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this is a detailed document it has all trigonometry nits and questions it also has topics which come under trigonometry
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Course Outcome : Utilize basic concepts of trigonometry to solve elementary engineering problems. Unit outcome:
a. Apply the concept of compound angle, allied angle, and multiple angles to solve the given
simple engineering problem(s). b. Apply the concept of Sub- multiple angle to solve the given simple engineering related
problem(s).
c. Employ concept of factorization and de-factorization formulae to solve the given simple
engineering problem(s).
d. Investigate given simple problems utilizing inverse trigonometric ratios.
Introduction: Trigonometry is a study of relationships in Mathematics involving lengths, heights and angles of different triangles. Presently Trigonometry finds wide applications in engineering faculties like Applied Mechanics, Electrical Technology, Basic Electronics, and Computer Engineering etc. Subtopic : Compound angles, Allied angles, Multiple and Submultiple angles. Significance : This sub-topic is used to find trigonometric ratios of angles other than standard angles. Some important formulae:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒
cosec θ = 1 sin θ sec θ = 1 cos θ
cot θ = 1 tan θ
tan θ = 𝑠𝑖𝑛 θ cos θ
cot θ =
𝑐𝑜𝑠 θ sin θ cosec θ × sin θ = 1 sec θ × cos θ = 1 cot θ × tan θ = 1 III. Measures of an angle : In practice we use two systems to measure the angle. a) Sexagesimal system (Degree) : In this system , the unit of measurement is degree. b) Circular systems (Radian) : In this system , the unit of measurement is radian. Relation between degrees and radians:
Notation : Any angle can be measured in degrees = θο^ and in radians = θ𝒸
angles
Compound angle:
Definition: The angle obtained by addition or subtraction of given angles is called compound
angle.
For e.g. A+B, A-B are called compound angles
e. g. A = 45, B = 30 then
A + B = 45 + 30 = 75 and A B = 45 30 = 15 are compound angles.
Trigonometric ratios of compound angles (Without proof) :
sin (A + B) = sin A cos B + cos A sin B
sin (A B) = sin A cos B cos A sin B
cos (A + B) = cos A cos B sin A sin B
cos (A B) = cos A cos B + sin A sin B
tan (A + B) =
tan A + tan B 1 tan A tan B
tan A tan B 1 + tan A tan B
Without using calculator, find the value of
Solution : 1) 15 = 45 30
sin 15 = Sin (45 30 )
= sin 45 cos 30 cos 45 sin 30
sin 15=
√𝟑−𝟏 𝟐√𝟐
cos 75 = cos (45+30)
= cos 45 cos 30 sin 45 sin 30
√3− 2√
2 , tan B =
3 find tan (A + B)
Solution : tan (A + B) =
tan A + tan B 1 tan A tan B
4 and tan (x^ ^ y) =
15 then show that tan 2x =
Solution : As 2x = (x + y) + (x y) L.H.S.= tan 2x = tan[(x + y) + (x y)]
=
tan (x + y) + tan (x y) 1 tan (x + y) tan (x y)
cos A =
13 sin B =^
∵ A and B are obtuse (More than 90º and less than 180º) A is the second quadrant, cos A = – 5/ B is the second quadrant, sin B = 3/
cos A =
13 sin B =^
sin (A + B) = sin A cos B + cos A sin B
=
− 65
tan 85 tan 40 1 + tan 85 tan 40
1 tan 2 tan 1 + tan 2 tan
cos 3 cos
3 , tan B =
4 , where 0 < A <^
2 , find sin (A + B)
13 , cos^ ^ =^
25 and^ ,^ ^ lies in the third quadrant, find sin (^ -^ )
13 and cos B =^
5 , then find sin (A + B) and the quadrant of angle A + B.
13 and cos B =^
5 , find tan (A + B).
Allied Angles
If the sum or difference of the measures of two angles is either zero or is an integral multiple of
90 , that is n
2 then these angles are called Allied angles.
For any angle , let be It’s allied angle then
+ = n
2 ^ ^ = n^ ^
2 –^ ^ allied angle of^
and – = n
2 ^ ^ = n^ ^
2 +^ ^ allied angle of^ .
For any angle of ; n
2 ^ ^ are allied angle of^ .
In general the above results can be written as
If n is an even integer
sin
n
2 ^ ^ ^ sin^ ^ cos^
n
2 ^ ^ =^ ^ cos^
If n is an odd integer
sin
n
2 ^ ^ =^ ^ cos^ ^ cos^
n
2 ^ ^ =^ ^ sin^
The algebraic sign is settled down by knowing the quadrant in which the angle (^) ^ n (^) 2 ^ ^ lies.
Note that sin n = 0
cos n = 1 If n is odd integer cos n = 1 If n is even integer
For the angles
2 ^ ^ i.e. (90^ +^ ) and^
2 ^ ^ i.e. (270^ ^ )
Make the changes as follows :
sin cos, tan cot, sec cosec and for the angles ( )
i.e. (180 ) and (2 ) i.e. (360 ) there is no change in trigonometric ratios.
i.e. sin sin, cos cos etc.
The algebraic sign is settled down by knowing the quadrant in which the angle lies.
e.g. sin (^)
For angle
2 ^ ^ sin changes to cos and angle^
2 ^ ^ lies in III
rd (^) - quadrant and in IIIrd
quadrant sin ratio is negative.
Solution : Given sin 420 cos 390 + cos ( 300 ) sin ( 330 )
sin (420) = sin (4 90 + 60) = sin 60 =
cos (390) = cos (360 + 30)
= cos (4 90 + 30) = cos 30 =
cos ( 300 ) = cos (300)
= cos (3 90 + 30) = sin 30 =
sin ( 330 ) = sin (330) = sin (360 30 ) = sin (90 4 30 ) = ( sin 30)
= sin 30 =
∴ sin 420 cos 390 + cos ( 300 ) sin ( 330 )
=
Without using calculator, find the value of
sin 210 2) cos 330 3) sec^2 ( 765 )
cot ( 710 ) 5) cosec ( 960 )
Without using calculator, find the value of
1)
sec^2 (135) cos ( 240 ) 2 sin (930) 2) Prove that : cos (570) sin (510) + sin ( 330 ) cos ( 390 ) = 0 3) tan (585) cot ( 495 ) cot (405) tan ( 495 ) 4) sin (150) tan (315) + cos (300) + sec^2 (3660) 5) cos ( 1125 ) + tan ( 945 ) + cos (495) 6) cos (225) cos (675) + sin (585) sin (315) 7) sin 660 sin 480 cos 120 sin 330
Trigonometric ratios of multiple angle: The integral multiple of A i.e. 2A, 3A, 4A, ........... are called multiple angles of A.
e.g. If A = 60 then 2A = 120, 3A = 180, 4A = 240 etc.
Trigonometric ratios of 2 :
=
2 tan 1 + tan^2
= 1 2 sin^2 = 2 cos^2 1 = 1−𝑡𝑎𝑛^2 𝜃 1+𝑡𝑎𝑛^2 𝜃 From the above we can deduce
tan 2 = 2 𝑡𝑎𝑛𝜃 1 −𝑡𝑎𝑛^2 𝜃 Trigonometric ratios of 3 :
sin 3 = 3 sin 4 sin^3
cos 3 = 4 cos^3 3 cos
tan 3 =
3 tan tan^3 1 3 tan^2
SOLVED EXAMPLES:
sin 3A =3 sin A 4 sin^3 A =3(0.4) 4(0.4)^3 = 1.2 0. = 0.
2 , find the value of cos 3A
Solution: Given cos A =
cos 3A=4 cos^3 A 3 cos A
=4 (^)
3 (^3)
1 cos 2 = 2 sin^2
1 + cos 2 = 2 cos^2
2 sin 2 cos 2 + sin 2 2 cos^2 2 + cos 2
=
sin 2 (2 cos 2 + 1) cos 2 (2 cos 2 + 1)
=
sin 2 cos 2 = tan 2 = R.H.S. EXERCISE:
If cos = 0.4, find cos 3
If sin A = 0.4, find cos 2A using multiple angle formula
If sin A =
2 , find sin 3A.
5 , find the value of sin 2A.
If A = 60 verify the result sin 3A = 3 sin A 4 sin^3 A
Prove that
sin 2 + cos 1 cos 2 + sin = cot^
1 + sin 2A + cos 2A 1 + sin 2A – cos 2A = cot A
Let A be the given angle then
4 ,^ .........^ are called submultiples angles of A.
e.g. If A = 60 then
4 = 15^ ...... etc.
Trigonometric ratios of any angle in terms of its sub-multiple angle:
=2 cos^2 ( 𝜃 2 )^ ^1 = 1 2 sin^2 ( 𝜃 2 )
𝜃 2 )
2 𝑡𝑎𝑛(𝜃 2 ) 1 −𝑡𝑎𝑛^2 (𝜃 2 )
Solved Examples:
, find the value of cos A.
Solution : We know that
cos A =
1 – tan^2 A/ 1 + tan^2 A/
=
2
sin 1 + cos = tan
Solution : L.H.S.=
sin 1 + cos
2 sin /2 cos / 2 cos^2 /
sin = 2 sin /2 cos / 1 + cos = 2 cos^2 /
2 sin /2 cos / 2 cos /2 cos /
=
sin / cos / = tan ( 𝜃 2 ) EXERCISE:
If = 45, find the value of cos ( 𝜃 2 )
If tan
find the value of sin
1 + sin cos 1 + sin + cos = tan /
Prove that
cos A 1 sin A
1 + tan
1 tan