Mathematics Marathon (1 Viewer)

[bored of sc]

Next question:

Simplify: tan²A(1 - sin²A)

Hence solve: 4tan²A(1 - sin²A) = 1 for 0^o < A < 360^o

 

[omniscience]

Next question:

Simplify: tan²A(1 - sin²A)

Hence solve: 4tan²A(1 - sin²A) = 1 for 0^o < A < 360^o

1.
tan^2(A) (1- sin^2(A)) = tan^2(A)(cos^2(A)) = sin^2(A)

2. 4tan^2(A)(1-sin^2(A)) = 1
4sin^2(A) = 1
sin^2(A) = 1/4
sin (A) = +_1/2
A = pi/6, 5pi/6, 7pi/6 or 11pi/12 (which are 30 degrees, 150 degrees, 210 degrees and 330 degrees)

My question: find the derivative of sec^2(x)

 

[bored of sc]

My question: find the derivative of sec^2(x)

I haven't done differentiation of trigonometric functions yet.

You're correct by the way.

 

[omniscience]

I haven't done diffentiation of trigonometric functions yet.

You're correct by the way.

Ok, then. I will make it easier for you to do it.

Differentiate tan x.

HENCE work the question that I posted above. It will be easy as hell.

 

[bored of sc]

Ok, then. I will make it easier for you to do it.

Differentiate tan x.

HENCE work the question that I posted above. It will be easy as hell.

So is it just sec²x?

 

[omniscience]

So is it just sec²x?

Yes. From there, you can work out what the integral of sec^2(x) is (i.e. tan(x) + c)

 

[kaz1]

1.
tan^2(A) (1- sin^2(A)) = tan^2(A)(cos^2(A)) = sin^2(A)

2. 4tan^2(A)(1-sin^2(A)) = 1
4sin^2(A) = 1
sin^2(A) = 1/4
sin (A) = +_1/2
A = pi/6, 5pi/6, 7pi/6 or 11pi/12 (which are 30 degrees, 150 degrees, 210 degrees and 330 degrees)

My question: find the derivative of sec^2(x)

sec²x = (cosx)^-2. Using the chain rule, this differentiates to

-2(cosx)^-2 * (-sinx)
= 2sinx / cos²x

If A and B are the zeroes of the function 3x²-5x-4=0 find the value of A³+B³

 

[kaz1]

Hang on, mate. You differentiated it, I asked you to integrate it.

1.
tan^2(A) (1- sin^2(A)) = tan^2(A)(cos^2(A)) = sin^2(A)

2. 4tan^2(A)(1-sin^2(A)) = 1
4sin^2(A) = 1
sin^2(A) = 1/4
sin (A) = +_1/2
A = pi/6, 5pi/6, 7pi/6 or 11pi/12 (which are 30 degrees, 150 degrees, 210 degrees and 330 degrees)

My question: find the derivative of sec^2(x)

.

 

[omniscience]

.

uh...

 

[zzdfa]

find 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ..... to infinity =D

 

[Mark576]

I haven't done diffentiation of trigonometric functions yet.

Ok, then. I will make it easier for you to do it.

Differentiate tan x.

HENCE work the question that I posted above. It will be easy as hell.

 

[untouchablecuz]

find 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ..... to infinity =D

....(infinity)
.........Σ [n/(2^n)]
.......n=1

 

[untouchablecuz]

find 1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ..... to infinity =D

....(infinity)
.........Σ [n/(2^n)]
.......n=1

<!-- / icon and title --> <!-- message --> Uluru is a large rock on flat ground in Central Australia. Three tourists A, B, and C are observing Uluru from the ground. A is due north of Uluru, C is due east of Uluru, and B is on the line-of-sight from A to C and between them. The angles of elevation to the summit of Uluru from A, B, and C are 26 degrees, 28 degrees, and 30 degrees, respectively. Determine the bearing of B from Uluru with working.

 

[Timothy.Siu]

....(infinity)
.........Σ [n/(2^n)]
.......n=1

huh, thats wrong
i think the answers 2

 

[untouchablecuz]

huh, thats wrong
i think the answers 2

LOL

I just put it in sigma notation.

Causeeeee I need someone to do my problem.

Anyone?

Please?

 

[lyounamu]

The answer for the infinity series is 2.

The sum -> 2 as x (numerator) -> infinity and y (denominator) -> infinity

 

[tommykins]

回复: Re: Mathematics Marathon

If A and B are the zeroes of the function 3x²-5x-4=0 find the value of A³+B³

A^3+B^3 = (A+B)(A^2+AB+B^2) = (A+B)[(A+B)^2-2AB+AB] = (A+B)[(A+B)^2 -AB]
A+B = 5/3, AB = -4/3

Thus A^3+B^3 = 185/27

 

[tommykins]

回复: Re: Mathematics Marathon

r isn't 1/2.

 

[Dota55]

Re: 回复: Re: Mathematics Marathon

r isn't 1/2.

oh....my....god. :jaw:

I blame lack of sleep.

 

[Mark576]

Re: 回复: Re: Mathematics Marathon

A^3+B^3 = (A+B)(A^2+AB+B^2) = (A+B)[(A+B)^2-2AB+AB] = (A+B)[(A+B)^2 -AB]
A+B = 5/3, AB = -4/3

Thus A^3+B^3 = 185/27

A³ + B³ = (A + B)(A² - AB + B²) = (A + B)[(A + B)² - 3AB]