Trigonometry for Specific Angles
In this section we will discuss about trigonometry for specific angles.
We have already learned that trigonometry is the study of relationships between the sides and angles of a triangle. The ratios of the sides in a right triangle with respect to some acute angles are called trigonometry for specific-angles. The angles 0
0, 30
0, 45
0, 60
0 and 90
0 are useful angles in trigonometry, and their numerical values are easy to remember.These are trigonometry for specific angles.
When two angles add up to 90
0, then any one angle is the complement of the other. Trigonometric ratios of complementary angles help in simplifying problems.
Trigonometric Ratios of 00
 |
| sin 00 = 0
cos 00 = 1
tan 00 = 0 |
csc 00 = Not defined
sec 00 = 1
cot 00 = undefined |
Trigonometric Ratios of 300
 |
| sin 300 = 1 / 2
cos 300 = √3 / 2
tan 300 = 1 / √3 |
csc 300 = 2
sec 300 = 2 / √3
cot 300 = √3 |
Trigonometric Ratios of 450
 |
| sin 450 = 1 / √2
cos 450 = 1 / √2
tan 450 = 1 |
csc 450 = √2
sec 450 = √2
cot 450 = 1 |
Trigonometric Ratios of 600
 |
| sin 600 = √3 / 2
cos 600 = 1 / 2
tan 600 = √3 |
csc 600 = 2 / √3
sec 600 = 2
cot 600 = 1 / √3 |
Some Solved Examples on trigonometry for specific angles:
1) Evaluate : ( sin
2 45
0 + cos
2 45
0) / tan
2 60
0
Solution :
( sin
2 45
0 + cos
2 45
0) / tan
2 60
0
= 1 / tan
2 60
0 [ since sin
2θ + cos
2θ = 1 ]
= 1/ (√3)
2
= 1 / 3
_______________________________________________________________
2) 2 sin
230
0tan 60
0 - 3 cos
260
0sec
20
Solution :
2 sin
230
0 - 3 cos
260
0sec
20
= 2 ( 1/ 2)
2 x √3 – 3 ( ½)
2 x (2 / √3)
2
= 2 x 1 / 4 x √ 3 – 3 x ¼ x 4 / 3
= √3 / 2 – 1
= ( √3 – 2 ) / 2
_______________________________________________________________
3) Solve tan 5A = 1 for 0
0 < A < 90
0
Solution :
tan 5A =1
5A = 45
A = 45/5
A = 9
0
_______________________________________________________________
4) Find the acute angles A and B, if sin(A + 2B) = √3 / 2 and
cos (A + 4B ) = 0, A>B.
Solution :
sin(A + 2B) = √3 / 2
sin(A + 2B) = sin 60
0
A + 2B = 60 ------> (1)
cos (A + 4B ) = 0
cos (A + 4B ) = cos 90
0
A + 4B = 90 -----------> (2)
Subtract equation (1) from (2) we get
2B = 30
B = 15
Equation (1)
A + 2(15) = 60
A + 30 = 60
A = 60 – 30
A = 30
Trigonometry
• SOHCAHTOA -Introduction to Trigonometry
• Trigonometric ratios and their Relation
• Trigonometry for specific angles
• Complementary angles in Trigonometry
• Trigonometric Equations
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