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sb3700 · « **Reply #210 on:** February 24, 2009, 04:52:33 pm »

For the second one:

RHS = $\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx}$
= $\frac{(\cfrac{cos^2x}{sinx} - \cfrac{sin^2x}{cosx})(sinxcosx)}{cosx-sinx}$
= $\frac{cos^3x - sin^3x}{cosx-sinx}$
= $\frac{(cosx-sinx)(cos^2x+sinxcosx + sin^2x)}{cosx-sinx}$ by difference of cubes factorisation
= $1 + sinxcosx$ = LHS

TrueTears · « **Reply #211 on:** February 24, 2009, 04:54:18 pm »

Quote from: sb3700 on February 24, 2009, 04:52:33 pm

For the second one:

RHS = $\frac{(cosxcotx - sinxtanx)(sinxcosx)}{cosx - sinx}$
= $\frac{(\cfrac{cos^2x}{sinx} - \cfrac{sin^2x}{cosx})(sinxcosx)}{cosx-sinx}$
= $\frac{cos^3x - sin^3x}{cosx-sinx}$
= $\frac{(cosx-sinx)(cos^2x+sinxcosx + sin^2x)}{cosx-sinx}$ by difference of cubes factorisation
= $1 + sinxcosx$ = LHS

ARGH! damn i did exactly what you have done, but difference of cubes, z.z.z.z didn't realise that at all omg. I am so bad. -_-

TrueTears · « **Reply #212 on:** February 24, 2009, 06:53:51 pm »

Let $sec \beta = b$ where $\beta \in [ \frac{\pi}{2},\pi]$ . Find, in terms of $\beta$ , 2 values of x in the range $[-\pi,\pi]$ which satisfies each of these equations:

a) $sec x = -b$

b) $cosec x = b$

Let $tan \gamma = c$ where $\gamma \in[ \pi,\frac{3\pi}{2}]$ . find in terms of $\gamma$ , 2 values of x in the range of $[0,2\pi]$ which satisfies

a) $cot x = c$

3. Given that $tan \alpha = p$ , where $\alpha$ is an acute angle, find each of the following in terms of $p$ .
a) $tan( \frac{3\pi}{2} + \alpha)$

sb3700 · « **Reply #213 on:** February 24, 2009, 09:49:45 pm »

Quote from: TrueTears on February 24, 2009, 06:53:51 pm

Let $sec \beta = b$ where $\beta \in [ \frac{\pi}{2},\pi]$ . Find, in terms of $\beta$ , 2 values of x in the range $[-\pi,\pi]$ which satisfies each of these equations:

a) $sec x = -b$

b) $cosec x = b$

As $\beta \in [\frac{\pi}{2},\pi]$ , then $b = sec\beta \le 0$
Choose $\alpha = \pi - \beta$ (so $\alpha \in [0, \frac{\pi}{2}]$ and $cosec\alpha = cosec\beta$ and $sec\alpha = -sec\beta$ )

a) $sec x = -b \ge 0$
So $x = \alpha, -\alpha$ , ie: quadrant 1 or quadrant 4 (note you can add/subtract 2 $\pi$ to make sure end result is in domain).
$x = \pi - \beta, -\pi + \beta$ as $\beta = \pi - \alpha$
$x = \pm(\pi - \beta)$

b) $cosecx = b \le 0$
So $x = -\pi + \alpha, -\alpha$
So $x = -\pi + \pi - \beta, -\pi + \beta$
So $x = -\beta, -\pi + \beta$

I might do the others later

TrueTears · « **Reply #214 on:** February 25, 2009, 02:40:09 pm »

thanks again sb3700, that really helped !

sb3700 · « **Reply #215 on:** February 25, 2009, 08:42:53 pm »

Quote from: TrueTears on February 24, 2009, 06:53:51 pm

3. Given that $tan \alpha = p$ , where $\alpha$ is an acute angle, find each of the following in terms of $p$ .
a) $tan( \frac{3\pi}{2} + \alpha)$

This one is easier:
$tan( \frac{3\pi}{2} + \alpha) = -cot(\alpha) = -\frac{1}{tan(\alpha)} = -\frac{1}{p}$

Quote from: TrueTears on February 24, 2009, 06:53:51 pm

Let $tan \gamma = c$ where $\gamma \in[ \pi,\frac{3\pi}{2}]$ . find in terms of $\gamma$ , 2 values of x in the range of $[0,2\pi]$ which satisfies

a) $cot x = c$

So $tan(\gamma) = c = cot(x) = tan(\frac{\pi}{2} - x)$
Then $\frac{\pi}{2} - x = -2\pi + \gamma$ or $-\pi + \gamma$ (Q1 or Q3)
So $x = \frac{5\pi}{2} - \gamma$ or $x = \frac{3\pi}{2} - \gamma$

TrueTears · « **Reply #216 on:** February 27, 2009, 04:24:38 pm »

thanks so much sb3700

TrueTears · « **Reply #217 on:** March 01, 2009, 06:26:01 pm »

Prove $\frac{cos3x}{sinx} + \frac{sin3x}{cosx} = 2cot(2x)$

TrueTears · « **Reply #218 on:** March 01, 2009, 06:41:46 pm »

and

Prove $cos4x = 8cos^4x - 8cos^2x +1$

Mao · « **Reply #219 on:** March 01, 2009, 06:44:57 pm »

Quote from: TrueTears on March 01, 2009, 06:26:01 pm

Prove $\frac{cos3x}{sinx} + \frac{sin3x}{cosx} = 2cot(2x)$

multiplying by $\sin (x)\cos(x)$ on both sides

$\implies \cos x \cos 3x + \sin x \sin 3x = \sin 2x \cot 2x$

$\implies \cos (x-3x) = \sin 2x\cdot \frac{\cos 2x}{\sin 2x}$

$\implies \cos (-2x) = \cos 2x$

$LHS=RHS,\; QED$

Quote from: TrueTears on March 01, 2009, 06:41:46 pm

and

Prove $cos4x = 8cos^4x - 8cos^2x +1$

$LHS = \cos 4x = 2\cos^2 2x - 1 = 2 (2\cos^2 x-1)^2 -1 = 2 (4\cos^4 x - 4\cos^2 x + 1) -1 = 8\cos^4 x - 8\cos^2 x + 1=RHS$

QED

TrueTears · « **Reply #220 on:** March 01, 2009, 06:46:58 pm »

Quote from: Mao on March 01, 2009, 06:44:57 pm

Quote from: TrueTears on March 01, 2009, 06:26:01 pm
Prove $\frac{cos3x}{sinx} + \frac{sin3x}{cosx} = 2cot(2x)$

multiplying by $\sin (x)\cos(x)$ on both sides

$\implies \cos x \cos 3x + \sin x \sin 3x = \sin 2x \cot 2x$

$\implies \cos (x-3x) = \sin 2x\cdot \frac{\cos 2x}{\sin 2x}$

$\implies \cos (-2x) = \cos 2x$

$LHS=RHS,\; QED$

nice thanks mao

TrueTears · « **Reply #221 on:** March 01, 2009, 06:50:47 pm »

and last one:
$\frac{1-cosx}{sinx} = tan{\frac{x}{2}}$

thanks heaps for your help!

Mao · « **Reply #222 on:** March 01, 2009, 06:56:17 pm »

Quote from: TrueTears on March 01, 2009, 06:50:47 pm

and last one:
$\frac{1-cosx}{sinx} = tan{\frac{x}{2}}$

thanks heaps for your help!

$LHS = \frac{1-\cos x}{\sin x} = \frac{1 - \left(1-2\sin^2 \frac{x}{2}\right)}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \frac{2\sin^2 \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} = \tan\frac{x}{2} = RHS$

QED

btw, what's with the posting questions in trios

you seem to be doing that a lot (and always in threes)

TrueTears · « **Reply #223 on:** March 01, 2009, 07:03:26 pm »

thanks again mao

lol something me /0 come up with, triforce FTW!! ganges + banks = banges

TrueTears · « **Reply #224 on:** March 02, 2009, 06:45:10 pm »

just a few more Q

1. Prove $\frac{sinx +1 - cosx}{sinx - 1 + cosx} = \frac{1+tan(\frac{x}{2})}{1-tan(\frac{x}{2})}$

2. $cot(x+y) = \frac{cotxcoty -1}{cotx + coty}$

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