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NDA Maths · Properties of Triangle

Triangle Identities — A+B+C = π, Half & Double Angle

Because the three angles of a triangle add to π, the usual trig identities collapse into special triangle forms — sin(B+C) becomes sin A, and tan A + tan B + tan C becomes their product.

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Why this matters

14 PYQs, 6 HARD. Every identity here is a consequence of A + B + C = π. Knowing the handful of derived forms — the half-angle complements, the tan-product identity, and the cos 2A sum — turns intimidating expressions into one-line simplifications.

Test yourself — a quick 18-question Properties of Triangle recall quiz, instant score.

Concept 1 of 4: Consequences of A + B + C = π

Since any one angle is π minus the other two, every sine/cosine of a sum of two angles rewrites in terms of the third — and the half-angle of two angles becomes the complementary half-angle of the third.

Definition

From A+B+C=πA + B + C = \pi:

  • sin⁡(B+C)=sin⁡A\sin(B+C) = \sin A, cos⁡(B+C)=−cos⁡A\cos(B+C) = -\cos A.
  • B+C2=π2−A2\dfrac{B+C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}, so sin⁡B+C2=cos⁡A2\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2} and tan⁡B+C2=cot⁡A2\tan\dfrac{B+C}{2} = \cot\dfrac{A}{2}.
  • Equations like sin⁡A=cos⁡B+cos⁡C\sin A = \cos B + \cos C are solved by replacing sin⁡A=sin⁡(B+C)\sin A = \sin(B+C) and using sum-to-product.

Half-angle complement

sin⁡B+C2=cos⁡A2,sin⁡(B+C)=sin⁡A\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2}, \qquad \sin(B+C) = \sin A

Worked example

In △ABC\triangle ABC, simplify cos⁡A+B2\cos\dfrac{A+B}{2}.

The same idea in a real exam question:

NDA · Apr 2017 · Q49Moderate

Example 1 · Properties of Triangle · Triangle Identities — A+B+C=π, Half-Angle, and Double-Angle

Consider the following for triangle ABCABC: 1. sin⁡ ⁣(B+C2)=cos⁡ ⁣(A2)\sin\!\left(\dfrac{B+C}{2}\right) = \cos\!\left(\dfrac{A}{2}\right) 2. tan⁡ ⁣(B+C2)=cot⁡ ⁣(A2)\tan\!\left(\dfrac{B+C}{2}\right) = \cot\!\left(\dfrac{A}{2}\right) 3. sin⁡(B+C)=cos⁡A\sin(B+C) = \cos A 4. tan⁡(B+C)=−cot⁡A\tan(B+C) = -\cot A Which of the above are correct?

sin⁡(B+C)=+sin⁡A\sin(B+C) = +\sin A, but cos⁡(B+C)=−cos⁡A\cos(B+C) = -\cos A

Both B+CB+C and AA sum to π\pi, so sin⁡(B+C)=sin⁡A\sin(B+C) = \sin A (same sign) but cos⁡(B+C)=−cos⁡A\cos(B+C) = -\cos A (opposite sign). Forgetting the minus on the cosine is the standard error.
Drill 3 more on consequences of a + b + c = π

Concept 2 of 4: The tan-Sum = tan-Product Identity

In a triangle the sum of the three tangents equals their product — a surprising identity that turns a sum into a product (and vice versa). The cotangents satisfy a companion identity.

Definition

For A+B+C=πA + B + C = \pi:

  • tan⁡A+tan⁡B+tan⁡C=tan⁡A tan⁡B tan⁡C\tan A + \tan B + \tan C = \tan A\,\tan B\,\tan C.
  • cot⁡Acot⁡B+cot⁡Bcot⁡C+cot⁡Ccot⁡A=1\cot A\cot B + \cot B\cot C + \cot C\cot A = 1.
  • tan⁡A2tan⁡B2+tan⁡B2tan⁡C2+tan⁡C2tan⁡A2=1\tan\dfrac{A}{2}\tan\dfrac{B}{2} + \tan\dfrac{B}{2}\tan\dfrac{C}{2} + \tan\dfrac{C}{2}\tan\dfrac{A}{2} = 1.

Sign reading: cot⁡Acot⁡Bcot⁡C>0\cot A\cot B\cot C > 0 forces all angles acute (acute triangle).

tan sum = tan product

tan⁡A+tan⁡B+tan⁡C=tan⁡A tan⁡B tan⁡C\tan A + \tan B + \tan C = \tan A\,\tan B\,\tan C

Worked example

In △ABC\triangle ABC, tan⁡A+tan⁡B+tan⁡C=6\tan A + \tan B + \tan C = 6. Find cot⁡Acot⁡Bcot⁡C\cot A\cot B\cot C.

The same idea in a real exam question:

NDA · Sep 2024 · Q28Moderate

Example 2 · Properties of Triangle · Triangle Identities — A+B+C=π, Half-Angle, and Double-Angle

In a triangle ABCABC, tan⁡A+tan⁡B+tan⁡C=k\tan A+\tan B+\tan C=k. What is the value of cot⁡Acot⁡Bcot⁡C\cot A\cot B\cot C?

The sum equals the PRODUCT, only in a triangle

tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C\tan A + \tan B + \tan C = \tan A\tan B\tan C holds because A+B+C=πA+B+C=\pi. It is NOT a general identity — don't apply it unless the three angles are a triangle's angles.
Drill 2 more on the tan-sum = tan-product identity

Concept 3 of 4: cos 2A Sums & Right-Angle Detection

Sums of cos 2A or sin²A over the three angles collapse to clean values that signal the triangle's type — in particular, a specific value of these sums means one angle is exactly 90°.

Definition

Standard triangle identities:

  • cos⁡2A+cos⁡2B+cos⁡2C=−1−4cos⁡Acos⁡Bcos⁡C\cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A\cos B\cos C.
  • sin⁡2A+sin⁡2B+sin⁡2C=2+2cos⁡Acos⁡Bcos⁡C\sin^2 A + \sin^2 B + \sin^2 C = 2 + 2\cos A\cos B\cos C; it equals 2 iff the triangle is right-angled (one cosine is 0).
  • Equivalently cos⁡2A+cos⁡2B+cos⁡2C=1\cos^2 A + \cos^2 B + \cos^2 C = 1 iff right-angled. A value of cos⁡2A+cos⁡2B+cos⁡2C=−1\cos 2A + \cos 2B + \cos 2C = -1 forces cos⁡Acos⁡Bcos⁡C=0\cos A\cos B\cos C = 0.
  • The plain double-angle forms these are built from — the ones the PYQ solutions actually use one angle at a time: cos⁡2A=1−2sin⁡2A=2cos⁡2A−1=cos⁡2A−sin⁡2A\cos2A=1-2\sin^2A=2\cos^2A-1=\cos^2A-\sin^2A, sin⁡2A=2sin⁡Acos⁡A\sin2A=2\sin A\cos A, sin⁡2A=1−cos⁡2A2\sin^2A=\tfrac{1-\cos2A}{2}. "If cos⁡A=35\cos A=\tfrac35, find cos⁡2A\cos2A" is 2⋅925−1=−7252\cdot\tfrac9{25}-1=-\tfrac7{25}, no triangle identity needed.

Right-angle signature

sin⁡2A+sin⁡2B+sin⁡2C=2  ⟺  right-angled\sin^2 A + \sin^2 B + \sin^2 C = 2 \iff \text{right-angled}

Worked example

If △ABC\triangle ABC has sin⁡2A+sin⁡2B+sin⁡2C=2\sin^2 A + \sin^2 B + \sin^2 C = 2, what can you conclude?

The same idea in a real exam question:

NDA · Sep 2021 · Q41Hard

Example 3 · Properties of Triangle · Triangle Identities — A+B+C=π, Half-Angle, and Double-Angle

Let ABCABC be a triangle. If cos⁡2A+cos⁡2B+cos⁡2C=−1\cos 2A+\cos 2B+\cos 2C=-1, then which one of the following is correct?
Drill 1 more on cos 2a sums & right-angle detection

Concept 4 of 4: Half-Angle Formulas & Sum-to-Product

The half-angle formulas express tan(A/2) through the sides and the inradius; sum-to-product turns cos A + cos B (and similar) into a product that simplifies using A + B + C = π. Both are workhorses for the harder identity questions.

Definition

  • Half-angle (sides): tan⁡A2=rs−a=(s−b)(s−c)s(s−a)\tan\dfrac{A}{2} = \dfrac{r}{s-a} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}, and sin⁡A2=(s−b)(s−c)bc\sin\dfrac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}}.
  • Sum-to-product: cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}; in a triangle cos⁡A+B2=sin⁡C2\cos\dfrac{A+B}{2} = \sin\dfrac{C}{2}.
  • Product-to-sum: sin⁡Xsin⁡Y=12[cos⁡(X−Y)−cos⁡(X+Y)]\sin X\sin Y = \tfrac12[\cos(X-Y) - \cos(X+Y)], useful for sin⁡A2sin⁡3A2\sin\tfrac{A}{2}\sin\tfrac{3A}{2}-type expressions.
  • Note tan⁡B2+cot⁡B2=2sin⁡B\tan\dfrac{B}{2} + \cot\dfrac{B}{2} = \dfrac{2}{\sin B}.
  • Difference of squares of sines: sin⁡(A+B)sin⁡(A−B)=sin⁡2A−sin⁡2B\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B; in a triangle sin⁡(A+B)=sin⁡C\sin(A+B)=\sin C, so sin⁡2A−sin⁡2B=sin⁡Csin⁡(A−B)\sin^2A-\sin^2B=\sin C\sin(A-B), and by the sine rule a2−b2=c (acos⁡B−bcos⁡A)a^2-b^2=c\,(a\cos B-b\cos A). The cosine sibling: cos⁡(A+B)cos⁡(A−B)=cos⁡2A−sin⁡2B\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B.

Half-angle tangent

tan⁡A2=rs−a=(s−b)(s−c)s(s−a)\tan\dfrac{A}{2} = \dfrac{r}{s-a} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}

Worked example

In △ABC\triangle ABC with C=60∘C = 60^\circ, simplify cos⁡A+cos⁡Bcos⁡A−B2\dfrac{\cos A + \cos B}{\cos\frac{A-B}{2}}.

The same idea in a real exam question:

NDA · Apr 2019 · Q42Moderate

Example 4 · Properties of Triangle · Triangle Identities — A+B+C=π, Half-Angle, and Double-Angle

If cos⁡A=34\cos A = \dfrac{3}{4}, then what is the value of sin⁡A2sin⁡3A2\sin\dfrac{A}{2}\sin\dfrac{3A}{2}?
Drill 4 more on half-angle formulas & sum-to-product

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

  • Consequences of A + B + C = π

    Half-angle complement

    sin⁡B+C2=cos⁡A2,sin⁡(B+C)=sin⁡A\sin\dfrac{B+C}{2} = \cos\dfrac{A}{2}, \qquad \sin(B+C) = \sin A
  • The tan-Sum = tan-Product Identity

    tan sum = tan product

    tan⁡A+tan⁡B+tan⁡C=tan⁡A tan⁡B tan⁡C\tan A + \tan B + \tan C = \tan A\,\tan B\,\tan C
  • cos 2A Sums & Right-Angle Detection

    Right-angle signature

    sin⁡2A+sin⁡2B+sin⁡2C=2  ⟺  right-angled\sin^2 A + \sin^2 B + \sin^2 C = 2 \iff \text{right-angled}
  • Half-Angle Formulas & Sum-to-Product

    Half-angle tangent

    tan⁡A2=rs−a=(s−b)(s−c)s(s−a)\tan\dfrac{A}{2} = \dfrac{r}{s-a} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}

Watch out for (2)

Test yourself on Properties of Triangle

15 past NDA questions from this chapter, timed at 19 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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