Author Topic: Help with Holiday Homework (Read 5045 times) Tweet Share

ChristineNguyen · « **Reply #15 on:** December 22, 2009, 01:35:41 pm »

Thanksss

ChristineNguyen · « **Reply #16 on:** December 22, 2009, 02:18:25 pm »

Hey just a few other questions

Given that tan(2x) = 4root2 / 7 where x --> [0, pie/4)
find the exact value of sin(x)

and

if cos(A) = sin(A-B)sin(B) prove that tan(A-B)tan(B) = 1/2

thanks

cipherpol · « **Reply #17 on:** December 22, 2009, 02:19:26 pm »

Hey kyzoo, how'd you get $tan(2x) - (tan(2x))(tan^2 x) -2 tanx = 0$ ?

TrueTears · « **Reply #18 on:** December 22, 2009, 02:22:42 pm »

Consider $X = 2x$

$\tan(X) = \frac{4\sqrt{2}}{7}$

Draw a right-angle triangle.

Opposite side is $4\sqrt{2}$ and the adjacent is $7$ .

The hypotenuse is given by $\sqrt{(4\sqrt{2})^2+7^2}$

$\therefore \sin(X) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\implies \sin(2x) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\frac{\sin(2x)}{\cos(2x)} = \frac{4\sqrt{2}}{7}$

$\implies \cos(2x) = \frac{\frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}}{\frac{4\sqrt{2}}{7}}$

I smell some double angle formulas and wishful thinking.

Can you do the rest?

kyzoo · « **Reply #19 on:** December 22, 2009, 02:33:19 pm »

Quote from: cipherpol on December 22, 2009, 02:19:26 pm

Hey kyzoo, how'd you get $tan(2x) - (tan(2x))(tan^2 x) -2 tanx = 0$ ?

$(tan(2x)) = \frac{2 tanx}{1-tan^2 x}$
$(tan(2x))(1-tan^2 x)=2 tan x$
$tan(2x) - (tan(2x))(tan^2 x) - 2 tan x = 0$

cipherpol · « **Reply #20 on:** December 22, 2009, 02:37:53 pm »

i must have been retarded, thanks :]

ChristineNguyen · « **Reply #21 on:** December 22, 2009, 04:31:06 pm »

sorry i dont get why you did the last part to find cos(2x)

TrueTears · « **Reply #22 on:** December 22, 2009, 04:33:05 pm »

Quote from: TrueTears on December 22, 2009, 02:22:42 pm

Consider $X = 2x$

$\tan(X) = \frac{4\sqrt{2}}{7}$

Draw a right-angle triangle.

Opposite side is $4\sqrt{2}$ and the adjacent is $7$ .

The hypotenuse is given by $\sqrt{(4\sqrt{2})^2+7^2}$

$\therefore \sin(X) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\implies \sin(2x) = \frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}$

$\frac{\sin(2x)}{\cos(2x)} = \frac{4\sqrt{2}}{7}$

$\implies \cos(2x) = \frac{\frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}}{\frac{4\sqrt{2}}{7}}$

I smell some double angle formulas and wishful thinking.

Can you do the rest?

$\tan(2x) = \frac{\sin(2x)}{\cos(2x)} = \frac{\frac{4\sqrt{2}}{\sqrt{(4\sqrt{2})^2+7^2}}}{\cos(2x)} = \frac{4\sqrt{2}}{7}$

ChristineNguyen · « **Reply #23 on:** December 22, 2009, 04:40:07 pm »

but it says to find the value of sin(x)

sorry

TrueTears · « **Reply #24 on:** December 22, 2009, 04:41:46 pm »

I know

That is the crux of the question, the rest is just simply playing around with double angle formulas, try it yourself

$\cos(2x) = 1-2\sin(x)^2$

ChristineNguyen · « **Reply #25 on:** December 22, 2009, 05:20:42 pm »

thanks i got it,
one quick question how do i get started on

evaluate P= sin6sin42sin66sin78

TrueTears · « **Reply #26 on:** December 22, 2009, 05:24:36 pm »

ChristineNguyen · « **Reply #27 on:** December 22, 2009, 05:30:32 pm »

LOL

how did you work those out?

TrueTears · « **Reply #28 on:** December 22, 2009, 05:49:43 pm »

From the Taylor Series

$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ where $x$ is in radians.

darkphoenix · « **Reply #29 on:** December 22, 2009, 11:32:04 pm »

Wow thanks ChristineNguyen for hijacking my topic! Haha jks

Um got another question:

Use De Moivre's theorem to solve the equation $z^2= -3-3i$ . (Give your answers in polar form.)

I tried working it out and got a different answer to what it was supposed to be..

Search the forums now!

We have moved!

Author Topic: Help with Holiday Homework (Read 5045 times) Tweet Share

ChristineNguyen

Re: Help with Holiday Homework

ChristineNguyen

Re: Help with Holiday Homework

cipherpol

Re: Help with Holiday Homework

TrueTears

Re: Help with Holiday Homework

kyzoo

Re: Help with Holiday Homework

cipherpol

Re: Help with Holiday Homework

ChristineNguyen

Re: Help with Holiday Homework

TrueTears

Re: Help with Holiday Homework

ChristineNguyen

Re: Help with Holiday Homework

TrueTears

Re: Help with Holiday Homework

ChristineNguyen

Re: Help with Holiday Homework

TrueTears

Re: Help with Holiday Homework

ChristineNguyen

Re: Help with Holiday Homework

TrueTears

Re: Help with Holiday Homework

darkphoenix

Re: Help with Holiday Homework

Recent Posts

User:
Password: