# Conditional Trigonometric Identities

In conditional trigonometric identities we will discuss certain relationship exists among the angles involved. We know some of the trigonometric identities which were true for all values of the angles involved. These identities hold for all values of the angles which satisfy the given conditions among them and hence they are called conditional trigonometric identities.

Such identities involving different trigonometrical ratios of three or more angles can be deduced when these angles are connected by some given relation. Suppose, if the sum of three angles be equal to two right angles then we can establish many important identities involving trigonometrical ratios of those angles. To establish such identities we require to use the properties of supplementary and complementary angles.

If A, B and C denote the angles of a triangle ABC, then the relation A + B + C = π enables us to establish many important identities involving trigonometric ratios of these angles The following results are useful to obtain the said identities.

If A + B + C = π, then the sum of any two angles is supplementary to the third i.e.,

(i) B + C = π - A or, C + A = π - B or A + B = π - C.

(ii) If A + B + C = π then sin (A + B) = sin (π - C) = sin C

sin (B + C) = sin (π - A) = sin A

sin (C + A) = sin (π - B) = sin B

(iii) If A + B + C = π then cos (A + B) = cos (π - C) = - cos C
cos (B + C) = cos (π - A) = - cos A
cos (C + A) = cos (π - B) = - cos B

(iv) If A + B + C = π then tan (A + B) = tan (π - C) = - tan C

tan (B + C) = tan (π - A) = - tan A

tan (C + A) = tan (π - B) = - tan B

(v) If A + B + C = π then $$\frac{A}{2}$$ + $$\frac{B}{2}$$ + $$\frac{C}{2}$$ = $$\frac{π}{2}$$

Hence, it is evident that the sum of any two of the three angles $$\frac{C}{2}$$, $$\frac{B}{2}$$, $$\frac{C}{2}$$  is complementary to the third.

i.e., $$\frac{A + B}{2}$$ = $$\frac{π}{2}$$ - $$\frac{C}{2}$$,

$$\frac{B + C}{2}$$ = $$\frac{π}{2}$$ - $$\frac{A}{2}$$

$$\frac{C + A}{2}$$ = $$\frac{π}{2}$$ - $$\frac{B}{2}$$

Therefore,

sin ($$\frac{A}{2}$$ + $$\frac{B}{2}$$) = sin $$\frac{π}{2}$$ - $$\frac{C}{2}$$ = cos $$\frac{C}{2}$$

sin ($$\frac{B}{2}$$ + $$\frac{C}{2}$$) = sin $$\frac{π}{2}$$ - $$\frac{A}{2}$$ = cos $$\frac{A}{2}$$

sin ($$\frac{C}{2}$$ + $$\frac{A}{2}$$) = sin $$\frac{π}{2}$$ - $$\frac{B}{2}$$ = cos $$\frac{B}{2}$$

cos ($$\frac{A}{2}$$ + $$\frac{B}{2}$$) = cos $$\frac{π}{2}$$ - $$\frac{C}{2}$$ = sin $$\frac{C}{2}$$

sin ($$\frac{B}{2}$$ + $$\frac{C}{2}$$) = cos $$\frac{π}{2}$$ - $$\frac{A}{2}$$ = sin $$\frac{A}{2}$$

sin ($$\frac{C}{2}$$ + $$\frac{A}{2}$$) = cos $$\frac{π}{2}$$ - $$\frac{B}{2}$$ = sin $$\frac{B}{2}$$

tan ($$\frac{A}{2}$$ + $$\frac{B}{2}$$) = tan $$\frac{π}{2}$$ - $$\frac{C}{2}$$ = cot $$\frac{C}{2}$$

tan ($$\frac{B}{2}$$ + $$\frac{C}{2}$$) = tan $$\frac{π}{2}$$ - $$\frac{A}{2}$$ = cot $$\frac{A}{2}$$

tan ($$\frac{C}{2}$$ + $$\frac{A}{2}$$) = tan $$\frac{π}{2}$$ - $$\frac{B}{2}$$ = cot $$\frac{B}{2}$$

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