Complementary angles in Trigonometry
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 (900 - A) = cos A tan(900 - A) = cot A sec(900 - A) = cosec A
cos (900 - A) = sin A cot(900 - A) = tan A cosec (900 - 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
________________________________________________________________
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
Trigonometry
• SOHCAHTOA -Introduction to Trigonometry
• Trigonometric ratios and their Relation
• Trigonometry for specific angles
• Complementary angles in Trigonometry
• Trigonometric Equations
Home Page