Trig Question

[Troyvdg]

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Which value of &#952; such that &#960;<&#952;<2&#960; makes the expression sin2(&#952😉&#8901;cos(&#952😉 largest?

A:&#960;
B:5&#960;/3 *answer
C:3&#960;/4
D:5&#960;/4
E:3&#960;/2

I have no idea how to solve trig problems like this. your help is greatly appreciated!

 

[qnzbabi91]

Don't worry on the real test they just ask you simple things like which is not a real trig function and sin, cos, and tan. i got a 27 on math and most of the questions were easy, algebra based. Focus more on rates and practice with math destroyer. But for that problem you have to use the trig identity that sin2x=2sinxcosx. And since pii is equal to radians find out what value makes them zero for instance sin 0 is 0 and cos 90 is 0. eliminate etc. also, realize that the highest sign or cos can be is 1.

 

[torobcheh21]

still don't get it...

 

[aqz]

Okay guys. Go learn your unit circles, it'll make trig much much easier.

The answer for this question is wrong though.

What qnz is saying is...these types of problems are more or less a "guess and check" kind of thing.

First you need to simplify everything, to make guess and check easier.

So what he did was....

sin(2x)cosx = (2sinxcosx)*cosx = 2sin(x)*cos^2(x)

He also knew that sin(pi) = 0 and cos(3pi/2) = 0, (this is where you need to memorize your unit circles), which makes choice A and choice E both 0.

So you have choice d and choice b. Let's look at the simplified equation again:

2*sin(x)*cos^2(x)

So as long as sin(x) isn't negative, the whole expression will be positive because cos^2 will always be positive.

But choice b is in the 4th quadrant where sin(x) is negative, and choice d is in the 3rd quadrant where sin(x) is still negative.

So unless it's a tie between choice A and choice E where the answer is 0, the answer has to be choice C .... but:

3pi/4 (choice c) lies in the 2nd quadrant...it can't even be right because it's not between pi and 2pi as the given condition in the problem. If the condition is changed to between 0 and 2pi or something, then 3pi/4 would be the right answer where sin(x) would be positive.

Where the hell did you get this problem from? -_-

 

[Troyvdg]

I got it from qvault.
Where's the best place to refresh on unit circles? I've tried khan academy, but I didn't find a video that covered unit circles to this extent..just the basic sin gives you y-coordinate, cos gives you x-coordinate type of stuff

 

[412smb412]

Okay guys. Go learn your unit circles, it'll make trig much much easier.

The answer for this question is wrong though.

What qnz is saying is...these types of problems are more or less a "guess and check" kind of thing.

First you need to simplify everything, to make guess and check easier.

So what he did was....

sin(2x)cosx = (2sinxcosx)*cosx = 2sin(x)*cos^2(x)

He also knew that sin(pi) = 0 and cos(3pi/2) = 0, (this is where you need to memorize your unit circles), which makes choice A and choice E both 0.

So you have choice d and choice b. Let's look at the simplified equation again:

2*sin(x)*cos^2(x)

So as long as sin(x) isn't negative, the whole expression will be positive because cos^2 will always be positive.

But choice b is in the 4th quadrant where sin(x) is negative, and choice d is in the 3rd quadrant where sin(x) is still negative.

So unless it's a tie between choice A and choice E where the answer is 0, the answer has to be choice C .... but:

3pi/4 (choice c) lies in the 2nd quadrant...it can't even be right because it's not between pi and 2pi as the given condition in the problem. If the condition is changed to between 0 and 2pi or something, then 3pi/4 would be the right answer where sin(x) would be positive.

Where the hell did you get this problem from? -_-

i got a problem with trig equations, i avoid them like the plague! make my life easier!!!! i have no idea what trig equations to memorize for this dam test! which do u recommend? thx!

 

[aqz]

Yeah I was thinking this was either a Kaplan or qvault question. Lots of mistakes.

Google search a unit circle picture. Make sure it has both radians/degrees and sin/cos/tan values. If it doesn't have sin/cos/tan values, figure them out and write them down on a new circle with the radians/degrees.

@SMB....the equations you should know are all in math destroyer but I'll list the important ones:

sin^2(x) + cos^2(x) = 1

tan^2(x) + 1 = sec^2(x) <----- For this one, just divide the above equation by cos^2(x).

sin(2x) = 2sinxcosx

cos(2x) = cos^2(x) - sin^2(x)

secx = 1/cosx

cscx = 1/sinx

And know the unit circle. That will allow you to not only memorize each trig value, but also the signs so you don't have to memorize if cos(-x) = cos(x) or whatever.

 

[mk4]

The OP stated sin2(&#952😉&#8901;cos(&#952😉

The problem should be written as sin^2(&#952😉&#8901;cos(&#952😉

The squared sine tells you that the angle should be in quadrant III or IV. The cosine would have to be in quadrant I or IV. Only quadrant IV is in common, and only answer B satisfies this.