Showing posts with label differentiation. Show all posts
Showing posts with label differentiation. Show all posts

## Solutions

Let ABCD be rectangle which is  inscribed in a given circle of radius ‘r’ And Let θ be the angle between side of rectangle and Diameter of given circle.

Therefore from right angled  Δ ABC ,

We have AB  = AC cosθ          ∵ AC = 2r
Let A(x) be the area of Rectangle ABCD
∴ A(x) = AB × BC
A = (2r cos θ)(2r sin θ )
A =  4r2 sin θ cos θ
A = 2r2  (2sin θ cos θ)
A = 2r2  (sin 2θ )

⇒ 2r2 2 (cos 2θ ) = 0 ,As r2 is constant
⇒cos 2θ = 0
⇒cos 2θ =cos (π/2)
⇒ θ = π/4

∴ A has Maximum value at θ = π/4

## Solution

Let us consider two numbers x and 16- x .
Then transforming our problem to mathematical form which says “sum of whose cube”  as follows
A (x) =   x3 + (16 - x)3…….. (1)
Differentiating both sides w .r. t  “x” , we get

X = 8
So  x  =  8 will be the 1st required numbers if Double derivatives of A  w. r. t  ‘x’ comes to be positive at x = 8.
Differentiate (2)  w. r. t. ‘x’  .

## FINAL WORDS

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## Application of Derivative

A piece of wire 28 cm long is to be cut into two pieces. One piece is to be made into a circle and another into a square. How should the wire be cut so that the combined area of the two figures is as small as possible?

Let the wire be cut at a distance of  x meter  from one end. Therefore then two pieces of wire be x m and (28-x) m.

## Calculate Dimension of Circle and Square

Now 1st part be turned into a square and  the 2nd part be be made into a circle.

Since 1st part of the wire is turned into square. then its perimeter will be x m.
So using formula of perimeter of square , we can calculate side of the square = x/4 m

## Calculate Areas of Circle and Square

Therefore Area of square = (x/4)(x/4) sq m

A1 = x2/16

And  when 2nd part of the wire is turned to circle, then its perimeter ( circumference ) will be 28 - x m. So using formula of perimeter of square , And if  "r" be  radius of the circle , Then
Circumference of circle =  2 π r =  (28-x)
∴  r = (28-x)/2π

We know that Area of Circle A2   = π r2

A2  π[(28-x)/2π]2

## To find value/s of x

Now to find the value of x for which this function A(x) is maximum or minimum ,put A(x) = 0

## To Test the Minimum Value of  Function

Now we have the value of "x" on which either A(x) have maximum or minimum value . To check the maximum or minimum value we have to find A''(x) as follows

So A''(x) has positive value Therefore A(x) shall have maximum value at x = 112/(π + 4)

Hence two pieces of wire should be of length x m and (28-x) m

These pieces should be of length 112/(π+4) and 28π/(π + 4)

## Verification

we can calculate the sum of these pieces , it must be 28 m

#### 1st part

112/(π+4) = 112/{(22/7)+4}=112×7/50 = 784/50

#### 2nd part

28π/(π + 4) = {28×22/7}/{(22/7)+4} = 88×7/50 = 616/50

#### Sum of Two Parts

112×7/50 + 28×7/50 = (784+616)/50

= 1400/50= 28 m

## Conclusion

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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[ 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

## 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

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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