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