Using this logic ,In right ΔABC
sine of ∠ A = P/H = BC/AC
cosine of ∠ A = B/H = AB/AC
tangent of ∠ A = P/B = BC/AB
cosecant of ∠ A = 1/ sine of A = AC/BC
secant of ∠ A = 1/cosine of A = AC/AB
cotangent of ∠ A = 1/tangent of A = AB/BC
The value of sin A or cos A never exceeds 1, whereas the value of sec A or cosec A is always greater than or equal to 1.
Trigonometric ratios of an acute angle in a right triangle express the relationship between the angle and the length of its sides.
The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.
Greek letter θ (theta) is also used to denote an angle.
Numerical: In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : sin A, cos A ,sin C, cos C.
We can find the third side, hypotenuse AC = √ AB2 + BC2 = √242 + 72 = 25
Sin A = P/H = BC/AC = 7/25
Cos A = B/H = AB/AC = 24/25
Sin C = P/H = AB/AC=24/25
Cos C = B/H =BC/AC = 7/25
Trigonometric Ratios of Some Specific Angles
Let’s see the values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90°.
From the table above you can observe that as ∠ A increases from 0° to 90°, sin A increases from 0 to 1 and cos A decreases from 1 to 0.
Numerical: In Δ ABC, right-angled at B, AB = 5 cm and ∠ ACB = 30°. Determine the lengths of the sides BC and AC.
Tan C =P/B =AB/BC
Or Tan 300 = 5cm/BC
1/√3 = 5cm /BC (using table, tan 300 =1/√3).
Or BC = 5√3 cm.
Now we can find the third side by Pythagoras Theorem. We can also find third side using trigonometric identities.
Sin C = P/H = AB/AC
Or Sin 300 = 5cm/AC
Or 1/2 = 5cm /AC (using table, Sin 300 =1/2).
Or AC = 10 cm.
Trigonometric Ratios of Complimentary Angles
Two angles are said to be complementary if their sum equals 900
sin (90° – A) = cos A,
cos (90° – A) = sin A,
tan (90° – A) = cot A,
cot (90° – A) = tan A,
sec (90° – A) = cosec A,
cosec (90° – A) = sec A,
sin ècos… and cos è sin… !!!! tan è cot …. And cot è tan.
Numerical: Evaluate tan 65°/ cot 25°
cot A = tan (90° – A)
So, cot 25° = tan (90° – 25°) = tan 65°
Thus tan 65°/ cot 25° = tan 65°/ tan 65° = 1
sin2 A + cos2 A = 1,
sec2 A – tan2 A = 1 for 0° ≤ A < 90°,
cosec2 A = 1 + cot2 A for 0° < A ≤ 90°.
Numerical: Express the ratios cos A, tan A and sec A in terms of sin A.
Solution: cos2 A + sin2 A = 1,
Or cos2 A = 1 – sin2 A, i.e., cos A = ±√ 1 − sin2 A
or cos A = √1 − sin2 A (Ignoring the negative value , since sin A is positive, Cos A will also be positive.)
tan A = sin A/cos A = sin A/√1 − sin2 A