 2sin(x+π/3).
  2. The equation becomes sin(x+π/3)=1/2. Thus x+π/3=π/6+2kπ or 5π/6+2kπ.
  3. The general families are x=−π/6+2kπ and x=π/2+2kπ.
  4. In the requested half-open interval, the values are 11π/6 and π/2. The amplitude bound also shows that replacing the right side by any number outside [−2,2] would give no real solution.

Answer x=π/2 or 11π/6 on [0,2π).

06 / UNDERSTAND & APPLY

6. Prove a common-domain identity without assuming the result

A reliable identity proof starts from one side and transforms it into the other using equalities valid on the stated domain. Alternatively transform both sides independently into the same expression. Starting by assuming the desired equality and deriving a true statement is not automatically a proof unless every step is reversible.

Common strategies are converting to sine and cosine, factoring with 1−sin²x or 1−cos²x, multiplying by a conjugate and finding a common denominator. Before canceling a factor, note where it vanishes. A simplified expression may extend beyond the original domain, but the original identity does not automatically gain those inputs.

One numerical counterexample disproves a claimed identity. Many matching numerical values do not prove it. Symbolic derivations or a complete structural argument are needed to establish the equality for all allowed x.

ORIGINAL WORKED EXAMPLE

A simplified formula has a larger natural domain

Show that (1−cos x)/sin x=sin x/(1+cos x) on the common domain and identify the exclusions.

  1. The original left side requires sin x≠0, excluding every integer multiple of π.
  2. Multiply its numerator and denominator by 1+cos x, which is nonzero on that original domain.
  3. The numerator becomes 1−cos²x=sin²x, so cancellation gives sin x/(1+cos x).
  4. The right side alone is also defined at even multiples of π, where it equals zero, but the original left side is undefined there. The equality as originally written remains restricted to x∉πℤ.

Answer The identity holds for x not an integer multiple of π; simplification does not restore the excluded inputs.

07 / UNDERSTAND & APPLY

7. General solutions require every periodic family

For |c|≤1, let α=arcsin c in [−π/2,π/2]. Then sin x=c has solutions x=α+2kπ or x=π−α+2kπ, k∈ℤ. At c=±1 the two families coincide, so do not double-count them.

For |c|≤1, let β=arccos c in [0,π]. Then cos x=c has solutions x=±β+2kπ. For any real c, tan x=c gives x=arctan c+kπ. The tangent domain excludes its poles automatically, but any additional original denominator restrictions still apply.

For a multiple-angle equation, solve for the entire angle first, including its period, then divide. For sin(3x)=1/2, there are two angle families and division by 3 also divides their period. Solving only 3x=π/6 would omit infinitely many solutions.

If an interval is specified, enumerate the general families within it, check whether endpoints are included and remove duplicates. A half-open interval such as [0,2π) avoids counting the same periodic endpoint twice. A closed interval includes both endpoints if both satisfy the equation.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Sine equation: two families in each periodThe graph y=sin x and level y=1/2 on the closed interval [0,2π] meet exactly at π/6 and 5π/6. Neither endpoint is a solution. Extending by period 2π gives the two complete families π/6+2kπ and 5π/6+2kπ. Quantitative axes show one complete period.Two sine families, then periodic repetitionxy1/2π/65π/6π2πsin x = 1/2 on [0, 2π]: exactly π/6 and 5π/6.On ℝ: x = π/6 + 2kπ or 5π/6 + 2kπ, k ∈ ℤ. Both families are needed.
A principal inverse-trigonometric value is only one part of the general solution.

08 / UNDERSTAND & APPLY

8. Factor before dividing by a trigonometric expression

An equation such as sin2x=sin x becomes sin x(2cos x−1)=0. Solve each factor separately. Dividing by sin x would lose every solution with sin x=0, which is a whole legitimate family.

If an equation is polynomial in sin x or cos x, set u equal to that function and solve the polynomial, then retain only u∈[−1,1]. A polynomial root outside that interval does not correspond to a real angle. For a polynomial in tan x, real u values are allowed but original tangent poles remain excluded.

Identities can bring an equation to one variable. For example, replace cos2x with 1−2sin²x when the other terms involve sin x. Expanding every angle indiscriminately can create more variables and hide a simple factorization.

When an equation has a product equal to zero, each factor must be defined before the zero-product rule is used. An undefined tangent factor multiplied by a zero sine is still undefined, not automatically zero.

ORIGINAL WORKED EXAMPLE

Factoring preserves the zero-sine family

Solve sin2x=sin x on 0≤x<2π.

  1. Use sin2x=2sin x cos x, then bring all terms to one side: sin x(2cos x−1)=0.
  2. The first factor gives sin x=0, so x=0 or π in the interval.
  3. The second gives cos x=1/2, so x=π/3 or 5π/3.
  4. All four values satisfy the original equation. Dividing by sin x at the start would have discarded 0 and π.

Answer x=0,π/3,π,5π/3.

ORIGINAL WORKED EXAMPLE

A quadratic root may not be a possible sine value

Solve 2sin²x−5sin x+2=0 over ℝ.

  1. Set u=sin x. The quadratic factors as (2u−1)(u−2)=0.
  2. The root u=2 is impossible for real sine. Retain only sin x=1/2.
  3. Therefore x=π/6+2kπ or x=5π/6+2kπ, k∈ℤ.
  4. Both families lie in the original domain because the equation contains only sine and polynomial operations.

Answer x=π/6+2kπ or 5π/6+2kπ, k∈ℤ.

09 / UNDERSTAND & APPLY

9. Necessary transformations need a final original check

Squaring preserves equality but can lose sign information. If sin x+cos x=1, squaring gives 1+sin2x=1, hence sin2x=0. The squared equation also accepts inputs where the original sum is −1. Recheck every candidate in the original equation or retain the sign condition throughout.

For rational expressions in sine and cosine, t=tan(x/2) gives sin x=2t/(1+t²) and cos x=(1−t²)/(1+t²). This converts many equations to rational algebra, but excludes odd multiples of π where t is undefined. Check those inputs separately in the original equation.

After solving for t, recover x=2arctan t+2kπ, not merely x=2arctan t if all real solutions are requested. Any denominators introduced or cleared during the rational calculation must also be checked.

For equations involving tan and cot, a sine-cosine rewrite often exposes their common domain and makes the algebra shorter. Multiplying by sin x cos x is reversible only after confirming it is nonzero on that original domain.

ORIGINAL WORKED EXAMPLE

Squaring creates two false candidates in one period

Solve sin x+cos x=1 on 0≤x<2π.

  1. Squaring gives 1+sin2x=1, so sin2x=0 and candidates are x=0,π/2,π,3π/2.
  2. At 0 and π/2 the original sum is 1, so both are valid.
  3. At π and 3π/2 the original sum is −1, so both must be rejected.
  4. An alternative amplitude form √2sin(x+π/4)=1 yields the same two solutions without introducing the negative-sum candidates.

Answer x=0 or π/2.

ORIGINAL WORKED EXAMPLE

A denominator restriction is part of the solution

Solve tan x+cot x=2 over ℝ.

  1. The original domain requires sin x≠0 and cos x≠0. Rewrite the sum as (sin²x+cos²x)/(sin x cos x)=1/(sin x cos x).
  2. Thus sin x cos x=1/2, or sin2x=1.
  3. The solutions are 2x=π/2+2kπ, giving x=π/4+kπ.
  4. At these angles sine and cosine are both nonzero and have the same sign; tan x=cot x=1, verifying the original sum.

Answer x=π/4+kπ, k∈ℤ.

10 / UNDERSTAND & APPLY

10. Solve on one period, then repeat with the correct boundaries

Trigonometric inequalities are interval problems. Locate equality points and domain exclusions on one full period, determine the sign between consecutive boundaries and then repeat. Isolated sample values identify a sign only after continuity on that interval is established.

For sin x≥1/2 on [0,2π), the solution is [π/6,5π/6]. For tan x>1, the solution is (π/4+kπ,π/2+kπ), k∈ℤ. The tangent pole is always excluded, even if an algebraic manipulation suggests a closed endpoint.

An amplitude equation A sin x+B cos x=c with R=√(A²+B²)>0 has no real solutions if |c|>R. In one full half-open period of the resulting nonconstant wave, it has one solution if |c|=R and two if |c|<R. If a nonunit frequency is present, count its periods within the requested interval.

The case A=B=0 is separate: 0=c gives every real x when c=0 and none otherwise. Parameter values that collapse a trigonometric expression to a constant can change both its period and its root count.

11 / UNDERSTAND & APPLY

11. Original practice: identities and exact values

Give the domain whenever a denominator or tangent appears. Determine half-angle signs from the actual angle interval.

ORIGINAL PRACTICE · 1

Simplify 1−sin²x.

Reveal answer & reasoning

cos²x for all real x.

This is a direct rearrangement of the unit-circle identity.

ORIGINAL PRACTICE · 2

Simplify (sec²x−1)/tan²x on its original domain.

Reveal answer & reasoning

1, where cos x≠0 and sin x≠0.

The numerator is tan²x, but cancellation requires that denominator to be nonzero.

ORIGINAL PRACTICE · 3

Evaluate cos(5π/12).

Reveal answer & reasoning

(√6−√2)/4.

Use cos(π/4+π/6)=cosπ/4 cosπ/6−sinπ/4 sinπ/6.

ORIGINAL PRACTICE · 4

Convert sin5x+sin x to a product.

Reveal answer & reasoning

2sin3x cos2x.

Use the half-sum 3x and half-difference 2x in the sum-to-product identity.

ORIGINAL PRACTICE · 5

Convert 2sin3x cos x to a sum.

Reveal answer & reasoning

sin4x+sin2x.

Apply 2sin u cos v=sin(u+v)+sin(u−v).

ORIGINAL PRACTICE · 6

Find the range of 3sin x+4cos x.

Reveal answer & reasoning

[−5,5].

Its amplitude is √(9+16)=5, and both endpoints are attained by an appropriate phase.

ORIGINAL PRACTICE · 7

Find the range of sin⁴x+cos⁴x.

Reveal answer & reasoning

[1/2,1].

It equals 1−(1/2)sin²2x, and sin²2x spans [0,1].

ORIGINAL PRACTICE · 8

If 2π<x<3π and cos x=1/2, choose the sign of cos(x/2).

Reveal answer & reasoning

Negative.

The half-angle lies in (π,3π/2), quadrant III. Thus cos(x/2)=−√[(1+1/2)/2]=−√3/2.

12 / UNDERSTAND & APPLY

12. Original practice: complete solution sets

All unrestricted solution families use k∈ℤ. For interval answers, respect the exact inclusion of endpoints.

ORIGINAL PRACTICE · 1

Solve cos x=−1/2 on [0,2π).

Reveal answer & reasoning

x=2π/3 or 4π/3.

Cosine is negative in quadrants II and III, with reference angle π/3.

ORIGINAL PRACTICE · 2

Solve tan2x=1 over ℝ.

Reveal answer & reasoning

x=π/8+kπ/2.

First 2x=π/4+kπ, then divide the entire equation by 2.

ORIGINAL PRACTICE · 3

Solve sin3x=0 on [0,2π).

Reveal answer & reasoning

x=0,π/3,2π/3,π,4π/3,5π/3.

The general solution is x=kπ/3. The interval allows k=0,…,5 and excludes k=6.

ORIGINAL PRACTICE · 4

Solve sin x cos x=0 over ℝ.

Reveal answer & reasoning

x=kπ/2.

Either sine is zero at kπ or cosine is zero at π/2+kπ; their union is all integer multiples of π/2.

ORIGINAL PRACTICE · 5

Solve sin x+sin3x=0 over ℝ.

Reveal answer & reasoning

x=kπ/2.

The sum is 2sin2x cos x. Zeros of cos x are already among the zeros of sin2x, so remove duplicate families.

ORIGINAL PRACTICE · 6

Solve sin x≥0 on [−π,π].

Reveal answer & reasoning

{−π}∪[0,π].

Sine is negative on (−π,0), zero at −π,0,π and nonnegative on [0,π]. The isolated included endpoint −π must be retained.

ORIGINAL PRACTICE · 7

For which real c does sin x+√3cos x=c have a real solution?

Reveal answer & reasoning

−2≤c≤2.

The amplitude is 2, so the expression attains exactly the interval [−2,2].

ORIGINAL PRACTICE · 8

Solve sin x=1−cos x on [0,2π).

Reveal answer & reasoning

x=0 or π/2.

Rearrange to sin x+cos x=1 and check the two solutions. A half-angle substitution also gives t=0 or 1; the omitted x=π is not a solution.

13 / UNDERSTAND & APPLY

13. Every transformation has a job and a domain

Use Pythagorean identities to reduce powers, addition formulas to combine angles, product-sum formulas to expose factors and amplitude forms to isolate one wave. An identity is useful when it simplifies the structure of the particular problem, not merely because it is available.

For equations, keep all periodic families and check the requested interval. Factor before dividing, reject polynomial sine/cosine values outside [−1,1], and test candidates after squaring. A half-angle substitution has an omitted angle class that must be checked separately.

For inequalities, split at equality points and poles, and preserve strict versus non-strict endpoints. For parameter problems, include the values that make an amplitude zero or a frequency vanish instead of applying a nonconstant formula blindly.

The final audit asks three questions: is every reported input in the original domain, does it satisfy the original equation, and has every possible periodic or factored branch been considered? Those checks make a short trigonometric answer complete.

KEEP THE BIG IDEAS

Quick revision

  • Identities hold on their common domains; simplification does not restore excluded inputs.
  • Addition formulas arise from rotations and generate many other identities.
  • Amplitude gives exact range and a direct equation strategy.
  • Complete solutions require every periodic family and original-domain checks.
  • Factoring, squaring and half-angle substitutions each need their own lost-or-extra-solution audit.
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