Some exercises

**Thaurus** · March 29th 2010, 11:09 AM

If someone could help me slove a few exercises ,I realy need them?
I would need them by Wednesday.
And sorry me english is bad.
Here they are:
http://img337.yfrog.com/img337/2927/mata2b.png

Thanks!

**Deadstar** · March 29th 2010, 11:44 AM

Will help with a couple... Please show any working so we know what you have done. I'll give you the answers as well but you'll need to some of your own working...

First question.
$(1-\sin(\alpha)-\cos(\alpha))(1-\sin(\alpha)+\cos(\alpha)) = 1 - 2\sin(\alpha) + \sin^2(\alpha) - \cos^2(\alpha)$ by expanding.

Then use $1 = \sin^2(\alpha) + \cos^2(\alpha)$ .

ANSWER: 2

Second question.

Use the fact that $a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)$ and replace $a$ and $b$ with $\sin(3\alpha)$ and $\cos(3\alpha)$ .

ANSWER: $1 + \tan^2(\alpha)$ (however this can be further simplified, I'll leave that up to you, use the hint in the next question)

Third question.

$\frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha)$ , hence for this question... $\frac{\sin(\alpha)}{\cos(\alpha)} = \tan(\alpha) = -2$ so $\sin(\alpha) = -2\cos(\alpha)$ . Substitute in the equation.

ANSWER: $-\frac{1}{3}$

**Soroban** · March 29th 2010, 12:34 PM

Hello, Thaurus!

$(3)\;\;\frac{(1-\sin\alpha - \cos\alpha)(1 - \sin\alpha + \cos\alpha)}{\sin\alpha(\sin\alpha - 1)}$

The numerator is:

. . $1-\sin\alpha + \cos\alpha - \sin\alpha + \sin^2\!\alpha - \sin\alpha\cos\alpha - \cos\alpha + \sin\alpha\cos\alpha - \cos^2\!\alpha$

. . . . $=\;\;(1-2\sin\alpha + \sin^2\!\alpha) -\cos^2\!\alpha \;\;=\;\;(1-\sin\alpha)^2 - (1-\sin^2\!\alpha)$

. . . . $=\;\;(1-\sin\alpha)^2 - (1-\sin\alpha)(1 + \sin\alpha) \;\;=\;\;(1-\sin\alpha)\bigg[(1-\sin\alpha) - (1 + \sin\alpha)\bigg]$

. . . . $=\;\;-2\sin\alpha(1-\sin\alpha)$

The fraction becomes: . $\frac{-2\sin\alpha(1-\sin\alpha)}{\sin\alpha(\sin\alpha - 1)} \;\;=\;\;\frac{2\sin\alpha(\sin\alpha-1)}{\sin\alpha(\sin\alpha-1)} \;\;=\;\;2$

$(2)\;\;\frac{\sin^4\!3u - \cos^4\!3u}{\sin^2\!3u - \cos^2\!3u} + \tan^2\!3u$

$\text{Factor: }\;\frac{(\sin^2\!3u - \cos^2\!3u)\overbrace{(\sin^2\!3u + \cos^2\!3u)}^{\text{This is 1}}}{\sin^2\!3u - \cos^2\!3u} + \tan^2\!3u$

. . . . $=\;\;1 + \tan^2\!3u \;\;=\;\;\sec^2\!3u$

**Deadstar** · March 29th 2010, 12:37 PM

Originally Posted by Soroban

$\text{Factor: }\;\frac{(\sin^2\!3u - \cos^2\!3u)\overbrace{(\sin^2\!3u + \cos^2\!3u)}^{\text{This is 1}}}{\sin^2\!3u - \cos^2\!3u} + \tan^2\!3u$

. . . . $=\;\;1 + \tan^2\!3u \;\;=\;\;\sec^2\!3u$

Is it not just $\tan^2{\alpha}$ ? (or $u$ as you have it)

**Deadstar** · March 29th 2010, 01:00 PM

Originally Posted by Soroban

. . $1-\sin\alpha + \cos\alpha - \sin\alpha + \sin^2\!\alpha - \sin\alpha\cos\alpha - \cos\alpha + \sin\alpha\cos\alpha - \cos^2\!\alpha$

. . . . $=\;\;(1-2\sin\alpha + \sin^2\!\alpha) -\cos^2\!\alpha \;\;=\;\;(1-\sin\alpha)^2 - (1-\sin^2\!\alpha)$

. . . . $=\;\;(1-\sin\alpha)^2 - (1-\sin\alpha)(1 + \sin\alpha) \;\;=\;\;(1-\sin\alpha)\bigg[(1-\sin\alpha) - (1 + \sin\alpha)\bigg]$

Just a thought...

A much simpler way (i think) to compute the above is...

$1-\sin\alpha + \cos\alpha - \sin\alpha + \sin^2\!\alpha - \sin\alpha\cos\alpha - \cos\alpha + \sin\alpha\cos\alpha - \cos^2\!\alpha$

. . . . $=\;\;1-2\sin\alpha + \sin^2\!\alpha -\cos^2\!\alpha \;\;$

$= \sin^2(\alpha) + \cos^2(\alpha) - 2\sin(\alpha) + \sin^2(\alpha) - \cos^2(\alpha)$

$= -2\sin(\alpha) + 2\sin^2(\alpha)$

$= -2\sin(\alpha)(1 - \sin(\alpha))$

**Soroban** · March 29th 2010, 05:44 PM

Hello again, Thaurus!

1690. Evaluate: . $\frac{\sin\alpha + \cos\alpha}{\cos\alpha - \sin\alpha}\,\:\text{ where }\:\tan\alpha\,=\,-2$

We have: . $\frac{\sin\alpha + \cos\alpha}{\cos\alpha - \sin\alpha}$

Divide top and bottom by $\cos\alpha\!:\quad \frac{\dfrac{\sin\alpha}{\cos\alpha} + \dfrac{\cos\alpha}{\cos\alpha}}{\dfrac{\cos\alpha} {\cos\alpha} - \dfrac{\sin\alpha}{\cos\alpha}} \;=\; \frac{\tan\alpha + 1}{1 - \tan\alpha}$

Substitute $\tan\alpha = \text{-}2\!:\;\;\frac{(\text{-}2)+1}{1-(\text{-}2)} \;=\;-\frac{1}{3}$

$2.\;\;\frac{\tan135^o}{\tan225^o}\cdot\sin300^o - \cos^2(\text{-}300^o)$

We have: . $\frac{-1}{+1}\left(-\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)^2\;\;=\;\;\frac{\sqrt{3}} {2} - \frac{1}{4}$

**Thaurus** · March 30th 2010, 07:08 AM

Thanks!
I did it myself , am I am right?

**Deadstar** · March 30th 2010, 07:32 AM

Originally Posted by Thaurus

Thanks!
I did it myself , am I am right?

Yep thats right!

If it was an assessment question you might want to show a few more steps of working (in my opinion!). Something like...
$1 + \cot^2(x) = 1+ \frac{1}{\tan^2(x)} = 1 + \frac{\cos^2(x)}{\sin^2(x)} = \frac{1}{\sin^2(x)}$

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