Showing posts with label cbse. Show all posts
Showing posts with label cbse. Show all posts

HOW TO PROVE TRIGONOMETRIC IDENTITIES || TRIGONOMETRY


Proof of trigonometric identities , trigonometric identities problems, proving trigonometric identities formulas,these trigonometric identities of class 10, fundamental trigonometric identities,trigonometric identities class 11 and its formation with the help of some examples.

How to prove Identity


cos 6x = 32cos6 x - 48.cos4 x   + 18.cos2 x  - 6.cos2 x  - 1

Proof

1st of all  rewrite 3x as 3.2x

L.H.S. = cos 6x =  cos (3.2x) 

Now using the result cos 3θ = 4cos3 θ - 3 cos θ  -----(1)

Replacing θ as 2x in (1), we get 

L.H.S. = 4cos3 2x - 3 cos 2x  -----------(2)


Now using the result  1+ cos 2θ = 2 cos2 θ 

                                   ⇒ cos 2θ = 2 cos2 θ -1

Replacing cos 2x = 2 cos2 x -1 in (2), we get 

L.H.S.= 4 {2cos2 x -1}3 - 3 {cos2 x  -1}


Now using the result {a - b }3 = {a}3 - b }3  -  3{a }2 .b   + 3(a). b2

  cos 6x    = 4[ {2cos2 x  }3 - { 1 }3  -  3{2cos2 x  }2 .1   +3.(2cos2 x) .12 ] - 3 . {cos2 x  -1}


Taking the product of powers to simplify it

cos 6x  =   4[ 8cos6 x  - 1 - 12cos4 x  + 6cos2 x]  - 3{2cos2 x-1}

Multiply by 4 in 1st term and multiply by -3 in 2nd term

 cos 6x  = 32cos6 x  - 4 - 48cos4 x  + 24cos2 x  - 6cos2 x + 3

Adding the like powers terms and arranging in descending order

cos 6x   = 32cos6 x - 48cos4 x  + 18cos2 x  - 6cos2 x  - 1

Hence the Proof



Prove the Identity 

tan (2x) =  2tan x  1 - tan2 x 

Proof

We know that 


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

Put A = B  = x in above formula . then it becomes

tan (x+x) =  tan x +  tan x1 - tan x tan x 


tan (2x) =  2tan x  1 - tan2 x 
Hence the Proof


Prove that sin 2x = 2sin x cos x

Proof


As we know that sin (A + B) = sin A cos B + cos A sin B..  ...(1)

Put A = B  = x in ...   (1)

sin (x + x) = sin x cos x + cos x sin x

sin (2x) = sin x cos x +  sin x cos x

sin (2x) = 2 sin x cos x

Hence the Proof


Prove that cos 2x = cos2 x - cos2 x

Proof


As we know that cos (A + B) = cos A cos B - sin A sin B..  ...(1)
Put X = A = B in (1) , we get

cos (x + x) = cos x cos x - sin x sin x

cos 2x = cos2 x - sin2 x   

cos 2x = cos2 x - sin2 x   


Hence the Proof


                                                                                                                                                                   

 Using the result 
1+cos 2θ = 2cos2 θ
cos 2θ = 2cos2 θ -1 -------------(1)
Replacing θ with 2x in eq (1)
1+ cos 4x = 2cos2 2x
cos 4x = 2cos2 2x -1

Again using  cos 2θ = 2cos2 θ -1

cos 4x = 2Sq(2cos2 x -1) -1

It is the square of 2cos2 x -1

cos 4x = 2Sq(2cos2 x -1) -1

cos 4x = 2(4cos4 x +1 - 4cos2 x) -1

cos 4x = 8cos4 x +2 - 8cos2 x -1

cos 4x =  8cos4 x - 8cos2 x +1

Hence the Proof


What is the value of sin3x?



To find the value of sin 3x ,  use this formula which contain sin (A+B)
therefore sin (A+B) = sin A cos B cos A sin B——-(1)
put A = 2x and B = x in (1)
then Sin 3x = sin 2x cos x + cos 2x sin x

As we know that cos 2x = 1 - 2sin3 x and sin 2x = 2 sin x cos x


sin 3x = (2 sin x cos x) cos x + (1 - 2sin3 x ) sin x
sin 3 x = 2 sin x cos2 x + sin x -  2sin3 x

As we know that cos2 x = 1sin2 x

sin 3x= 2 sin x (1-sin3 x) + sin x - 2sin3 x
sin 3x = 2 sin x -2 sin3 x + sin x - 2sin3 x
sin 3x = 3 sin x - 4 sin3 x

Similarly we prove that cos 3x= 4 cos3 x - 3 cos x
For learning and memorising more trigonometric formulas

Conclusion




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HOW TO LEARN INTEGRATION FORMULAE/FORMULAS USING TRICKS

Let us learn and remember most Important formulas of Integration , tips and tricks to learn algebraic ,most important differentiation questions for plus 2 maths, indefinite integration tricks and shortcuts trigonometric and by parts formulas in an easy and short cut manners.


Trigonometric Formulae



  1  ∫ sin x dx           =  - cos x +c  


where "c" is called constant of Integration.

The integration of sin x is  - cos x ,then divide it with the derivative of its angle. 


If we have to find the integration of  sin 2x , then we shall find it as


Step1    1st find the integration of sin x which is - cos x .

Step2    Divide it with the derivative of 2x ,which is 2, so 


∫ sin 2x dx     =  - ( cos 2x) 2 + c ,
   ∫ sin 8x dx      =  - ( cos 8x) 8 + c ,

∫ sin  3x4  dx  =   - ( cos  3x4 )   3 4  + c ,
Therefore   ∫ sin nx dx =  - ( cos nx) n +c ,






2  ∫ cos x dx          =    sin x +c  

The integration of cos x is  sin x ,then divide it with the derivative of its angle.
If we have to find the integration of  cos 2x , then we shall find it as

Step 1   1st find the integration of cos x which is sin x .

Step 2   Divide it with the derivative of 2x ,which is 2, so 


∫ cos 2x dx     =  ( sin 2x) 2 + c ,

∫ cos 8x dx     =  ( sin 8x) 8 + c ,

∫ cos  3x4  dx  = ( cos  3x4 )   3 4  + c ,

Therefore   ∫ cos nx dx =   ( sin nx) n +c ,



3  ∫ tan x dx = log |sec x| +c or - log |cos x| + c


The integration of tan x is   log |sec x| +c   or -log |cos x| +c  , then divide it with the derivative of its angle.


If we want to find the integration of tan  x2  . The integration of  tan  x2 is  log |sec x2 | +c , then divide it with the derivative of its angle.

Step 1

Find the integration of  tan  x2 ,which is log |sec  x2 | or  - log |cos  x2 |,


Step 2
 Divide it with the derivative of angle x2 ,which is 2x.

Therefore
∫ tan  x2  dx = -(1/2x) log |cos x2 | +c or (1/2x)log |sec x2 | + c

4  ∫ cot x dx = log |sin x| +c or - log |cosec x| + c


 The integration of cot x is   -log |cosec x| +c   , then divide it with the derivative of its angle.



If we want to find the integration of  cot x2  . Then integration of  cot  x2 is  -log |cosec x2 | +c  or  log |sin x2 | +c  , then divide it with the derivative of its angle.

Step 1 

 Find the integration of  cot  x2 ,which is -log |coec  x2 | or  log |sin x2 |,

Step 2

  Divide it with the derivative of angle x2 , which is 2x.   

Therefore      ∫ cot  x2  dx = -(1/2x) log |cosec x2 | +c or (1/2x ) log |sin x2 | + c



 5  ∫ sec x dx         =  log |sec x - tan x | +c 


If we want to integrate sec√x .Then 1st of all we apply the formula of integration of sec(any angle) then divide with the formula of integration of √x,So we have


 ∫  sec√x dx = ( log |sec√x - tan √x | )(2√x) +c


6 ∫ cosec x dx = - log |cosec x - cot x | +c

If we want to integrate cosec√x .Then 1st of all we apply the formula of integration of cosec (any angle) then formula of integration of √x,So we have



 ∫  cosec√x dx = - (log | cosec√x -co√x | )(2√x) +c
  

7  ∫ sec2 x dx = tan x + c


Because the derivative of tan x is sec 2 x , So the Antiderivative or Integration of sec 2 x  is tan x .




∫ sec 2 √x dx     =  (2√x ) tan √x  + c

∫ cosec 2 dx = - cot x +c



Because the derivative of  cot x is  - cosec 2 x , So the  Anti derivative or Integration of  cosec 2 x is - cot x .









 8 ∫ sec x tan x dx = sec x +c



Because the derivative of sec x is sec x tan x ,Therefore the integration of tan x sec x is sec 
x .


If we want to integrate sec√x .tan √x .Then its  integration  will be sec √x,


    sec √x tan √x   dx      =  √x   sec √x + c  


9 ∫ cosec x cot x dx = - cosec x +c


Because the derivative of cosec x is    - cosec x cot x , Therefore the integration of tan x sec x is sec x .


∫ cosec √x cot √x dx = - ( 2 √x  cosec √x ) +c 


 Also Read   WHAT IS SET, TYPES OF SETS, UNION INTERSECTION AND VENN DIAGRAMS


Algebraic Formulae


1 ∫ (constant) dx = (constant ) x +c

Integration of constant function is the constant function itself multiplied by the variable .

∫ 5 dx   = 5x  +c

2  ∫  xn  dx  = xn+1  n+1dx  + c ,

∫ x3   dx  =  x4  4  + c ,


HOW TO LEARN INTEGRATION  FORMULAE/FORMULAS USING TRICKS

   
To find the integration of function where variable "x" or f(x) has power 'n' , where "n" is any real number, we shall increase the power of "x"  by 1 and divide it with increased power.
e.g 

 ∫  x2  dx   =  {1/(2+1)} x2+1  + c

∫  (x ) 2/3   dx     =   (x ) (2/3)+1  (2/3)+1 + c ,

                  =   3(x ) 5/3  5 + c ,

∫  (ax+b ) n   dx     =   (ax+b ) n+1  a(n+1) + c ,

∫  (3x + 7 ) 2   dx     =   (3x+7 )2+1  3(2+1) + c ,

                              =   (3x+7 )3  9 + c ,

If we have to integrate sum of two functions ,then we shall integrate it separately as follows

4  ∫  [ f(x) + g(x)] dx = ∫f(x)dx +  g(x) dx + c


∫  [{ x2}3  + (2x) ]dx = ∫ { x2}3 dx +  ∫ (2x) dx

  =∫   x6dx +  ∫ 2x dx =  
 x  6+1  6+1 +(2/2)x2  + c

 =   x  7  7 x2  + c

∫  {4x2  + 3x }dx = 4x2   + ∫ 3x dx
                         =  4×(1/3)x3  (3/2)x2 +c


 ∫ [ 6x / 3x2] dx = log |  3x2 | + c

HOW TO LEARN INTEGRATION  FORMULAE/FORMULAS USING TRICKS




Memorize  these integration formulas along with differentiation in Hindi




Integration By Parts 

∫ [ f(x) g(x)] dx = f(x) ∫ g(x) dx -  {f '(x) ∫ g(x) dx}dx + c


    
∫ x sin x dx = x ∫ sin x dx - x' { ∫ sin x dx}dx + c
= x(-cos x) -  (-cos x)dx + c
 = -x cos x - sin x +c

∫ log x dx =  ∫ log x.1 dx 
= log x  - f '(log x) ( x )dx + c
= log x .1  - ∫(1/x)  x dx + c  
= log x  - ∫ 1 dx + c 
= log x  - x  + c


 Integration   Exponential Function

1  ∫  ex  dx   =  ex  + c

2 ∫  ax  dx = ax / log a    + c  

3 ∫ log x dx = x log x - x + c

4 ∫ (1/x ) dx = ln |x | + c




Exponential and Derivative Mixed Formula

   ex  [ f(x) + f '(x)] dx =ex  f(x) + c


  ex  [ sin x + cos x] dx = ex  sin x + c

Conclusion


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