Question

(1+cos(y))/(1+sec(y)) in simplest numerical form?

Answer 1

\(\bf \cfrac{1+cos(y)}{1+sec(y)}\qquad {\color{brown}{ sec(\theta)=\cfrac{1}{cos(\theta)}}}\quad thus\implies \cfrac{1+cos(y)}{1+\frac{1}{cos(y)}}\)

what would that simplify to?

Answer 2

That is where I hit the wall xD wondering what exactly to do with the denominator

Answer 3

well \(\bf 1+\cfrac{1}{cos(y)}\implies \cfrac{1}{1}+\cfrac{1}{cos(y)}\implies \cfrac{}{{\color{brown}{ lcd?}}}\)

Answer 4

what do you think the LCD or GCF would be?

Answer 5

so cosy

Answer 6

Now theres 1/cosy + 1/cosy?

Answer 7

yeap... thus one sec

Answer 8

\(\bf \cfrac{1+cos(y)}{1+sec(y)}\qquad {\color{brown}{ sec(\theta)=\cfrac{1}{cos(\theta)}}}\quad thus\implies
\\ \quad \\
\cfrac{1+cos(y)}{\frac{cos(y)+1}{cos(y)}}\implies 1+cos(y)\cdot \cfrac{cos(y)}{cos(y)+1}\implies \cfrac{1+cos(y)}{1}\cdot \cfrac{cos(y)}{cos(y)+1}
\\ \quad \\
\cfrac{\cancel{ cos(y)+1 }}{1}\cdot \cfrac{cos(y)}{\cancel{ cos(y)+1 }}\)

Answer 9

hmm m issing a few ... anyhow \(\bf \cfrac{1+cos(y)}{1+sec(y)}\qquad {\color{brown}{ sec(\theta)=\cfrac{1}{cos(\theta)}}}\quad thus\implies \cfrac{1+cos(y)}{1+\frac{1}{cos(y)}}
\\ \quad \\
\cfrac{1+cos(y)}{\frac{cos(y)+1}{cos(y)}}\implies 1+cos(y)\cdot \cfrac{cos(y)}{cos(y)+1}\implies \cfrac{1+cos(y)}{1}\cdot \cfrac{cos(y)}{cos(y)+1}
\\ \quad \\
\cfrac{\cancel{ cos(y)+1 }}{1}\cdot \cfrac{cos(y)}{\cancel{ cos(y)+1 }}\)

Answer 10

Thanks I see the step I was missing lol stay tuned I believe I have another question...should I make a new post or post it here?

Answer 11

yes... thus if I dunno.. someone else may, and we can revise each other :)

Answer 12

Gotcha xD soooo this one is basically the same thing but with (2+tan^2x)/(sec^2x) =(f(x))^2

Answer 13

hmmm are you supposed to simplify \(\bf \cfrac{2+tan^2(x)}{sec^2(x)}?\)

Answer 14

Yeah and that equals (f(x))^2

Answer 15

yeap pretty much \(\bf \cfrac{2+tan^2(x)}{sec^2(x)}\implies \cfrac{2+\frac{sin^2(x)}{cos^2(x)}}{\frac{1}{cos^2(x)}}\implies
\cfrac{\frac{2cos^2(x)+sin^2(x)}{cos^2(x)}}{\frac{1}{cos^2(x)}}
\\ \quad \\

\cfrac{2cos^2(x)+sin^2(x)}{\cancel{ cos^2(x) }}\cdot \cfrac{\cancel{ cos^2(x) }}{1}\)

Answer 16

LCD for the 2 in the numerator? I'm beginning to see thats where I stop thinking xD

Answer 17

well the LCD atop is \(cos^2(x)\) so 2 is really 2/1

so \(cos^2(x) \div 1 = cos^2(x)\) then that * the 2 atop is \(2cos^2(x)\)

Answer 18

Ahh, but now this answer needs some sort of numerical value or further simplification, I havent tried it with the f(x)^2

Answer 19

Still no luck with that either, your results?

Answer 20

hmm ok so \(\bf \cfrac{2+tan^2(x)}{sec^2(x)}\implies \cfrac{2+\frac{sin^2(x)}{cos^2(x)}}{\frac{1}{cos^2(x)}}\implies
\cfrac{\frac{2cos^2(x)+sin^2(x)}{cos^2(x)}}{\frac{1}{cos^2(x)}}
\\ \quad \\

\cfrac{2cos^2(x)+sin^2(x)}{\cancel{ cos^2(x) }}\cdot \cfrac{\cancel{ cos^2(x) }}{1}\implies 2cos^2(x)+sin^2(x)
\\ \quad \\
cos^2(x)+cos^2(x)+sin^2(x)\qquad {\color{brown}{ sin^2(\theta)+cos^2(\theta)=1}}\qquad thus
\\ \quad \\
cos^2(x)+1\)

Answer 21

it could simplify to that

Answer 22

Nope :\

Answer 23

Not sure how to do this one lol

Answer 24

hmm can you post a screenshot of the material?.... maybe it has some hint

Answer 25

Yeah one sec

Answer 26

hmmm the most I can think of it could come up to .... would be \(\bf cos^2(x)+1=[f(x)]^2\implies \pm \sqrt{cos^2+1}=f(x)\)

Answer 27

Negative on both according to WeBWork lol

Answer 28

ohhh ahemm smokes... just notice.... there's a -1 there

Answer 29

You're right xD

Answer 30

Its just cosx

Answer 31

\( \bf \cfrac{2+tan^2(x)}{sec^2(x)}-1\implies \cfrac{2+\frac{sin^2(x)}{cos^2(x)}}{\frac{1}{cos^2(x)}}-1\implies
\cfrac{\frac{2cos^2(x)+sin^2(x)}{cos^2(x)}}{\frac{1}{cos^2(x)}}-1
\\ \quad \\

\cfrac{2cos^2(x)+sin^2(x)}{\cancel{ cos^2(x) }}\cdot \cfrac{\cancel{ cos^2(x) }}{1}-1\implies 2cos^2(x)+sin^2(x)-1
\\ \quad \\
cos^2(x)+cos^2(x)+sin^2(x)\qquad {\color{brown}{ sin^2(\theta)+cos^2(\theta)=1}}\qquad thus
\\ \quad \\
cos^2(x)\cancel{ +1-1 }\)

Answer 32

well \(\bf cos^2(x)+cos^2(x)+sin^2(x)-1\qquad {\color{brown}{ sin^2(\theta)+cos^2(\theta)=1}}\qquad thus
\\ \quad \\
cos^2(x)\cancel{ +1-1 }\)

Answer 33

Just cosx the (f(x)) is squared so they were canceled

Answer 34

and yes... once you take the square root... leaves you with cos(x) only

Answer 35

This one seems easy its on the interval [0,2pi], I tried sqrt3/2 but no luck

Answer 36

Ahh wait it has to be a multiple of pi

Answer 37

\(\bf sin^2(t)=\cfrac{3}{4}\implies sin(t)=\sqrt{\cfrac{3}{4}}\implies sin^{-1}[sin(t)]=sin^{-1}\left(\cfrac{\sqrt{3}}{2}\right)
\\ \quad \\
\measuredangle t=sin^{-1}\left(\cfrac{\sqrt{3}}{2}\right)\)

Answer 38

which should be pi/3 correct?

Answer 39

yes

Answer 40

or 2pi/3

Answer 41

Hmm let me include the whole problem and maybe youll get further than I

Answer 42

k

Answer 43

I've tried pi/3, 2pi/3, and 1/3

Answer 44

Not sure how long you will be on, I'll be back in 5-15. Let me know what you get.

Answer 45

your answer is correct, \(\frac{\pi}{3},\frac{2\pi}{3}\) are both correct HOWEVER
the instructions ask to only give the rational, leaving the \(\pi\) out, since is already understood

thus \(\bf \frac{1}{3},\frac{2}{3}\)

Answer 46

I'm still here for now and yeah just tried lol nothing for that either >.>

Answer 47

hmmm

Answer 48

\(\bf sin^2(t)=\cfrac{3}{4}\implies sin(t)=\pm\sqrt{\cfrac{3}{4}}\implies sin^{-1}[sin(t)]=sin^{-1}\left(\cfrac{\pm\sqrt{3}}{2}\right)
\\ \quad \\
\measuredangle t=sin^{-1}\left(\cfrac{\pm\sqrt{3}}{2}\right)\to
\large
\begin{cases}
\frac{\pi}{3}\\
\frac{2\pi}{3}\\
\frac{4\pi}{3}\\
\frac{5\pi}{3}
\end{cases}\)

Answer 49

the \(\Large \pm\) of the root...would end up including all angles in each quadrant

Answer 50

Ahh do + and - both?

Answer 51

So*

Answer 52

so, you'd need to include all those, the + ones and the - ones

Answer 53

Yeah no luck lol not sure why but none