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Showing posts sorted by relevance for query trigonometry. Sort by date Show all posts

## Memorise A B AND C D Formulas In Trigonometry in an Easy Manner

Welcome to this post of learning trigonometric formulas  .Most of the Students or Mathematics Learner ,most of the time confuse to remember or memorise A  B and C  D formulas, They  mixed A B and C D formulas with each other and could not reproduce what they have learnt . So today we going to learn new techniques to learn "How to memorise AB and CD formulas" forever. Before this we must have knowledge of different trigonometric values of different angles in different quadrants.
First of all have a quick look at some of  these formulas

## AB Formulas

S N Fomula Result
1 sin (A + B) sin A cos B + cos A sin B
2 sin (A - B) sin A cos B - cos A sin B
3 cos (A + B) cos A cos B - sin A sin B
4 cos (A - B) cos A cos B + sin A sin B

To clear all your doubts on   " How to Calculate Different Trigonometric values in different quadrants "  in an easy Method. Click on the  above  links  .

## Tricks to Learn    A  B   Formulae  For  sine  angles

When angles are added  i. e  Sin  ( A+B )
When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Sine of  added angles, then it can be done like this.

Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Cosine of angle A and multiply with Sine of angle B. i.e. start with sine and ends with sine and in middle both the terms are cosine ,and angles start  A then B again A then again B.

## Sin (A+B) = Sin A Cos B + Cos A Sin B

When angles are subtracted    i. e  Sin  ( A-B )

When Angles are subtracted and  their Trigonometric Ratios is taken , and if we have to take the  Sine of  subtracted  angles, then it can be done like this

Start with  sine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign    Start with Cosine of angle A and multiply with Sine of angle B. i.e  start with sine and ends with sine and in middle both the terms are cosine ,and angles start  with A then B again A then again B.

## Tricks to Learn    A  B   Formulae For  Cosine  angles

When angles are added   i. e  Cos  ( A+B )

When Angles are added and then their Trigonometric Ratios is taken , and if we have to take the  Cosine  of  added angles, then it can be done like this.

Start with  cosine of angle A  and multiply it with cosine of  angle B and in other part i. e.  After  -ve sign      Start with Sine of angle A and multiply with Sine of angle B. i.e. 1st   and 2nd terms are   cosine and  3rd and 4th terms    are sine , Angles start with   A then B again A then again B.

"Here  Sum of cosine of  Two angles  is equal to difference of  product of  cosines of both the angles    and product of sine of both the angles ".

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

When angles are subtracted    i. e  Cos  ( A-B )
When Angles are subtracted and  then their Trigonometric Ratios is taken , and if we have to take the  cosine of  subtracted  angles, then it can be done like this.

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

Start with  cosine of  angle A  and multiply it with cosine of  angle B and in other part i. e.  After  +ve sign      Start with Sine of angle A and multiply with Sine of angle B. i. 1st   and 2nd terms are   cosine and  3rd and 4th  terms    are sine , and angles start with   A then B again A then again B.

"Here  Difference  of cosine of  Two angles  is equal to the  Sum  of  product of  cosines of both the angles    and product of sine of both the angles" .

## Step 2.2

For subtraction of Sine Formula start with cosine of 1st angle as mentioned in step 1 and multiply it with sine of  2nd angle as mentioned in step 1.

Step 3.1

Step 3.2

## For subtraction of cosine Formula start with sine of 1st angle as mentioned in step 1 and multiply it with sine   of  2nd angle as mentioned in step 1,and do not forget to multiply it with -ve sign.

or

If you do not want to multiply it with -ve sign  ,then you can change 2nd angle (D-C)/2 instead of (C-D)/2

## CD Formulas

S N Fomula Result
1 sin C + sin D 2 sin {(C+D)/2} cos {(C-D)/2}
2 sin C - sin D 2 cos {(C+D)/2} sin {(C-D)/2}
3 cos C + cos D 2 cos {(C+D)/2} cos {(C-D)/2}
4 cos C - cos D -2 sin {(C+D)/2}  sin {(C-D)/2}
or cos C - cos D 2 sin {(C+D)/2}   sin {(D-C)/2}

Want to learn Matrix , Different Types of matrices And Determinants.

Want to learn How to Solve Linear Equations of Two and Three Variables  || Matrix Method to Solve System of Linear Equations.

## Conclusion

Thanks for devoting your valuable time for the post Easy Tricks to Memorise A B and C D Formulae in Trigonometry and Trigonometry 's shortcut formulas of this blog ,trigonometry formulas for class 11 ncert,trigonometric functions class 11 notes, trigonometry class 11 tricks,,trigonometry formulas list,. If you found this this blog/post of your concern, Do Follow me on my blog and share this post with your friends . We shall meet again in next post ,till then Good Bye.

How to Memorise A   B and C   D  formulas  easily ,watch this video

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## 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[ 8cos 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  = 32cos x  - 4 - 48cos x  + 24cos² x  - 6cos² x + 3

Adding the like powers terms and arranging in descending order

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

Hence the Proof

## 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

## 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

## 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

Hence the Proof

## Prove that cos 4x = 8 cos⁴ x - 8 cos² x + 1

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 = 2(2cos2 x -1)² -1

It is the square of 2cos2 x -1

cos 4x = 2(2cos2 x -1)² -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 = Sin (2x+x)
Sin 3x = sin 2x cos x + cos 2x sin 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 can prove that cos 3x= 4 cos³ x - 3 cos x
For learning and memorising more trigonometric formulas

## Conclusion

In this post I have discussed trigonometric identities ,trigonometric identities problems, proving trigonometric identities formulas . If this post helped you little bit, then please share it with your friends to benefit them, comment your views on it to boost me and to do better, and also follow me on my Blog .We shell meet in next post till then Bye

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## HOW TO MEMORISE DIFFERENT VALUES OF TRIGONOMETRIC ANGLES

Welcome to this post of learning trigonometric formulas in an easy way with me. Because 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 are  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. MEMORISING TRIGONOMETRIC ANGLES

Table for  Trigonometric Angles  in 1st Quadrant
S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (π/2 - x)   cos x
2 Cos (π/2 - x)   sin x
3 tan (π/2 - x)   cot x
4 Cot (π/2 - x)    tan x
5 Sec (π/2 - x)    Cosec x
6 Cosec (π/2 - x)     Sec x

(3)  When angle involve  2π ,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.

Table for  Trigonometric Angles  in 1st Quadrant
S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (2π + x)  Sin x
2 Cos (2π + x)  cos x
3tan (2π + x)   tan x
4Cot (2π + x)   Cot x
5Sec (2π + x)   Sec x
6Cosec (2π + x)    Cosec x

## 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. MEMORISING TRIGONOMETRIC ANGLES

Table for  Trigonometric Angles  in 2nd Quadrant
S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (π/2 + x)   cos x
2 Cos (π/2 + x)   -sin x
3 tan (π/2 + x)   -cot x
4 Cot (π/2 + x)    -tan x
5 Sec (π/2 + x)    cosec x
6 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.

Table for  Trigonometric Angles  in 2nd Quadrant
S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (π - x)   Sin x
2 Cos (π - x)   - cos x
3 tan (π - x)    -tan x
4 Cot (π - x)    -Cot x
5 Sec (π - x)    -Sec x
6 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.

Table for  Trigonometric Angles  in 3rd Quadrant

S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (3π/2 - x)   -Cos x
2 Cos (3π/2 - x)   -Sin x
3 tan (3π/2 - x)    cot x
4 Cot (3π/2 - x)    tan x
5 Sec (3π/2 - x)    -Cosec x
6 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.

Table for  Trigonometric Angles  in 3rd Quadrant
S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (π + x)   -Sin x
2 Cos (π + x)   - cos x
3 tan (π + x)    tan x
4 Cot (π + x)    Cot x
5 Sec (π + x)    -Sec x
6 Cosec (π + x)     -Cosec x

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

## 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.

Table for  Trigonometric Angles  in 4th Quadrant

S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (3π/2 + x)   -Cos x
2 Cos (3π/2 + x)   Sin x
3 tan (3π/2 + x)    -cot x
4 Cot (3π/2 + x)    -tan x
5 Sec (3π/2 + x)    Cosec x
6 Cosec (3π/2 + x)    -Sec 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.

Table for  Trigonometric Angles  in 4th Quadrant

S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (2π - x)   -Sin x
2 Cos (2π - x)   cos x
3 tan (2π - x)    -tan x
4 Cot (2π - x)    -Cot x
5 Sec (2π - x)    Sec x
6 Cosec (2π - x)     -Cosec x

### When Angle do not involve π

Table for  Trigonometric Angles  in 4th Quadrant
S
N

Trigonometric

Angles
Value of T  Angles
1 Sin (- x)   -Sin x
2 Cos (- x)   cos x
3 tan (- x)   -tan x
4 Cot (- x)    -Cot x
5 Sec (- x)    Sec x
6 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

Thanks for devoting your valuable time for the post memorising different values of trigonometric angles in different quadrants of this blog. If you found this  blog/post of your concern, Do Follow me on my blog and share  post with your friends . We shall meet again in next post ,till then Good Bye.

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