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1

Arjuna JEE 2
...


2
5 −1
• sin18 = sin = cos72 = cos
=
10
5
4

• sin3 = 3sin  − 4sin 
3

• cos3 = 4cos3  − 3cos 
• tan3 =


3
5 +1
= sin54 = sin =
5
10
4

5
• sin15 = cos75 = sin = cos
12
12
3 −1
6− 2
=
=
4
2 2

3tan  − tan 2 
1 − 3tan 2 

• cos36 = cos

Hamaray Sath(8) Yaar

1
• Y1 : sin  sin(60 − )sin(60 + ) = sin3
4
1
• Y2 : cos  cos(60 − )cos(60 + ) = cos3
4
• Y3 : tan  tan(60 − ) tan(60 + ) = tan3

• cos15 = sin 75 = cos

3 +1
6+ 2
=
4
2 2

5
• tan15 = tan = cot 75 = cot
=2− 3
12
12

5
• cot15 = cot = tan 75 = tan
=2+ 3
12
12
=

• Y4 : cot  cot(60 − )cot(60 + ) = cot 3
• Y5 : tan  + tan(60 + ) + tan(120 + ) = 3tan 3
• Y6 : cot  + cot(60 + ) + cot(120 + ) = 3cot 3
• Y7 : cot + tan = 2cosec2
• Y8 : cot  − tan  = 2cot 2
Golden Point
...
+ cos n
r =1

▪
▪
▪
▪
▪
▪

sin  = 0,  ∈ n
cosec  is ND,  ∈ n
cot  is ND,  ∈ n
tan  = 0,  ∈ n
cos  = 1,  ∈ (Even ), cos  = –1,  ∈ (Odd )
sec  = 1,  ∈ (Even ), cos  = –1,  ∈ (Odd )

▪

sin θ = 1, θ ∈

▪

n

sin(r) = sin  sin 2 sin 3
...
5 = cot
= 2 −1
8
8

3
• cot 22
...
5 = tan
= 2 +1
8
8
• tan 22
...
cos2n+1 =


5
= sin
12
12

sin 2
2n sin 

▪

( 4n + 1) 

2
( 4n − 1) 
sin θ = –1, θ ∈
2
( 4n + 1) 
cosec θ = 1, θ ∈
2
( 4n − 1) 
cosec θ = –1, θ ∈
2
( 2n + 1) 
cos θ = 0, θ ∈
2
( 2n + 1) 
cot θ = 0, θ ∈
2
( 2n + 1) 
sec , tan  = ND, θ ∈
2

3

Conditional Identities
Note: If A + B + C = 
• sin (A + B) = sin ( – C) = sin C
• cos (B + C) = cos ( – A) = – cos A
• tan (C + A) = tan ( – B) = – tan B

Splitting the nth term as Difference of Two Terms
Observe!!

C
 A+ B 
 −C 
 C
= cos 
= cos  −  = sin
• cos 


2
 2 
 2 
2 2
A
 A+ B 
 − A
  A
= sin 
= cos  −  = − cos
• sin 


2
 2 
 2 
2 2
1
...

3
...

5
...

7
...
sin (a + (n – 1)d)
  a + a + (n − 1)d 
nd 
  sin 2 
sin 
2


= 


d
sin


2
S = cos a + cos (a + d) + cos (a + 2d)
...

Hathiyaar 3:
If the argument of sin and cos are different or we are
given a quadrant in sin/cos, then we make a perfect
square
...


PW Web/App - https://smart
...
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