complementary angles in trigonometry

Complementary angles in Trigonometry

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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

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