Author Topic: help on trig plz (Read 1316 times) Tweet Share

hard · « **on:** March 23, 2009, 09:13:40 pm »

i need help on 3 and 4. they're bugging me.

Over9000 · « **Reply #1 on:** March 23, 2009, 09:34:13 pm »

3.a) for now
$sinx=-cosx$
$sin^{2}(x)=cos^{2}(x)$
$sin^{2}(x)=1-sin^{2}(x)$
$2sin^{2}x=1$
$sin^{2}x=\frac{1}{2}$
$sinx=\pm\frac{\sqrt{2}}{2}$
$x=\frac{\pi}{4}, \frac{3\pi}{4}, \frac{7\pi}{4}$

hard · « **Reply #2 on:** March 23, 2009, 09:38:11 pm »

Quote from: Over9000 on March 23, 2009, 09:34:13 pm

3.a) for now
$sinx=-cosx$
$sin^{2}(x)=cos^{2}(x)$
$2sin^{2}x=1$
$sin^{2}x=\frac{1}{2}$
$sinx=\pm\frac{\sqrt{2}}{2}$
$x=\frac{\pi}{4}, \frac{3\pi}{4}$ ..... and so on

how did you do step 2 and 3?

Over9000 · « **Reply #3 on:** March 23, 2009, 09:39:50 pm »

soz, forgot a step, let me add it in
k, so square both sides so you can use a trig identity, then collect the $sin^{2}(x)$ 's

Glockmeister · « **Reply #4 on:** March 23, 2009, 09:41:15 pm »

hard · « **Reply #5 on:** March 23, 2009, 09:47:01 pm »

oh sweet thanks yer i got it.

Over9000 · « **Reply #6 on:** March 23, 2009, 09:53:47 pm »

For 4a)
$sin3x=3cos3x [0,\frac{2\pi}{3})$
$sin^{2}3x=9cos^{2}3x$
$sin^{2}3x=9(1-sin^{2}x)$
$sin^{2}3x=9-9sin^{2}x$
$10sin^{2}3x=9$
$sin^{2}3x=\frac{9}{10}$
$sin3x=\pm \frac{3}{\sqrt{10}}$
$3x=sin^{-1}(\pm\frac{3}{\sqrt{10}})$
$3x=1.25$
$x=0.42$
Domain is between [0,2.09)
Therefore this is the only allowed value

hard · « **Reply #7 on:** March 23, 2009, 10:08:15 pm »

cheeer cuzz

Over9000 · « **Reply #8 on:** March 23, 2009, 10:11:18 pm »

Ill do the last question because its usually the hardest just so you can do the rest.
$3sin(\frac{{\pi}x}{3})-cos(\frac{{\pi}x}{3})=0$
$9sin^{2}(\frac{{\pi}x}{3})- cos^{2}(\frac{{\pi}x}{3})=0$
$9sin^{2}(\frac{{\pi}x}{3})- (1- sin^{2}(\frac{{\pi}x}{3}))=0$
$10sin^{2}(\frac{{\pi}x}{3})-1=0$
$10sin^{2}(\frac{{\pi}x}{3})=1$
$sin^{2}(\frac{{\pi}x}{3})=\frac{1}{10}$
$sin(\frac{{\pi}x}{3}) = \frac{\sqrt{10}}{10}$
$\frac{{\pi}x}{3} = sin^{-1}(\frac{\sqrt{10}}{10})$
$\frac{{\pi}x}{3} = 0.32$
${\pi}x=0.965$
$x=0.31$
Now domain is $[-3,6]$
so $\pi - 0.31 = 2.83$
$.31-3.14= 3.45$
$.31+3.14= -2.83$
$2.83-3.14= 5.97$
$2.83+3.14= -0.31$

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Author Topic: help on trig plz (Read 1316 times) Tweet Share

hard

help on trig plz

Over9000

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hard

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Glockmeister

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hard

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