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# trick trig..

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Hi friends,

Could someone kindly help me with this one:

If Sin$$\theta$$=$$-6k\over{k^2+9}$$ and $$\theta \varepsilon (90;270)$$, and k is > 0 and < 3, determine in terms of k:

1) Cos$$\theta$$

2) $$tan\theta +{1\over{Cos\theta}}$$

The "k" being greater and smaller confuses me...Thank you all for helping, which always is appreciated!!

Apr 15, 2021
edited by juriemagic  Apr 15, 2021
edited by juriemagic  Apr 15, 2021
edited by juriemagic  Apr 15, 2021
edited by juriemagic  Apr 15, 2021

#1
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I am very confused too, but I'll try my best.

question 1

cos(x) = sqrt(1-sine(x)^2)

cos(x) = sqrt(1 - (-6k/(k^2+9))^2)

cos(x) = sqrt(1-(36k^2)/(k^4+18k^2+81))

cos(x) = sqrt((k^4-18k^2+81)/(k^4+18^2+81))

cos(x) = (k^2-9)/(k^2+9)

question 2

tan(x)+1/cos(x) = tan(x)+1/cos(x)

tan(x)+1/cos(x) = sine(x)/cos(x) + 1/(cos(x))

tan(x)+1/cos(x) = (sine(x)+1)/cos(x)

tan(x)+1/cos(x) = (-6k/(k^2+9)+1)/((k^2-9)/(k^2+9))

tan(x)+1/cos(x) = ((k^2-6k+9)/(k^2+9))/((k^2-9)/(k^2+9))

tan(x)+1/cos(x) = ((k^2-6k+9)*(k^2+9))/((k^2+9)*(k^2-9))

tan(x)+1/cos(x) = (k^2-6k+9)/(k^2-9)

tan(x)+1/cos(x) = (k-3)(k-3)/(k+3)(k-3)

tan(x)+1/cos(x) = (k-3)/(k+3)

I am about 99% that I am wrong, but I hope this helped. :))

=^._.^=

Apr 15, 2021
edited by catmg  Apr 15, 2021
#2
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Hi Juriemagic and  Catmg ,

$$sin\theta=\frac{-6k}{k^2+9}$$

0

The denominator will always be positive, the numerator will always be negative so sin(theta) will always be negative.

We are told that sin(theta) is in the 2nd or third quadrant.  Since it is negative it must be in the 3rd quadrant.

Cos(theta) is also negative in the 3rd quadrant.

By inspection I am happy that  the value of sin(theta) is between 0 and 1. I can expand on this if you want.

So theta can be any value in the 3rd quadrant      [180, 270)

$$sin^2\theta=\frac{(-6k)^2}{(k^2+9)^2}=\frac{36k^2}{(k^2+9)^2}\\ cos^2\theta=1-sin^2\theta\\ cos^2\theta=1-\frac{36k^2}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^2+9)^2-36k^2}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^4+18k^2+81)-36k^2}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^4-18k^2+81)}{(k^2+9)^2}\\ cos^2\theta=\frac{(k^2-9)^2}{(k^2+9)^2}\\ cos\theta=\frac{\pm(k^2-9)}{(k^2+9)}\\$$

Now 0

$$cos\theta=\frac{k^2-9}{k^2+9}\\$$

Catmg got the second one correct too.  Great work Catmg.

To be honest I usually do these more visually.

I'll show you the pic I would normally use:

I'd just read all the ratios off my pic. Edit:   some has not displayed properly ... problems with LaTex.  Hopefully it won't matter

Ask  questions  if you want though. Coding:

sin^2\theta=\frac{(-6k)^2}{(k^2+9)^2}=\frac{36k^2}{(k^2+9)^2}\\
cos^2\theta=1-sin^2\theta\\
cos^2\theta=1-\frac{36k^2}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^2+9)^2-36k^2}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^4+18k^2+81)-36k^2}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^4-18k^2+81)}{(k^2+9)^2}\\
cos^2\theta=\frac{(k^2-9)^2}{(k^2+9)^2}\\
cos\theta=\frac{\pm(k^2-9)}{(k^2+9)}\\

cos\theta=\frac{k^2-9}{k^2+9}\\

Apr 15, 2021
edited by Melody  Apr 15, 2021
#3
+1

Ohhh, I've never thought of using a triangle to solve these problems.

It seems a lot easier to do it like that.

Thank you. :))

=^._.^=

catmg  Apr 15, 2021
#4
+2

Yes, I think it is.

You can just use any right-angled triangle (and ignore the minus signs)

But you have to think about the minus afterwards to get the right quadrants.

Melody  Apr 16, 2021