Trig Question

[samoth]

Can someone simplify:

v = bw sin[(pi/2)[1+(1/4)cos(4wt)] - b(pi/2)w sin(4wt), where b,w = constants.

I know the answer, but I'm missing some damn trig identity that gets me there.

TYIA

 

[samoth]

Can I set sin [(pi/2)[1+cos(something)]] to cos [ncos(something)]? Like, take the sin (pi/2) and make it a cosine?

 

[samoth]

Strange, the [(pi/2)[1+(1/4)cos(4wt)] doesn't show up in the plat preview option. I wonder if you can make a post entirely in [] and have it not show up in the plat preview?

 

[jerkbox]

dude, you're expecting a lot from a community of beings who are slightly more intelligent than apes

 

[samoth]

dude, you're expecting a lot from a community of beings who are slightly more intelligent than apes

I thought they genetically engineered apes to be able to do algebra? I think they're called "juicers" or some scientifically technical slang like that.

 

[jerkbox]

I thought they genetically engineered apes to be able to do algebra? I think they're called "juicers" or some scientifically technical slang like that.

me hungry. me eat now.

 

[all the whey]

 

[UA_Iron]

I should ask my old roommate. Undergrad in engineering math/mathematics (3.7 cumulative GPA), completing a masters in Mathematics currently.

 

[heatherrae]

I went to law school so I wouldn't have to do any more math.

 

[redguru]

Can you give me the answer so I can figure out which trig identity you need?

 

[all the whey]

I sat by and befriended an asian chick in college.

 

[NickyE3]

subnettin and summarizing an IP network is about all the math i deal with anymore, maybe some dB's and mW's every now and then

 

[samoth]

Can you give me the answer so I can figure out which trig identity you need?

The problem is finding and graphing the speed, so the answers are:

...

=> v = bw cos [(pi/8)cos(4wt)] - bw(pi/2)sin(4wt)

and thus

Abs (v) = bw[cos^2[(pi/8)cos(4wt)+(pi^2/4)sin^2(4wt)]^(1/2)

What I can't figure out is how it goes from a sine and cosine in the first term to two cosines, to a cosine square and a cosine. I'm pretty sure it has to do with setting sin(pi/2) => cos. And the cos^2 and cos come from squaring, so I guess I know that part. The only thing I can't quite get is going from sin(x+cos) to cos(cos). The minus sign seperates coordinates in the cylindrical system, so there can't be any combining there until Abs(v) is taken.

Hmm... I wonder if the Ti-200 can simplify like Maple can...

 

[redguru]

The problem is finding and graphing the speed, so the answers are:

...

=> v = bw cos [(pi/8)cos(4wt)] - bw(pi/2)sin(4wt)

and thus

Abs (v) = bw[cos^2[(pi/8)cos(4wt)+(pi^2/4)sin^2(4wt)]^(1/2)

What I can't figure out is how it goes from a sine and cosine in the first term to two cosines, to a cosine square and a cosine. I'm pretty sure it has to do with setting sin(pi/2) => cos. And the cos^2 and cos come from squaring, so I guess I know that part. The only thing I can't quite get is going from sin(x+cos) to cos(cos). The minus sign seperates coordinates in the cylindrical system, so there can't be any combining there until Abs(v) is taken.

Hmm... I wonder if the Ti-200 can simplify like Maple can...

Ok, that's easy.

sin[(pi/2) -x] = cos x is the first trig identity you need for the first term.

so the first term goes from sin[(pi/2)(1+1/4cos(4wt))] to

sin [pi/2 + pi/8 cos(4wt)] = cos (-pi/8 cos (4wt))

cos(-x) = cos(x)

cos (-pi/8 cos (4wt)) = cos (pi/8 cos (4wt))

 

[redguru]

The problem is finding and graphing the speed, so the answers are:

...

=> v = bw cos [(pi/8)cos(4wt)] - bw(pi/2)sin(4wt)

and thus

Abs (v) = bw[cos^2[(pi/8)cos(4wt)+(pi^2/4)sin^2(4wt)]^(1/2)

What I can't figure out is how it goes from a sine and cosine in the first term to two cosines, to a cosine square and a cosine. I'm pretty sure it has to do with setting sin(pi/2) => cos. And the cos^2 and cos come from squaring, so I guess I know that part. The only thing I can't quite get is going from sin(x+cos) to cos(cos). The minus sign seperates coordinates in the cylindrical system, so there can't be any combining there until Abs(v) is taken.

Hmm... I wonder if the Ti-200 can simplify like Maple can...

I redid the first simplification in MathType

 

[redguru]

You should be then able to square the terms then take the square root to get |v|

 

[samoth]

sin[(pi/2) -x] = cos x is the first trig identity you need for the first term.

THAT'S what I needed. I've totally never seen that identity before. I know sin(pi/2) is 1, but I didn't think it went from two terms to just the cosine term. Thanks!

BTW, do you own MathType, too? I got tired of using MTLite, so I slurged and purchased it a couple years ago. (Although using the preview mode of Physics Forums works too . But that's not too efficient for writing papers, lol.)

 

[redguru]

THAT'S what I needed. I've totally never seen that identity before. I know sin(pi/2) is 1, but I didn't think it went from two terms to just the cosine term. Thanks!

BTW, do you own MathType, too? I got tired of using MTLite, so I slurged and purchased it a couple years ago. (Although using the preview mode of Physics Forums works too . But that's not too efficient for writing papers, lol.)

Yeah, I own mathtype.

 

[redguru]

 

[samoth]

What the crap... fine time for my Ti-200 to stop working. I guess it requires the 3V Li battery to be working. (Changed AAA's; no effect.)

 

[rykertest]

what just happened here?

 

[Nathan]

THAT'S what I needed. I've totally never seen that identity before. I know sin(pi/2) is 1, but I didn't think it went from two terms to just the cosine term. Thanks!

BTW, do you own MathType, too? I got tired of using MTLite, so I slurged and purchased it a couple years ago. (Although using the preview mode of Physics Forums works too . But that's not too efficient for writing papers, lol.)

A sine plot is simply a cosine plot shifted by pi/2. Picture their respective graphs. I wouldn't call it an identity so much as the definition of either sine or cosine.