Quick answer: When A + B + C = π A + B + C = \pi A + B + C = π tan A + tan B + tan C = tan A tan B tan C \tan A + \tan B + \tan C = \tan A\tan B\tan C tan A + tan B + tan C = tan A tan B tan C sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C \sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A\cos B\cos C cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C extreme values , write the expression in a single trigonometric function or use ∣ a sin x + b cos x ∣ ≤ a 2 + b 2 \lvert a\sin x + b\cos x\rvert \le \sqrt{a^2 + b^2} ∣ a sin x + b cos x ∣ ≤ a 2 + b 2
These identities are called conditional because they rely on a condition linking the angles. They are proved with the compound angle and product-to-sum formulas .
Identities when A + B + C = π
Identity Holds when tan A + tan B + tan C = tan A tan B tan C \tan A + \tan B + \tan C = \tan A\tan B\tan C tan A + tan B + tan C = tan A tan B tan C A + B + C = π A + B + C = \pi A + B + C = π cot A cot B + cot B cot C + cot C cot A = 1 \cot A\cot B + \cot B\cot C + \cot C\cot A = 1 cot A cot B + cot B cot C + cot C cot A = 1 A + B + C = π A + B + C = \pi A + B + C = π tan A 2 tan B 2 + tan B 2 tan C 2 + tan C 2 tan A 2 = 1 \tan\frac{A}{2}\tan\frac{B}{2} + \tan\frac{B}{2}\tan\frac{C}{2} + \tan\frac{C}{2}\tan\frac{A}{2} = 1 tan 2 A tan 2 B + tan 2 B tan 2 C + tan 2 C tan 2 A = 1 A + B + C = π A + B + C = \pi A + B + C = π sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C \sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C A + B + C = π A + B + C = \pi A + B + C = π cos 2 A + cos 2 B + cos 2 C = − 1 − 4 cos A cos B cos C \cos 2A + \cos 2B + \cos 2C = -1 - 4\cos A\cos B\cos C cos 2 A + cos 2 B + cos 2 C = − 1 − 4 cos A cos B cos C A + B + C = π A + B + C = \pi A + B + C = π sin A + sin B + sin C = 4 cos A 2 cos B 2 cos C 2 \sin A + \sin B + \sin C = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2} sin A + sin B + sin C = 4 cos 2 A cos 2 B cos 2 C A + B + C = π A + B + C = \pi A + B + C = π cos A + cos B + cos C = 1 + 4 sin A 2 sin B 2 sin C 2 \cos A + \cos B + \cos C = 1 + 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} cos A + cos B + cos C = 1 + 4 sin 2 A sin 2 B sin 2 C A + B + C = π A + B + C = \pi A + B + C = π cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A\cos B\cos C cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C A + B + C = π A + B + C = \pi A + B + C = π
Proof of the tan identity
Step 1: A + B = π − C A + B = \pi - C A + B = π − C tan ( A + B ) = − tan C \tan(A + B) = -\tan C tan ( A + B ) = − tan C
Step 2: Expand:
tan A + tan B 1 − tan A tan B = − tan C \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C 1 − t a n A t a n B t a n A + t a n B = − tan C
Step 3: Cross-multiply and rearrange:
tan A + tan B + tan C = tan A tan B tan C \tan A + \tan B + \tan C = \tan A\tan B\tan C tan A + tan B + tan C = tan A tan B tan C
Proof of the sin 2A identity
Step 1: sin 2 A + sin 2 B = 2 sin ( A + B ) cos ( A − B ) = 2 sin C cos ( A − B ) \sin 2A + \sin 2B = 2\sin(A + B)\cos(A - B) = 2\sin C\cos(A - B) sin 2 A + sin 2 B = 2 sin ( A + B ) cos ( A − B ) = 2 sin C cos ( A − B )
Step 2: sin 2 C = 2 sin C cos C = − 2 sin C cos ( A + B ) \sin 2C = 2\sin C\cos C = -2\sin C\cos(A + B) sin 2 C = 2 sin C cos C = − 2 sin C cos ( A + B )
Step 3: Add: 2 sin C [ cos ( A − B ) − cos ( A + B ) ] = 2 sin C ⋅ 2 sin A sin B = 4 sin A sin B sin C 2\sin C[\cos(A - B) - \cos(A + B)] = 2\sin C \cdot 2\sin A\sin B = 4\sin A\sin B\sin C 2 sin C [ cos ( A − B ) − cos ( A + B )] = 2 sin C ⋅ 2 sin A sin B = 4 sin A sin B sin C
Maximum and minimum values
Expression Range a sin x + b cos x a\sin x + b\cos x a sin x + b cos x [ − a 2 + b 2 , a 2 + b 2 ] \left[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}\right] [ − a 2 + b 2 , a 2 + b 2 ] a sin x + b cos x + c a\sin x + b\cos x + c a sin x + b cos x + c [ c − a 2 + b 2 , c + a 2 + b 2 ] \left[c - \sqrt{a^2 + b^2}, c + \sqrt{a^2 + b^2}\right] [ c − a 2 + b 2 , c + a 2 + b 2 ] sin x cos x \sin x\cos x sin x cos x [ − 1 2 , 1 2 ] \left[-\frac{1}{2}, \frac{1}{2}\right] [ − 2 1 , 2 1 ] sin 2 x + cos 4 x \sin^2 x + \cos^4 x sin 2 x + cos 4 x [ 3 4 , 1 ] \left[\frac{3}{4}, 1\right] [ 4 3 , 1 ] a 2 sec 2 x + b 2 csc 2 x a^2\sec^2 x + b^2\csc^2 x a 2 sec 2 x + b 2 csc 2 x minimum ( a + b ) 2 (a + b)^2 ( a + b ) 2 a , b > 0 a, b > 0 a , b > 0 a tan x + b cot x a\tan x + b\cot x a tan x + b cot x a , b > 0 a, b > 0 a , b > 0 x x x minimum 2 a b 2\sqrt{ab} 2 ab
The last two come from the AM–GM inequality. The sin 2 x + cos 4 x \sin^2 x + \cos^4 x sin 2 x + cos 4 x
Method 1: reduce to one function
sin 2 x + cos 4 x = 1 − cos 2 x + cos 4 x \sin^2 x + \cos^4 x = 1 - \cos^2 x + \cos^4 x sin 2 x + cos 4 x = 1 − cos 2 x + cos 4 x u = cos 2 x ∈ [ 0 , 1 ] u = \cos^2 x \in [0, 1] u = cos 2 x ∈ [ 0 , 1 ]
1 − u + u 2 = ( u − 1 2 ) 2 + 3 4 1 - u + u^2 = \left(u - \frac{1}{2}\right)^2 + \frac{3}{4} 1 − u + u 2 = ( u − 2 1 ) 2 + 4 3
Minimum 3 4 \frac{3}{4} 4 3 u = 1 2 u = \frac{1}{2} u = 2 1 u = 0 u = 0 u = 0 u = 1 u = 1 u = 1
Method 2: the a sin x + b cos x bound
5 sin x + 12 cos x + 7 5\sin x + 12\cos x + 7 5 sin x + 12 cos x + 7 7 + 13 = 20 7 + 13 = 20 7 + 13 = 20 7 − 13 = − 6 7 - 13 = -6 7 − 13 = − 6 25 + 144 = 13 \sqrt{25 + 144} = 13 25 + 144 = 13
Method 3: AM–GM for positive terms
For acute x x x 9 tan 2 x + 4 cot 2 x ≥ 2 36 = 12 9\tan^2 x + 4\cot^2 x \ge 2\sqrt{36} = 12 9 tan 2 x + 4 cot 2 x ≥ 2 36 = 12 9 tan 2 x = 4 cot 2 x 9\tan^2 x = 4\cot^2 x 9 tan 2 x = 4 cot 2 x tan 2 x = 2 3 \tan^2 x = \frac{2}{3} tan 2 x = 3 2
Worked examples
Example 1: a triangle with tan values
In triangle ABC, tan A = 1 \tan A = 1 tan A = 1 tan B = 2 \tan B = 2 tan B = 2 tan C \tan C tan C
Using tan A + tan B + tan C = tan A tan B tan C \tan A + \tan B + \tan C = \tan A\tan B\tan C tan A + tan B + tan C = tan A tan B tan C 3 + tan C = 2 tan C 3 + \tan C = 2\tan C 3 + tan C = 2 tan C tan C = 3 \tan C = 3 tan C = 3
Check: tan ( A + B ) = 3 1 − 2 = − 3 = − tan C \tan(A + B) = \frac{3}{1 - 2} = -3 = -\tan C tan ( A + B ) = 1 − 2 3 = − 3 = − tan C
Example 2: the cos² identity
If A + B + C = π A + B + C = \pi A + B + C = π cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C \cos^2 A + \cos^2 B + \cos^2 C = 1 - 2\cos A\cos B\cos C cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C
With C = 90 ∘ C = 90^\circ C = 9 0 ∘ cos C = 0 \cos C = 0 cos C = 0 B = 90 ∘ − A B = 90^\circ - A B = 9 0 ∘ − A cos 2 A + sin 2 A + 0 = 1 \cos^2 A + \sin^2 A + 0 = 1 cos 2 A + sin 2 A + 0 = 1 1 − 0 = 1 1 - 0 = 1 1 − 0 = 1
Example 3: maximum of a product
Find the maximum of sin x cos x \sin x\cos x sin x cos x sin ( x + π 6 ) + cos ( x + π 6 ) \sin\left(x + \frac{\pi}{6}\right) + \cos\left(x + \frac{\pi}{6}\right) sin ( x + 6 π ) + cos ( x + 6 π )
sin x cos x = 1 2 sin 2 x \sin x\cos x = \frac{1}{2}\sin 2x sin x cos x = 2 1 sin 2 x 1 2 \frac{1}{2} 2 1 The second is a sin θ + b cos θ a\sin\theta + b\cos\theta a sin θ + b cos θ a = b = 1 a = b = 1 a = b = 1 2 \sqrt{2} 2
Summary table
Topic Key result A + B + C = π, tan ∑ tan A = ∏ tan A \sum\tan A = \prod\tan A ∑ tan A = ∏ tan A A + B + C = π, half-angle tan ∑ tan A 2 tan B 2 = 1 \sum\tan\frac{A}{2}\tan\frac{B}{2} = 1 ∑ tan 2 A tan 2 B = 1 A + B + C = π, sin 2A ∑ sin 2 A = 4 ∏ sin A \sum\sin 2A = 4\prod\sin A ∑ sin 2 A = 4 ∏ sin A Linear combination ∣ a sin x + b cos x ∣ ≤ a 2 + b 2 \lvert a\sin x + b\cos x\rvert \le \sqrt{a^2 + b^2} ∣ a sin x + b cos x ∣ ≤ a 2 + b 2 Positive terms AM–GM: p + q ≥ 2 p q p + q \ge 2\sqrt{pq} p + q ≥ 2 pq
Exam tips
Whenever a question mentions a triangle, remember that A + B + C = π A + B + C = \pi A + B + C = π ( 0 , π ) (0, \pi) ( 0 , π )
In max–min questions, first check whether the expression is of the form a sin x + b cos x + c a\sin x + b\cos x + c a sin x + b cos x + c
For squared terms, substitute u = sin 2 x u = \sin^2 x u = sin 2 x u = cos 2 x u = \cos^2 x u = cos 2 x u ∈ [ 0 , 1 ] u \in [0, 1] u ∈ [ 0 , 1 ]
With AM–GM, check that both terms are positive and that equality is actually attainable.
See all topics in the NIMCET Trigonometry chapter .
Common mistakes
Applying ∑ tan A = ∏ tan A \sum\tan A = \prod\tan A ∑ tan A = ∏ tan A
Forgetting the restriction u ∈ [ 0 , 1 ] u \in [0, 1] u ∈ [ 0 , 1 ] u = cos 2 x u = \cos^2 x u = cos 2 x
Using AM–GM on terms that can be negative.
Writing the maximum of a sin x + b cos x a\sin x + b\cos x a sin x + b cos x a + b a + b a + b
Frequently Asked Questions
What is a conditional trigonometric identity?
It is an identity that is true only when the angles satisfy a condition, such as A + B + C = π, rather than for all angles.
If A + B + C = π, what is tan A + tan B + tan C?
It equals tan A tan B tan C, provided none of the angles is a right angle.
What is the maximum value of a sin x + b cos x?
The maximum is √(a² + b²) and the minimum is −√(a² + b²).
What is the minimum value of sin²x + cos⁴x?
The minimum is 3/4, reached when cos²x = 1/2. The maximum is 1.
How is AM–GM used for trigonometric minimum values?
For positive terms p and q, p + q ≥ 2√(pq). For example, 9tan²x + 4cot²x ≥ 12 for acute x.
