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Trigonometric Functions of Sum and Difference of Two Angles (Part II)

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Trigonometric Functions of Sum and Difference of Two Angles (Part II) - Lesson Summary

Proving that cos 2A = cos2 A - sin2 A = 2 cos2 A - 1 = 1 - 2 sin2 A = 1 - tan 2 A 1 + tan 2 A  

cos (A + B) = cos A cos B - sin A sin B

Replacing B by A,

cos (A + A) = cos A cos A - sin A sin A 

cos 2A = cos2A - sin2 A

sin2 A + cos2 A = 1, sin2 A =1 - cos2 A

cos2 A = 1- sin2 A

cos 2A = cos2 A - sin2 A

= cos2 A - (1 - cos2 A)

= 2 cos2 A - 1

= 2(1 - sin2 A) - 1

= 1 - 2 sin2 A

Consider cos 2A = cos2A - sin2A

= cos2 A ( cos 2 A - sin 2 A cos 2 A )

=   cos 2 A cos 2 A - sin 2 A cos 2 A sec 2 A  

=       1 - tan 2 A sec 2 A

But sec2 A = 1 + tan2 A

cos 2A =      1 - tan 2 A 1 + tan 2 A  

Proving that  sin 2A = 2 sin A cos A =  2 tan A 1 + tan 2 A .

sin(A + B) = sin A cos B + cos A sin B

Replacing B by A, sin(A + A) = sin A cos A + cos A sin A

⇒ sin 2A = 2 sin A cos A = 2 Sin  A Cos A Cos A Cos  A   
= (2 Sin  A Cos  A  )cos2 A

=  2 tan A 1 + tan 2 A

sin 2A =  2 tan A 1 + tan 2 A    (∵ sec2A = 1 + tan2A)

Proving that tan 2A =  2 tan A 1 -  tan 2 A   

tan(A + B) = tan  A + tan B1 – tan  A tan B   

Replacing B by A, we get

tan(A + A) =  tan  A + tan A1 – tan  A tan A 

⇒ tan 2A =  2 tan A 1 -  tan 2 A  

Proving that sin 3A = 3 sin A - 4 sin3 A

Consider, sin 3A = sin(2A + A)

sin(A + B) = sin A cos B + cos A sin B

= sin 2A cos A + cos 2A sin A

= (2sin A cos A) cos A + (1 - 2 sin2 A) sin A        (∵sin 2A = 2 sin A cos A and cos 2A = 1 - 2 sin2A)

= 2 sin A cos2 A + (1 - 2 sin2 A) sin A

= 2 sin A (1 - sin2 A) + (1 - 2 sin2 A) sin A

= 2 sin A - 2 sin3 A + sin A - 2 sin3 A

⇒ sin 3A = 3 sin A - 4 sin3 A

Proving that cos 3A = 4 cos3 A - 3 cos A

LHS: cos 3A = cos (2A + A)

∴ Cos (A + B) = Cos A Cos B - Sin A Sin B

= cos 2A cos A - sin 2A sin A

Using Cos 2A = 2 Cos2 A - 1 and Sin 2A = 2 Sin A Cos A, we get

cos 3A = (2 cos2 A - 1) cos A - (2 sin A cos A) sin A

= (2 cos2 A - 1) cos A - 2 sin2 A cos A

= (2 cos2 A - 1) cos A - 2 cos A (1 - cos2 A)

= 2 cos3 A - cos A - 2 cos A + 2 cos3 A

Proving that tan 3A = 3 tan A - tan 3 A 1 - 3 tan 2 A

LHS : tan 3A = tan (2A + A)

tan(A + B)  =  tan  A + tan B1 – tan  A tan B     

tan(2A + A) =  tan  2A + tan A1 – tan  2A tan A   

                    = 2 tan A 1 - tan 2 A + tan A 1 - 2 tan A 1 - tan 2 A .tan A              ∵ tan 2A = 2 tan A 1 - tan 2 A   

                  =  2 tan A + tan A - tan 2 A 1 - tan 2 A - 2 tan 2 A

⇒ tan 3A   =  3 tan A - tan 3 A 1 - 3 tan 2 A

Proof of some more results:

sin(A + B) + sin(A - B) = 2 sin A cos B

LHS: sin(A + B) + sin(A - B)

= sin A cos B + cos A sin B + sin A cosB - cos A sin B

= 2 sin A cos B

∴ sin(A + B) + sin(A - B) = 2 sin A cos B


ii) sin(A + B) - sin(A - B) = 2 cos A sin B

LHS: sin(A + B) - sin(A - B)

= (sin A cos B + cos A sin B) - (sin A cos B - cos A sin B)

= sin A cos B + cos A sin B - sin A cos B + cos A sin B

= 2 cos A sin B

∴ sin(A + B) - sin(A - B) = 2 cos A sin B

iii) cos(A + B) + cos(A - B) = 2 cos A cos B

LHS: cos(A + B) + cos(A - B)

= cos A cos B - sin A sin B + cos A cos B + sin A sin B

= 2 cos A cos B

∴ cos(A + B) + cos(A - B) = 2 cos A cos B


iv) cos(A + B) - cos(A - B) = -2 sin A sin B

LHS : cos (A + B) - cos(A - B)

= (cos A cos B - sin A sin B) - (cos A cos B + sin A sin B)

= cos A cos B - sin A sin B - cos A cos B - sin A sin B

= -2 sin A sin B

∴ cos(A + B) - cos(A - B) = -2 sin A sin B

v) Let A + B = C and A - B = D

Solving for A and B, we get

A =  C + D 2    and B =   C  – D 2

Substituting these values in the identities of result 13 one by one, we get

sin (A + B) + sin (A - B) = 2 sin A cos B

⇒ sin C + sin D = 2 sin   C + D 2 . cos   C  – D 2

sin (A + B) - sin (A - B) = 2 cos A sin B

⇒ sin C - sin D = 2 cos   C + D 2 . sin   C  – D 2

cos (A + B) + cos (A - B) = 2 cos A cos B

⇒ cos C + cos D = 2 cos   C + D 2 . cos   C  – D 2

cos (A + B) - cos (A - B) = -2 sin A sin B

⇒ cos C - cos D = -2 sin   C + D 2 . sin   C  – D 2

sin (A+B) sin (A - B) = sin2 A - sin2 B = cos2 B - cos2 A

vi) Consider, sin (A + B) sin (A - B)

= (sin A cos B + cos A sin B) (sin A cos B - cos A sin B)

= (sin2 A cos2 B - sin A cos A sin B cos B + sin A cos A sin B cos B - cos2 A sin2 B)

= sin2 A cos2 B - cos2 A sin2 B

= sin2 A (1 - sin2 B) - (1 - sin2 A) sin2 B

= sin2 A - sin2 A sin2 B - sin2 B + sin2 A sin2 B

= sin2 A - sin2 B

= sin2 A - sin2 B = (1 - cos2 A) - (1 - cos2 B) = cos2 B - cos2 A

vii) cos (A + B) cos (A - B) = cos2 A - sin2 B = cos2 B - sin2 A

Consider, cos (A + B) cos (A - B)

= (cos A cos B - sin A sin B) (cos A cos B + sin A sin B)

= cos2 A cos2 B - cos A cos B sin A sin B + sin A sin B cos A cos B - sin2 A sin2 B

= cos2 A cos2 B - sin2 A sin2 B

= cos2 A (1 - sin2 B) - (1 - cos2 A) sin2 B

= cos2 A - cos2 A sin2 B - sin2 B + cos2 A sin2 B

= cos2 A - sin2 B

cos2 A - sin2 B = (1 - sin2 A) - (1 - cos2 B)

= cos2 B - sin2 A


viii) tan( π 4 + A) =  1 + tan A 1 – tan A   =  cos A + sin A cos A – sin A  

LHS = tan( π 4 + A) = (tan  π 4 + tan A)/(1 - tan π 4 .tan A)

= 1 + tan A 1 – tan A 

= 1 + Sin A Cos A 1 – Sin A Cos A    (Hence tan A = Sin A Cos A )

= Cos A + Sin A Cos A – Sin A  

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