Trigonometrical ratios
SinA = p/h ,
cosA = b/h ,
tanA = p/b ,
cotA = b/p ,
SecA= h/b and
cosecA = h/p
Where, p = perpendicular, b = base and h =
hypotenuse and h = p² + b² .
2. sinA * cosecA = 1,
cosA * secA = 1,
tanA * cotA = 1.
tanA =sinA/cosA ,
cotA = cosA/sinA
3. sin²A + cos²A = 1,
sec²A – tan²A = 1,
cosec²A – cot²A = 1
4. The trigonometric values of different degree
(0° - 180°) are listed below:
Angles Sin Cos Tan
0° 0 1 0
30° 1/2 √3/2 1/√3
45° 1/√2 1/√2 1
60° √3/2 1/2 √3
90° 1 0 ∞
120° √3/2 -1/2 -√3
135° 1/√2 -1/√2 -1
150° 1/2 -√3/2 -1/√3
180° 0 -1 0
5.With angle is (-A),
sin(-A) = -sinA,
cos(-A) = cosA.
tan(-A) = -tanA, etc.
6. With angle is (90° - A) and (90° + A)
sin(90° - A) = cosA
sin(90°+A) = cosA
cos(90° - A) = sinA
cos(90° + A) = -sinA
tan(90° - A) = cotA
tan(90°+A) = -cotA etc.
7. With angle (180° - A) and (180° + A)
sin(180° - A) = sinA
sin(180°+A) = -sinA
cos(180° - A) = -cosA
cos(180° + A) = -cosA
tan(180° - A) = -tanA
tan(180°+A) = tanA etc.
8. When angle is (270° - A) and (270° + A)
sin(270° - A) = -cosA
sin(270°+A) = -cosA
cos(270° - A) = -sinA
cos(270° + A) = sinA
tan(270° - A) = cotA
tan(270°+A) = -cotA etc.
9. With angle (360° - A) and (360° + A)
sin(360° - A) = -sinA
sin(360°+A) = sinA
cos(360° - A) = cosA
cos(360° + A) = cosA
Tan(360° - A) = -tanA
tan(360°+A) = tanA etc.
Trigonometric Ratios of compound angles:
1. sin(A + B) = sinA.cosB + cosA.sinB
2. cos(A + B) = cosA.cosB – sinA.sinB
3. tan(A + B) = (tanA+tanB)/(1-tanA.tanB)
4. cot(A + B) = (cotA.cotB+1)/(cotB-cotA)
5. sin (A – B) = sinA.cosB - cosA.sinB
6. cos(A – B) = cosA.cosB + sinA.sinB
7. tan(A – B) =(tanA-tanB)/(1+tanA.tanB)
8. cot(A - B) = (cotA.cotB-1)/(cotB+cotA)
Trigonometric ratios of multiple angles:
1.
(i)sin2A = 2sinA.cosA
(ii)cos2A = cos²A – sin²A
= 2cos²A – 1
= 1 – 2sin²A
(iii)tan2A = 2tanA/(1-tan²A)
(iv)sin2A = 2tanA/(1+tan²A)
(v)cos2A = (1-tan²A)/(1+tan²A)
2.
(i) sin3A = 3sinA – 4sin³A
(ii) cos3A = 4cos³A – 3cosA
(iii) tan3A = (3tanA-tan³A)/(1-3tan²A)
Trigonometric ratios of Sub-multiple angles:
1.
(i)sinA = 2sin(A/2).cos(A/2)
(ii)cosA = cos²(A/2) –sin²(A/2)
= 2cos²(A/2) – 1
= 1 – 2sin²(A/2)
(iii)tanA =2tan(A/2)/(1-tan²(A/2))
(iv) sinA = 3sin(A/2) – 4sin³(A/2)
(v) cosA = 4cos³(A/2) – 3cos(A/2)
(vi) tanA = (3tan(A/2)-tan³(A/2))/(1-3tan²(A/2))
Transformation of Trigonometric Formulae:
1.2sinA.cosB = sin(A + B) + sin(A – B)
2. 2cosA.sinB = sin(A+B) – sin(A – B)
3. 2cosA.cosB = cos(A + B) + cos(A – B)
4. 2sinA.sinB = cos(A + B) – cos(A + B)
5. sinC + sinD = 2sin{(C+D)/2} cos{(C-D)/2}
6. sinC – sinD = 2cos{(C+D)/2} sin{(C-D)/2}
7. cosC + cosD = 2cos{(C+D)/2}.cos{(C-D)/2}
8. cosC – cosD = -2sin{(C+D)/2}.sin{(C-D)/2}
= 2sin{(C+D)/2}.sin{(D-C)/2}
Conditional Trigonometric Identities:
If A + B + C = π
1.sinA = sin[ π– (B + C)] = sin(B + C)
2. cosA = cos[π – (B + C)] = -cos(B + C)
3. tanA = tan[π – (B + C)] = -tan(B + C)etc.
If, A+B+C = π
1.sin(C/2) = sin[π/2– (A/2+B/2)] = cos(A/2+B/2)
2.cos(C/2)= cos[π/2– (A/2+B/2)] = sin(A/2+B/2)
3.tan(C/2) = tan[π/2– (A/2+B/2)] = cot(A/2+B/2)
To prove the identities involving sines and cosines we use the
following algorithm.
Step I: Convert the sum of first two terms as product by using one of the following formulae:
a. sinC + sinD = 2sin{(C+D)/2} cos{(C-D)/2}
b. sinC – sinD = 2cos{(C+D)/2} sin{(C-D)/2}
c. cosC + cosD = 2cos{(C+D)/2}.cos{(C-D)/2}
d. cosC – cosD = -2sin{(C+D)/2}.sin{(C-D)/2}
= 2sin{(C+D)/2}.sin{(D-C)/2}
Step II: In the product obtain in step II replace the sum of two angles in terms of the third by using the given relation.
Step III: Expand the third term by using one of the following
formulas:
sin 2θ = 2 sin θ cos θ,
cos 2θ = 2 cos²θ - 1
cos 2θ = 1 - 2 sin²θ etc.
Step IV: Take the common factor outside.
Step V: Express the trigonometric ratio of the single angle in terms of the remaining angles.
Step VI: Use one of the formulas given in step I to convert the sum into product.
Examples on identities involving sines and
cosines:
1. If A + B + C = π prove that, sin 2A + sin 2B + sin 2C = 4 sinA sinB sinC.
Solution:
L.H.S. = (sin 2A + sin 2B) + sin 2C
= 2 sin[(2A+2B)/2] cos[(2A-2B)/2] + sin 2C
= 2 sin (A + B) cos (A - B) + sin 2C
= 2 sin (π - C) cos (A - B) + sin 2C, [Since, A + B + C = π ⇒ A + B = π - C]
= 2 sin C cos (A - B) + 2 sin C cos C,
= 2 sin C [cos (A - B) + cos C], taking common 2 sin C
= 2 sin C [cos (A - B) + cos {π - (A + B)}],
= 2 sin C [cos (A - B) - cos (A + B)],
= 2 sin C [2 sin A sin B], [Since cos (A - B) - cos (A + B) = 2 sinA sin B]
= 4 sin A sin B sin C. Proved.
2. If A + B + C = π prove that, cos 2A + cos 2B - cos 2C = 1- 4 sin A sin B cos C.
Solution:
L.H.S. = cos 2A + cos 2B - cos 2C
= (cos 2A + cos 2B) - cos 2C
= 2 cos cos - cos 2C
= 2 cos (A + B) cos (A- B) - cos 2C
= 2 cos (π - C) cos (A- B) - cos 2C, [Since we know A + B + C= π , ⇒A + B = π – C]
= - 2 cos C cos (A - B) – (2 cos C - 1), [Since cos (π - C) = - cos C]
= - 2 cos C cos (A - B) - 2 cos C + 1
= - 2 cos C [cos (A - B) + cos C] + 1
= -2 cos C [cos (A - B) - cos (A + B)] + 1,
[Since cos C = - cos(A + B)]
= -2 cos C [2 sin A sin B] + 1,
[Since cos (A - B) - cos (A + B) = 2 sin A sin B]
= 1 - 4 sin A sin B cos C. Proved.
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3. Prove:
cos (A – 120°) + cos A + cos (A + 120°) = O
L.H.S.
cos (A – 120°) + cos A + cos (A + 120°)
= {cos (A – 120°)} + cos A + {cos (A + 120°)}
= (cosA.cos120°+sinA.sin120°) + cos A +
(cosA.cos120°– sinA.sin120°)
=cos(A - 120°) + cos A + cos(A+120°)
= cos(A - 120°) + cos(A+120°) + cos A
= 2cosA.cos120°+cosA
= -cosA+cosA
=0
= R.H.S.
Height and Distance
Angle of elevation: The angle of elevation is
defined as angle between the line of sight and
horizontal line made by the observer when the
observer observes the object above the
horizontal line.
Qn.1
A man observes the top of a pole 52cm height
situated in front of him and finds the angle of
elevation to be 30° .If the distance between
man and pole is 86m .Find the height of that
man.
Qn. 2
The angle of elevation from the roof of a house
to the top of atop of tree is found to be 30°. If
the height of the house and tree are 8 m and
20m respectively, find the distance between the
house and the tree.
