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Trigonometry (Sum and Differece Formulas)

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The students will learn

  • trigonometric ratios of the sum or difference of two arcs,
  • half-angle formulas,

  • conversion and inverse conversion formulas

given examples with solutions

1.  cos(a+b) = cos a. cos b - sin a. sin b

2.  cos(a-b) = cos a. cos b + sin a. sin b

3.  sin(a+b) = sin a. cos b + cos a. sin b

4.  sin(a-b) = sin a. cos b - cos a. sin b

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«Trigonometry (Sum and Differece Formulas)»

Sum and Difference Formulas in Trigonometry

Sum and Difference Formulas in Trigonometry

Introduction The students will learn

Introduction

The students will learn

  • trigonometric ratios of the sum or difference of two arcs,
  • half-angle formulas,
  • conversion and inverse conversion formulas .
Trigonometric Ratios of  The Sum or Difference of Two Arcs Theorem 1.4 Theorem 1.5 Theorem 1.6 Theorem 1.7

Trigonometric Ratios of The Sum or Difference of Two Arcs

Theorem 1.4

Theorem 1.5

Theorem 1.6

Theorem 1.7

Theorem 1.4 If a and b are two real numbers, then Proof-1 1. cos(a+b) = cos a . cos b - sin a . sin b Example-1 Proof-2 2. cos(a-b) = cos a . cos b + sin a . sin b Example-2 Proo f -3 3. sin(a+b) = sin a . cos b + cos a . sin b Example-3 Proof-4 4. sin(a-b) = sin a . cos b - cos a . sin b Example-4 BACK Theorem 1.5

Theorem 1.4

If a and b are two real numbers, then

Proof-1

1. cos(a+b) = cos a . cos b - sin a . sin b

Example-1

Proof-2

2. cos(a-b) = cos a . cos b + sin a . sin b

Example-2

Proo f -3

3. sin(a+b) = sin a . cos b + cos a . sin b

Example-3

Proof-4

4. sin(a-b) = sin a . cos b - cos a . sin b

Example-4

BACK

Theorem 1.5

Proof-1  cos(a+b) = cos a . cos b - sin a . sin b Full Screen Since the lenghts of arc AP and AP’ are equal then  AR  =  P’Q  . If  is the distance between two points (x 1 ,y 1 ) and (x 2 ,y 2 ),  then the distance formula will be as follow s    then  AR  =  P’Q  AR  2 =  P’Q  2  (cos(a+b)-1) 2 + (sin(a+b)- 0) 2 = (cos a – cos b) 2 + (sin b + sin a) 2  (cos(a+b)-1) 2 + (sin(a+b)- 0) 2 = (cos a – cos b) 2 + (sin b + sin a) 2  cos 2 (a+b) – 2cos(a+b) + 1 + sin 2 (a+b) = cos 2 a –2cosa.cosb + cos 2 b + sin 2 b + 2sina. sinb + sin 2 a   2 – 2cos(a+b) = 2 – 2cosa.cosb + 2sinb.sina    cos(a+b) = cosa. cosb – sina. sinb  BACK Example-1

Proof-1 cos(a+b) = cos a . cos b - sin a . sin b

Full Screen

Since the lenghts of arc AP and AP’ are equal then  AR  =  P’Q  .

If is the distance between two points (x 1 ,y 1 ) and (x 2 ,y 2 ),

then the distance formula will be as follow s then

 AR  =  P’Q  AR  2 =  P’Q  2

(cos(a+b)-1) 2 + (sin(a+b)- 0) 2 = (cos a – cos b) 2 + (sin b + sin a) 2

  • (cos(a+b)-1) 2 + (sin(a+b)- 0) 2 = (cos a – cos b) 2 + (sin b + sin a) 2

cos 2 (a+b) – 2cos(a+b) + 1 + sin 2 (a+b) = cos 2 a –2cosa.cosb + cos 2 b + sin 2 b + 2sina. sinb + sin 2 a

2 – 2cos(a+b) = 2 – 2cosa.cosb + 2sinb.sina

cos(a+b) = cosa. cosb – sina. sinb

BACK

Example-1

Full screen BACK

Full screen

BACK

Example-1: cos 75 o = ? Solution-1

Example-1:

cos 75 o = ?

Solution-1

Solution-1: cos 75º = cos ( 45º + 30º )  = cos 45º.cos 30º - sin 45º. sin 30º  =  = BACK

Solution-1:

cos 75º = cos ( 45º + 30º )

= cos 45º.cos 30º - sin 45º. sin 30º

=

=

BACK

Proof-2   cos(a-b) = cos a . cos b + sin a . sin b Since  cos(a+b) = cosa . cosb – sina . s inb  is valid far all real numbers a and b, t hen  cos(a+(-b)) = cosa . cos(-b) – sina . sin(-b)   cos(a-b) = cosa . cosb + sina . sinb  BACK Example-2

Proof-2 cos(a-b) = cos a . cos b + sin a . sin b

Since cos(a+b) = cosa . cosb – sina . s inb is valid far all real numbers a and b,

t hen cos(a+(-b)) = cosa . cos(-b) – sina . sin(-b)

cos(a-b) = cosa . cosb + sina . sinb

BACK

Example-2

Example-2: cos 15 o = ? Solution-2

Example-2:

cos 15 o = ?

Solution-2

Solution-2: cos 1 5 º = cos ( 45 º  - 30 º )  = cos 45 º .cos 30 º  + sin 45 º . sin 30 º  =  = BACK

Solution-2:

cos 1 5 º = cos ( 45 º - 30 º )

= cos 45 º .cos 30 º + sin 45 º . sin 30 º

=

=

BACK

Proof-3   sin(a+b) = sin a . cos b + cos a . sin b We can write sine in terms of cosine by using reduction formulas. Hence, sin(a+b) = sin a . cos b + cos a. sin b  BACK Example-3

Proof-3 sin(a+b) = sin a . cos b + cos a . sin b

We can write sine in terms of cosine by using reduction formulas. Hence,

sin(a+b) = sin a . cos b + cos a. sin b

BACK

Example-3

Example-3: sin 105 o = ? Solution-3

Example-3:

sin 105 o = ?

Solution-3

Solution-3: sin  10 5º = cos ( 60 º +  45 º )  = sin  60 º.cos 45 º +  cos  60 º. sin 45 º  =  = BACK

Solution-3:

sin 10 5º = cos ( 60 º + 45 º )

= sin 60 º.cos 45 º + cos 60 º. sin 45 º

=

=

BACK

Proof-4   sin(a+b) = sin a . cos b - cos a . sin b Since the equality sin(a+b) = sin a . cos b + cos a. sin b is valid for all real numbers a and b, by writing –b instead of b, we get sin(a+ (-b) ) = sin a . cos (- b ) + cos a. s in (- b ) then  sin(a+b) = sin a . cos b - cos a  . sin b  BACK Example-4

Proof-4 sin(a+b) = sin a . cos b - cos a . sin b

Since the equality sin(a+b) = sin a . cos b + cos a. sin b is valid for all real numbers a and b, by writing –b instead of b, we get

sin(a+ (-b) ) = sin a . cos (- b ) + cos a. s in (- b ) then

sin(a+b) = sin a . cos b - cos a . sin b

BACK

Example-4

Example-4: sin 15 o = ? Solution-4

Example-4:

sin 15 o = ?

Solution-4

Solution-4: sin  1 5º = cos ( 60 º -  45 º )  = sin  60 º.cos 45 º -  cos  60 º. sin 45 º  =  = BACK

Solution-4:

sin 1 5º = cos ( 60 º - 45 º )

= sin 60 º.cos 45 º - cos 60 º. sin 45 º

=

=

BACK

Theorem 1.5 1. If a, b and a+b real numbers different from  /2+k  , k  Z, then Proof-5 Example-5 2. If a, b and a-b real numbers different from  /2+k  , k  Z, then Proof-6 Example- 6 BACK Theorem 1.6

Theorem 1.5

1. If a, b and a+b real numbers different from  /2+k  , k  Z, then

Proof-5

Example-5

2. If a, b and a-b real numbers different from  /2+k  , k  Z, then

Proof-6

Example- 6

BACK

Theorem 1.6

Proof-5 BACK Example-5

Proof-5

BACK

Example-5

Example-5: Let  and  be two acute angles.  If sin  = 3/5 and cos  = 5/13, then find tan(  +  ). Solution-5

Example-5:

Let  and  be two acute angles. If sin  = 3/5 and cos  = 5/13, then find tan(  +  ).

Solution-5

Solution-5: BACK

Solution-5:

BACK

Proof-6 2. If we write –b instead of b in the previous equality then, BACK Example-6

Proof-6

2. If we write –b instead of b in the previous equality then,

BACK

Example-6

Example-6: In triangle ABC, if m(CAD)=  , then find tan  . Solution-6

Example-6:

In triangle ABC, if m(CAD)= , then find tan .

Solution-6

Solution-6: tan  = tan (a – b) BACK

Solution-6:

tan  = tan (a – b)

BACK

Theorem 1.6 1. If a, b and a+b real numbers different from k  , k  Z, then Proof-7 Example-7 2. If a, b and a-b real numbers different from k  , k  Z, then Proof-8 Example-8 BACK Theorem 1.7

Theorem 1.6

1. If a, b and a+b real numbers different from k  , k  Z, then

Proof-7

Example-7

2. If a, b and a-b real numbers different from k  , k  Z, then

Proof-8

Example-8

BACK

Theorem 1.7

Proof-7 BACK Example-7

Proof-7

BACK

Example-7

Example-7: In the adjoining figure  AB  = 4,  BC  = 3,  AD  = 12 and m(ABC)=90º, m(CAD)=90º, m(CED)=90º. Find cot  , if m(DCE) =  . Solution-7

Example-7:

In the adjoining figure  AB  = 4,  BC  = 3,  AD  = 12 and m(ABC)=90º, m(CAD)=90º, m(CED)=90º. Find cot  , if m(DCE) =  .

Solution-7

Solution-7:  AC  = 5 and  CD  = 13.  Put m(ACB) = a and m(ACD) = b then,    = 180  - (a + b)  cot  = cot (180  - (a + b))  = - cot (a + b)  BACK

Solution-7:

 AC  = 5 and  CD  = 13.

Put m(ACB) = a and m(ACD) = b then,

 = 180  - (a + b)

cot  = cot (180  - (a + b))

= - cot (a + b)

BACK

Proof-8 2. If we write –b instead of b in the previous equality then, BACK Example-8

Proof-8

2. If we write –b instead of b in the previous equality then,

BACK

Example-8

Example-8: Adjoining figure consists of three equivalent squares. Find cot  , if m(CAE)=  Solution-8

Example-8:

Adjoining figure consists of three equivalent squares. Find cot  , if m(CAE)= 

Solution-8

Solution-8: Let x be the lenght of the sides of each equivalent square. If m(BAC) = 45º, and m(BAE) =   then,    =  - 45  cot  = cot (  - 45  )  =  = BACK

Solution-8:

Let x be the lenght of the sides of each equivalent square.

If m(BAC) = 45º, and m(BAE) =  then,

 =  - 45 

cot  = cot (  - 45  )

=

=

BACK

Theorem 1.7 1. If a, b and a+b real numbers different from k  , k  Z, then Proof-7 Example-7 2. If a, b and a-b real numbers different from k  , k  Z, then Proof-8 Example-8 BACK

Theorem 1.7

1. If a, b and a+b real numbers different from k  , k  Z, then

Proof-7

Example-7

2. If a, b and a-b real numbers different from k  , k  Z, then

Proof-8

Example-8

BACK


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Предмет: Математика

Категория: Презентации

Целевая аудитория: 9 класс

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Trigonometry (Sum and Differece Formulas)

Автор: Balkhozhayeva Aliya Buribayevna

Дата: 30.01.2016

Номер свидетельства: 285244


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