Trigonometric Ratio and Identity

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Session 2
Trigonometric Ratio and Identity 2
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Topics
Multiple Angles
Sub Multiple Angles
Conditional Identities
Relation between sides and interior angle of
polygon
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Multiple Angles
J009
An angle of the form of 2A, 3A, 4A etc are
called multiple angles of A
Trigonometric ratios of 2A in terms of A
? sin 2A 2sinAcosA
? cos2A cos2A sin2A
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Multiple Angles
J009
sin2A and cos2A in terms of tanA
? sin 2A 2sinAcosA
Dividing Nr and Dr by cos2A we get
Similarly
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Illustrative Problem
J009
Solution
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Illustrative Problem
J009
Prove that
Solution
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Illustrative Problem
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Prove that
Solution
Ans.
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Multiple Angles
Trigonometric ratios of 3A in terms of A
? sin3A 3sinA 4 sin3A
Proof
sin3A sin(2AA)
sin2AcosA cos2AsinA
2sinAcosAcosA (12sin2A)sinA
2sinAcos2A sinA 2sin3A
2sinA(1-sin2A) sinA 2sin3A
3sinA 4 sin3A
Similarly
? cos3A 4cos3A 3cosA
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Illustrative Problem
J009
Solution
L.H.S sinA.sin(60oA).sin(60oA)
sinA sin260osin2A
sin(AB).sin(A-B) sin2Asin2B
Proved
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Illustrative Problem
J009
Solution
L.H.S sin 5? sin(3? 2?)
sin 3?cos2? cos3?sin2?
(3sin? 4sin3?) (1 2sin2?) (4cos3?
3cos?) (2sin?cos?)
(3sin? 4sin3?) (1 2sin2?) cos?(4cos2?
3) (2sin?cos?)
(3sin? 4sin3?) (1 2sin2?) 2sin?
(1sin2?) 4(1sin2?) 3
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Illustrative Problem
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Solution
L.H.S. (3sin? 4sin3?) (1 2sin2?)
2sin? (1sin2?) 4(1sin2?) 3
(3sin? 4sin3?) (1 2sin2?) (2sin?
2sin3?) (14sin2?)
3sin? 6sin3? 4sin3? 8 sin5? 2sin?
8sin3? 2sin3? 8sin5?
5sin? 20sin3? 16sin5? Proved
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Sub - Multiple Angle
An angle of the form of A/2, A/3, A/4 etc are
called sub - multiple angles of A
J0010
Trigonometric ratios of A in terms of A/2
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Sub - Multiple Angle
Trigonometric ratios of A in terms of A/2
J0010
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Sub - Multiple Angle
Trigonometric ratios of A in terms of A/3
J0010
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Illustrative Problem
Find the value of
J0010
(a) sin18o (b) cos18o (c) tan18o
Solution
(a) Let ? 18o ? 5? 90o
? 2? 3? 90o
? 2? 90o 3?
? sin2? sin(90o 3? )
? sin2? cos3?
? 2sin?cos? 4cos3? 3cos?
? 2sin? 4cos2? 3
? 2sin? 4(1-sin2?) 3
? 2sin? 4 4sin2? 3
? 4sin2? 2sin? 1 0
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Illustrative Problem
Find the value of
J0010
(a) sin18o (b) cos18o (c) tan18o
? 4sin2? 2sin? 1 0
This is a quadratic equation in sin?
Why ??
? sin? is positive
(b) cos218o 1-sin218o
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Illustrative Problem
Find the value of
J0010
(a) sin18o (b) cos18o (c) tan18o
Again ? 18o lies in the first quadrant
? cos18o gt 0
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Illustrative Problem
J0010
Show that
sin59o sin13o sin49o sin23o cos 5o
Solution
L.H.S
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Illustrative Problem
Show that
J0010
sin59o sin13o sin49o sin23o cos5o
Solution
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Illustrative Problem
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Solution
? from (1) and (2)
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Illustrative Problem
J0010
Solution
Proved.
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Conditional Identities
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Identities with certain given condition.
Conditions ABC ?, ?/2, 2?
Four Types of Problems
(1) Identities involving sines or cosines of
multiples and submultiples of the angle
(2) Identities involving squares of sines or
cosines of angle
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Conditional Identities
J0011
(3) Identities involving tangents or co-tangents
of angles
(4) Identities which can be solved by
transforming the given identities in
trigonometric form.
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Illustrative Problem
(1) Identities involving sines or cosines of
multiples and sub multiples of the angle
involved.
J0011
If ABC ?, prove that
cos2A cos2B cos2C 1-4sinAsinBcosC
Solution
L.H.S
(cos2A cos2B) cos2C
2cos(AB) cos(AB) (2cos2C 1)
2cos(? C) cos(AB) 2cos2C 1
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Illustrative Problem
J0011
Solution
2cosC cos(AB) 2cos2C 1
1-2cosC cos(A B) cos (?(AB))
1-2cosC cos(A B) cos (AB)
1 4sinAsinBcosC
Proved.
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Illustrative Problem
J0011
Solution
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Illustrative Problem
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Solution
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Illustrative Problem
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Solution
Proved.
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Illustrative Problem
J0011
(2) Identities involving squares of sines or
cosines of angle
Solution
L.H.S
cos2Acos2Bcos2C2cosAcosBcosC
cos2A(1sin2B)cos2C2cosAcosBcosC
1(cos2Asin2B)cos2C2cosAcosBcosC
1cos(AB)cos(AB)cos2C2cosAcosBcosC
1cos(?C)cos(AB)cos2C2cosAcosBcosC
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Illustrative Problem
J0011
Solution
1cos(?C)cos(AB)cos2C2cosAcosBcosC
1cosCcos(AB)cos2C2cosAcosBcosC
1cosCcos(AB)cos(AB)2cosAcosBcosC
12cosAcosBcosC 2cosAcosBcosC
1
Proved.
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Illustrative Problem
(3) Identities involving tangents or co-tangents
of angles
J0011
If ABC ?, prove that
cotBcotCcotCcotAcotAcotB 1
Solution
Proved.
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Illustrative Problem
(4) Identities which can be solved by
transforming the given identities in
trigonometric form.
J0011
Solution
Let xtanA, ytanB, ztanC
? tanAtanBtanBtanCtanCtanA 1
? tanB(tanAtanC) 1 tanAtanC
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Illustrative Problem
J0011
Solution
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Illustrative Problem
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Solution
putting the values of tanA, tanB, tanC
Proved.
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Relation Between side and interior angles of a
polygon
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No of Sides n
Let O be a point inside the polygon
Polygon n triangles
sum of all the angle of the triangle
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Relation Between side and interior angles of a
polygon
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Sum of all the angle of the triangle
Sum of all the angle of the triangle Sum of
all the angles at O Sum of all interior angle
of polygon

• Sum of all interior angle of polygon 3600
  nx1800

• Sum of all interior angle of polygon
• nx1800 - 3600

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Relation Between side and interior angles of a
polygon
J0011

• Sum of all the angles at O
• nx1800 - 3600

? Sum of interior angle of polygon
If the polygon is regular
Each interior angle
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Illustrative Problem
The angle in one regular polygon is to that in
another as 32 also the number of sides in the
first is twice that in second how many sides
the polygons have ?
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Solution
Let no. of sides of one regular polygon x
? No. of sides of other polygon 2x
No. of sides 4 ,8
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Class Exercise - 1
If cotA tanA 4, the value of cot2A is
Solution -
cot2A 2
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Class Exercise - 2
Solution -
LHS sin12 sin48 sin54
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Class Exercise - 3
Solution -
?2ab 2(sinx siny)(cosx cosy)
2(sinx cosx sinx cosy siny cosx siny cosx)
2 sinx cosx 2 siny cosy 2(sinx cosy cosx
siny)
sin2x sin2y 2 sin(x y)
2 sin(x y) . cos(x y) 2 sin(x y)
?2ab 2 sin(x y)1 cos(x y) ... (i)
and a2 b2 (sinx siny)2 (cosx cosy)2
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Class Exercise - 3
Solution -
sin2x sin2y 2 sinx siny cos2x cos2y
2 cosx cosy
2 2 cos(x y) 2(1 cos(x y) ... (ii)
From (i) and (ii)
sin(x y) LHS
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Class Exercise - 4
If a cos2? and bsin2? c has ? and ? as its
solution, then prove that
Solution -
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Class Exercise - 4
If a cos2? and bsin2? c has ? and ? as its
solution, then prove that
Solution -
? ? and ?are the roots of equation (i),
? tan? and tan? are the roots of equation (ii).
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Class Exercise - 5
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Class Exercise - 5
Solution -
(1 tanx)(1 3 tan2x) (9 tanx 3 tan3x)(1
tanx)
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Class Exercise - 5
Solution -
3 tan4x 6 tan2x 8 tanx 1 0 ... (ii)
??,?,? and ? are the roots of equation (i),
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Class Exercise - 6
Prove that
cos3A cos3A sin3A sin3A cos3 2A.
Solution -
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Class Exercise - 6
Prove that
cos3A cos3A sin3A sin3A cos3 2A.
Solution -
RHS
cos32A
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Class Exercise - 7
Solution -
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Class Exercise - 7
Solution -
RHS
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Class Exercise - 8
Solution -
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Class Exercise - 8
Solution -
LHS (cotB cotC)(cotC cotA)(cotA cotB)
cosecA cosecB cosecC
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Class Exercise - 9
Solution -
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Class Exercise - 9
Solution -
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Class Exercise - 10
The number of sides in two regular polygons are 5
4 and the difference between their angles is
9. Find the number of sides in the polygons.
Solution -
Let the number of sides of regular polygons are
5x and 4x respectively.
Each interior angle of regular polygons is
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Class Exercise - 10
The number of sides in two regular polygons are 5
4 and the difference between their angles is
9. Find the number of sides in the polygons.
Solution -
? Sides are 10 and 8.