2u/3u math questions (1 Viewer)

addoil · Oct 2, 2012

(1)
How to use identities or other method to integrate:
Cot(x) - cosec^2(x) dx

(2)
Solve:
3^(2x+1) - 3^x = 10

(3)
If:
N = x^(n-1) + x^(n-2)y + x^(n-3)y^2 +...+ xy^(n-2) + y^(n-1)
Simplify N in terms of x and y.

The answer is N = [ x^n -y^n ] / [ x-y ]
Why cant it use the binomial expansion concept and just be (x+y)^(n-1) ?

(4)
How to prove sin-1(x) + cos-1(x) = pi/2 using the triangle?

(5)
Find the smallest positive solution of cos3x = sin2x

I tried using the double angle identities but don't think they work

(6)
(using 0 = pheta) Using t formula, solve:
2sin0 + 4cos0 = 3, pheta is between 0 and 2pi

I found the 2 exact values of t (0.7595... & -0.1881...) but don't know how to apply the t=tan0/2) from there

(7)
Using pheta = ( pi/15 )(6n +- 1),
Solve;
2cos5pheta = 1, pheta is between 0 and pi/2

(8)
How does
[ -4+ x^2 ] / [ x(x^2 - 4)^(1/2) ]
get to
[ x^2 - 4 ] / x
?

Sorry for all the questions

SpiralFlex · Oct 2, 2012

$1.\; \int (\cot x - \csc^2x) \; dx$

$=\int ( \frac{\cos x}{\sin x}- \csc^2x) \; dx$

$=\ln |\sin x|+\cot x+C$

$2. \; 3^{2x+1} - 3^x = 10$

$3^{2x}*3^1-3^x-10=0$

$3*3^{2x}-3^x-10=0$

$$Let$\; u = 3^x$

$3u^2-u-10=0$

$(3u+5)(u-2)=0$

$u=2$

$3^x=2$

$x\ln 3=\ln 2$

$x=\frac{\ln 2}{\ln 3}$

$3. \; N = x^{n-1} + x^{n-2}y + x^{n-3}y^2 +...+ xy^{n-2} + y^{n-1}$

We will use

$(1+a+a^2+...+a^{n-1})(a-1)=a^n-1$

Let $a=\frac{y}{x}$

$(1+\frac{y}{x}+\frac{y^2}{x^2}+...+\frac{y^{n-1}}{x^{n-1}})(\frac{y-x}{x}=(\frac{y^n-x^n}{x^n})$

Factor x to the power of n-1

$x^{n-1} + x^{n-2}y + x^{n-3}y^2 +...+ xy^{n-2} + y^{n-1}=\frac{-(x^n-y^n)}{-(x-y)}$

$\therefore N = \frac{x^n-y^n}{x-y}$

$4. \; $Let$\; y = \sin^{-1}x$

$x = \sin y$

$x = \cos((\frac{\pi}{2}) - y)$

$\cos^{-1}x = \frac{\pi}{2}- y$

$\cos^{-1}x= \frac{\pi}{2}- \sin^{-1}x$

$\sin^{-1}x+\cos^{-1}x= \frac{\pi}{2}$

$5. \; 2\sin x\cos x= \cos(2x + x)$

$= \cos 2x \cosx - \sin2x \sinx$

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

$=2\cos^3 x - \cos x - 2(1- \cos^2x)\cosx$

$= \cos x(4\cos^2 x - 3)$

$\therefore 2\sin x \cos x= \cos x (1 - 4\sin^2 x)$

$\cos x (4\sin^2 x + 2\sin x - 1) = 0$

$\cos x = 0, \sin \frac{-1+\sqrt{5}}{4}$

$x=18,90...$

$\therefore \;$smallest is 18$$

$6.\; \frac{\pi}{15}(6n \pm 1)$

Let n = 0 (rejecting the minus sign in this one, 1,

$\theta = \frac{\pi}{15}, \frac{\pi}{3}, \frac{7\pi}{15}$

$2\sin \theta + 4\cos \theta = 3,$

$$Let$\; t = \tan \frac{\theta}{2}$

$7. \; 2(\frac{2t}{1+t^2}+4(\frac{1-t^2}{1+t^2})=3$

$4t+4-4t^2=3+3t^2$

$7t^2-4t-1=0$

$t=\frac{4+\sqrt{16+28}{14}}$

$\tan \frac{\theta}{2}=\frac{2+\sqrt{11}}{7}$

$\theta = 2*\tan^{-1}(\frac{2+\sqrt{11}}{7})$

OR

$\theta = 2(\pi+\tan^{-1}(\frac{2-\sqrt{11}}{7}))$

$8.\;\frac{x^2-4}{x\sqrt{x^2 - 4}}*\frac{\sqrt{x^2 - 4}}{\sqrt{x^2 - 4}}$

$=\frac{(x^2-4)\sqrt{x^2 - 4}}{x(x^2 - 4)}$

$=\frac{\sqrt{x^2-4}}{x}$

Timske · Oct 2, 2012

its 3^(2x)*3 for the second one spi ??

SpiralFlex · Oct 2, 2012

Oops I forgot to do a question

SpiralFlex · Oct 2, 2012

Timske said:
its 3^(2x)*3 for the second one spi ??

Thanks lololol

SpiralFlex · Oct 2, 2012

WTH? My solutions are gone for binomial!

2u/3u math questions (1 Viewer)

addoil

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SpiralFlex

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Timske

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SpiralFlex

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SpiralFlex

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SpiralFlex

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