Showing posts with label trigonometry. Show all posts
Showing posts with label trigonometry. 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 MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES

HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES


Hello Friends welcome to this post of learning trigonometric formulas in an easy way with me . AS most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise value of different trigonometric angles in different quadrants and could not reproduce what they have learnt . So today we going to learn new techniques to learn "How to memorise different values of trigonometric angles in various quadrants" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.

When angle lies in 1st Quadrant

(1) When angle lies in 1st quadrant, then all the t- Ratios have positive values. As in 1st quadrant all the three  sides Perpendicular ,base and hypotenuse of  right angled triangle are positive.
      
(2) When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan  x to cot x , cos x to sin x ,cosec c to sec x and sec  x to cosec x.

HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES IN DIFFERENT QUADRANTS

sin (π/2 - x)     =  cos x
cos (π/2 - x)    =   sin x
tan (π/2 - x)     =  cot x
cot (π/2 - x)     =  tan x
sec (π/2 - x)     =  cosec x
cosec (π/2 - x) =  sec x

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x  to sin x  , cos x  to cos x  , tan x  to tan  x and so on.

sin (2π + x)      =    sin x
cos (2π x)    =     cos x
tan (2π x)      =    tan x
cot (2π x)     =     cot x
sec (2π x)     =     sec x
cosec (2π x) =     cosec x


ALSO READ   MATRIX , DIFFERENT TYPES OF MATRICES AND DETERMINANTS .

When angle lies in 2nd  Quadrant


(1)  when angle lies in 2nd quadrants ,then only two t - ratios sin x and it reciprocal cosec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 2nd  quadrant two  out of the three  sides Perpendicular and  hypotenuse of  right angled triangle are positive  while base is negative. So in all those  t -ratios ,when base involves  they will be negative. So  cos x, tan x ,cot x ,sec x involve with -ve value of base therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x to cosec x.
HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES IN DIFFERENT QUADRANTS


sin (π/2 + x)     =  sin x
cos (π/2 + x)    = -cos x
tan (π/2 + x)     =- cot x
cot (π/2 + x)     = -tan  x
sec (π/2 + x)     =  -cosec x
cosec (π/2 + x) = sec x

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (π - x)      =   sin x
cos (π - x)    =     cos x
tan (π - x)      = - tan x
cot (π - x)     =  - cot x
sec (π - x)     =   sec x
cosec (π - x) =  cosec x


When angle lies in 3rd Quadrant


(1)  when angle lies in 3rd quadrants ,then only two t - ratios tan x and it reciprocal cot x shall have +ve values and remaining t-Ratios shall have -ve values. As in 3rd  quadrant two  out of the three  sides Perpendicular and  base of  right angled triangle are negative  and hypotenuse is positive. So in all those  t -ratios ,when one of perpendicular or base  involves  they will be negative. So  sin x, cos x , sec x ,cosec x involve with -ve value of base or perpendicular therefore these t- ratios shall be negative. And tan x and cot x involves with both -ve values of perpendicular and base so they are positive in 3rd quadrant.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.
HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES IN DIFFERENT QUADRANTS

sin (3π/2 - x)     = - cos x
cos (3π/2 - x)    = -  sin x
tan (3π/2 - x)     =    cot x
cot (3π/2 - x)     =    tan x
sec (3π/2 - x)     = - cosec x
cosec (3π/2 - x) = - sec x


(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (π + x)      =   -sin x
cos (π x)    =    - cos x
tan (π x)      =   tan x
cot (π x)     =    cot x
sec (π x)     =  - sec x
cosec (π x) =  - cosec x

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Let us understand these learning of trigonometric formulae with the help of this video

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When angle lies in 4th Quadrant


(1)  when angle lies in 4th quadrants ,then only two t - ratios cos x and it reciprocal sec x shall have +ve values and remaining t-Ratios shall have -ve values. As in 4th quadrant two out of the three sides Base and hypotenuse of right angled triangle are positive and perpendicular is negative. So in all those t -ratios ,when perpendicular involves they will be negative. So sin x, tan x ,cot x ,cosec x involve with -ve value of perpendicular therefore these t- ratios shall be negative.

(2)  When angle involve  π/2 ,then T - Ratios on the left hand side  will changes accordingly , if  t-Ratios start with "co" , then co will be removed and if it is  without "co" then co will be added at the beginning .i.e. sin x will changes to cos x, tan x to cot x and sec x  to cosec x.
HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES IN DIFFERENT QUADRANTS

sin (3π/2 + x)     =   cos x
cos (3π/2 + x)    = - sin x
tan (3π/2 + x)     = - tan x
cot (3π/2 + x)     = - cot  x
sec (3π/2 + x)     =  - sec x
cosec (3π/2 + x) =  cosec x

So if we want to calculate sin 300° ,sin 240° and sin 330°  then it can be find out as follows


sin 300° = sin (270° + 30° ) = - cos 30° = -√3/2
sin 330° = sin (360° - 30° )  = - sin 30° = -1/2
sin 240° = sin (270° - 30° ) = - cos 30° = -√3/2

and if we want to calculate cos 300° , cos 240° and   cos 330°  then it can be find out as follows
cos 300° = cos (270° + 30° ) =  sin 30° = 1/2
cos 330° = cos (360° - 30° )  =  cos 30° = √3/2
cos 240° = cos (270° - 30° ) = - sin 30° = 1/2

(3)  When angle involve  π ,then T - Ratios on the left hand side  will NOT changes , put the same  t -ratios which is on the left hand side on right hand side i.e.  sin x to sin x , cos x to cos x , tan x to tan x  and so on.

sin (2π-x)      =   - sin x
cos (2π-x)    =      cos x
tan (2π-x)      =  - tan x
cot (2π-x)     =   - cot x
sec (2π-x)     =     sec x
cosec (2π-x) =  -  cosec x

sin (-x)      =   - sin x
cos (-x)    =      cos x
tan (-x)      =  - tan x
cot (-x)     =   - cot x
sec (-x)     =     sec x
cosec (-x) =  -  cosec x



GENERALISATION OF THE FORMULAE



So when an angle involves integral multiple of π , i,e -3π, -2π, -π, 2π, 3π, 4π then of the T-Ratios will change , But + or - sign can be added at beginning , e. g . sin(nπ + x) may change to + sin x or - sin x ,similarly cos (nπ + x) may change to + or - cos x depending upon the quadrant in which angle lies.


So if we want to find the value of sin 1110° ,then it can be written as sin (3×360° + 30°) = sin 30° = 1/2 . (As the angle is lying in 1st Quadrant )

Similarly if we want to find the value of sin 1050° ,then it can be written as sin (3×360° - 30°) = - sin 30° = - 1/2. (As the angle is lying in 4th Quadrant )


and when an angle involves odd integral multiple of π/2 i.e. (2n+1)π/2 , i . e -7π/2 , -5π/2 , -3π/2 , π/2 , 3π/2 , 5π/2.

Conclusion



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