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3 years ago

CosA + cosB + cosC = 1 + 4sinA/2 sinB/2 sinC/2

1 answer:

6 0

$\cos(A) + \cos( B ) + \cos(C) \\ = 2 \cos( \frac{A + B}{2} ) \cos( \frac{ A- B}{2} ) + 1 - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2\cos( \frac{\pi - C}{2} ) \cos( \frac{A - B}{2} ) - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2 \sin \frac{C}{2} \cos( \frac{A - B}{2} ) - 2 { \sin }^{2} \frac{C}{2} \\ = 1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \sin \frac{C}{2}] \\ = 1 + 2 \sin \frac{C}{2} [ \cos( \frac{A - B}{2} ) - \cos( \frac{A + B}{2} )] \\ = 1 + 2 \sin\frac{C}{2} \times 2 \sin( \frac{ \frac{A + B}{2} +\frac{A - B}{2} }{2} ) \sin( \frac{ \frac{A + B}{2} - \frac{A - B}{2} }{2} ) \\ = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

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How to integrate cos^-2x . Question from Alevel Pure Maths 3

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3 years ago

Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges

sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

$f(x)=a^{x}\\ f'(x)=a^xln(a)$

The process will help us to understand what is happening, at first we rewrite the function:

$f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\$

And then, we use the chain rule to differentiate:

$f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)$

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b) (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

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3 years ago

There are 50 kids at the park 30 of the girls what percent of the kids are boys

Flauer [41]

Answer:

40 %

Step-by-step explanation:

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3 years ago

A random sample of 100 students was taken. Of the 100 students, 50 liked vanilla ice cream, 37 liked chocolate ice cream, and 13

tino4ka555 [31]

Answer:

444

Step-by-step explanation:

since 37/100 students liked chocolate you need to scale that data to the new data size and find thirty seven percent of 1200 which is 444

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3 years ago

Determine if the graph is symmetric about the x-axis, the y-axis, or the origin.

Alexandra [31]

Answer: y-axis.

Explanation:

The equation r = 3 - 4 sin θ may be graphed using cartesian coordinates (i.e. x and y) or polar coordinates.

The logical way is to graph in polar coordinates. I did it using a graphing software and obtained the attached graph.

Also, I attached the graph in cartesian coordinates.

Here you have a table that you can build by your self and use the pairs to draw the graph (either cartesian or polar coordinates).

 θ (rad) r = 3 - 4 sin θ 

0 3 
π/8 2.61731657 
 π/4 2.29289322 
3π/8 2.07612047
 π/2 2
 5π/8 2.07612047 
3π/4 2.29289322 
7π/8 2.61731657 
π 3 
9π/8 3.38268343 
5π/4 3.70710678 
11π/8 3.92387953 
3π/2 4 
13π/8 3.92387953 
7π/4 3.70710678 
15π/8 3.38268343 
2π 3 

In both of the attached graphs you can see that the left side is equal to the right side, but the upper side is not equal to the bottom side.

Therefore, the graph is symmetric about the y-axis.

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3 years ago

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