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2 years ago

What is the best method to solve griedent and midpoint in math40!! point

Mathematics

1 answer:

Aleks [24]2 years ago

6 0

I usually add x and divide by too and add y and divide by 2

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Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

======================================================

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

2 years ago

(6d^(2)c)/(3c) what is the excluded value

podryga [215]

Answer:

The excluded value is c=0

Step-by-step explanation:

The excluded value is when the denominator goes to zero

(6d^(2)c)/(3c)

3c = 0

c =0

The excluded value is c=0

4 0

2 years ago

Find the derivative of y= 1 / x

givi [52]

Answer:

dy/dx = -1/ $x^{2}$

Step-by-step explanation:

y = 1/x

dy/dx = d/dx(1/x)

=> dy/dx = d/dx( $x^{-1}$ )

=> dy/dx = - $x^{-2}$

=> dy/dx = -1/ $x^{2}$

6 0

2 years ago

The number of subsets that can be created from the set {1, 2, 3} is:

kkurt [141]

Answer:3

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

PLEASE HELP ME PLZ PLZ PLZ

Oksana_A [137]

Answer:

Step-by-step explanation:

pythagoras theorem = a* + b* = c* ( by '*' i mean squared)

2* + b* = square root twenty

4 + b* = 20

b* = 16

b = 4

4 0

2 years ago

What is the best method to solve griedent and midpoint in math40!! point​

What is the best method to solve griedent and midpoint in math40!! point