Complementary angles in trigonometry : Two angles are said to be complementary, if their sum is 90^{0}.
It follows from the above definition that θ and ( 90 - θ ) are complementary angles in trigonometry for an acute angle θ
In ΔABC, ∠B = 90^{0}
∴ ∠A + ∠C = 90^{0}
∠C =90^{0} - ∠A
For the sake of easiness in this derivation, we will write ∠C and ∠A as C and A respectively
Thus
C = 90^{0} - A
sin A = BC / AC cosec A = AC / BC
cos A = AB / AC sec A = AC / AB
tan A = BC / AB cot A = AB / BC and
sin C = sin (90^{0} - A ) = AB / AC; cosec C = cosec (90^{0} - A) = AC / AB
cos C = cos (90^{0} - A) = BC / AC; sec C = sec (90^{0} - A) = AC / BC
tan C = tan (90^{0} - A) = AB / BC; cot C = cot (90^{0} - A) = BC /AB
sin (90^{0} - A) = cos A tan(90^{0} - A) = cot A sec(90^{0} - A) = cosec A
cos (90^{0} - A) = sin A cot(90^{0} - A) = tan A cosec (90^{0} - A) = sec A
This means, for example
sin 70^{0} = cos 20^{0}
The cofunction of the sine is the cosine. And 20° is the complement of 70°. Some Solved Examples on complementary angles in trigonometry :
1) Evaluate : cos 37^{0} / sin 53^{0} Solution :
cos 37^{0}/sin 53^{0} = cos( 90 – 53 )/sin 53 = sin 53^{0}/sin 53^{0} = 1
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2) Show that : ( cos 70^{0}) / (sin 20^{0}) + (cos 59^{0}) / sin 31^{0} - 8 sin^{2} 30^{0} = 0 Solution : Consider
( cos 70^{0}) / (sin 20^{0}) + (cos 59^{0}) / sin 31^{0} - 8 sin^{2} 30^{0}
= [ cos ( 90^{0} - 20^{0})] / [sin 20^{0}] + [ cos(90^{0} - 31^{0})] / sin 31^{0} - 8 x(1/2) ^{2}
= sin 20^{0} / sin 20^{0} + sin 31^{0} / sin 31^{0} - 8 x 1/ 4
= 1 + 1 – 2
= 0
∴ ( cos 70^{0}) / (sin 20^{0}) + (cos 59^{0}) / sin 31^{0} - 8 sin^{2} 30^{0} = 0