Complementary angles in Trigonometry

Complementary angles in trigonometry : Two angles are said to be complementary, if their sum is 90⁰.
It follows from the above definition that θ and ( 90 - θ ) are complementary angles in trigonometry for an acute angle θ
In ΔABC, ∠B = 90⁰
∴ ∠A + ∠C = 90⁰
∠C =90⁰ - ∠A
For the sake of easiness in this derivation, we will write ∠C and ∠A as C and A respectively
Thus
C = 90⁰ - 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 (90⁰ - A ) = AB / AC; cosec C = cosec (90⁰ - A) = AC / AB
cos C = cos (90⁰ - A) = BC / AC; sec C = sec (90⁰ - A) = AC / BC
tan C = tan (90⁰ - A) = AB / BC; cot C = cot (90⁰ - 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 70⁰ = cos 20⁰
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 37⁰ / sin 53⁰
Solution :
cos 37⁰/sin 53⁰ = cos( 90 – 53 )/sin 53 = sin 53⁰/sin 53⁰ = 1
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2) Show that : ( cos 70⁰) / (sin 20⁰) + (cos 59⁰) / sin 31⁰ - 8 sin² 30⁰ = 0
Solution : Consider
( cos 70⁰) / (sin 20⁰) + (cos 59⁰) / sin 31⁰ - 8 sin² 30⁰
= [ cos ( 90⁰ - 20⁰)] / [sin 20⁰] + [ cos(90⁰ - 31⁰)] / sin 31⁰ - 8 x(1/2) ²
= sin 20⁰ / sin 20⁰ + sin 31⁰ / sin 31⁰ - 8 x 1/ 4
= 1 + 1 – 2
= 0
∴ ( cos 70⁰) / (sin 20⁰) + (cos 59⁰) / sin 31⁰ - 8 sin² 30⁰ = 0

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Trigonometry

• SOHCAHTOA -Introduction to Trigonometry
• Trigonometric ratios and their Relation
• Trigonometry for specific angles
• Complementary angles in Trigonometry
• Trigonometric Equations

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