Trig Equation Involving a Multiple of an Angle

[sweiss]

Question is:

Evaluate sin 2x, if cot x = -2/9 and π/2 < x < π.
sin 2x _______

(in the problem, x is theta...I couldn't find a way to utilize that notation here.)

So far this is what I have worked through...

tan x = -9/2
(-9/r)^2 + (2/r)^2 = 1
85/r^2 =1
r = + - sqrt 85

since x is between π/2 and π, x must be in Quadrant II, making sin x = 9sqrt85/85

I'm not sure if what I have worked through is even moving in the correct direction for this problem. I know that the 2 in front of theta changes the solution to the problem, but am unsure where to go from here. Also, I am definitely aware that I could be completely off base with my attempts so far. Thank you so much for any assistance that can be provided. I am happy to clarify if anything within this post doesn't make sense!

Thank you,

S. Weiss

 

[Deleted member 4993]

Question is:

Evaluate sin 2x, if cot x = -2/9 and π/2 < x < π.
sin 2x _______

(in the problem, x is theta...I couldn't find a way to utilize that notation here.)

So far this is what I have worked through...

tan x = -9/2
(-9/r)^2 + (2/r)^2 = 1
85/r^2 =1
r = + - sqrt 85

since x is between π/2 and π, x must be in Quadrant II, making sin x = 9sqrt85/85

I'm not sure if what I have worked through is even moving in the correct direction for this problem. I know that the 2 in front of theta changes the solution to the problem, but am unsure where to go from here. Also, I am definitely aware that I could be completely off base with my attempts so far. Thank you so much for any assistance that can be provided. I am happy to clarify if anything within this post doesn't make sense!

Thank you,

S. Weiss

Excellent work so far.

sin x = 9/sqrt(85) ...............................correct

Now then

cos (x) = ? .......................................(remember x is in 2nd. quadrant)

sin(2x) = 2 * sin(x) * cos(x) = ?

 

[Steven G]

I would draw the triangle in quad II with the opposite being 9, the adjacent being -2 ant the hypotenuse being sqrt(9^2+(-2)^2) = sqrt(85).
Now that you have the complete triangle labelled and know that sin(2x) = 2sin(x)cos(x) the rest should follow. Always draw the triangle!

 

[sweiss]

I would draw the triangle in quad II with the opposite being 9, the adjacent being -2 ant the hypotenuse being sqrt(9^2+(-2)^2) = sqrt(85).
Now that you have the complete triangle labelled and know that sin(2x) = 2sin(x)cos(x) the rest should follow. Always draw the triangle!

Drawing the triangle makes sense, but why is sin2(theta) = 2sin(theta)cos(theta) ?

 

[Deleted member 4993]

Drawing the triangle makes sense, but why is sin2(theta) = 2sin(theta)cos(theta) ?

That is one of the fundamental trigonometric identities (may be not so fundamental) derived from:

sin (x + y) = sin(x) * cos(y) + sin(y) * cos(x)

putting x = y, we get:

sin(2x) = 2 * sin(x) * cos(x)

 

[sweiss]

Right after I posted a reply, I realized that is what you were referencing! I’m away from my desk right now, but will try applying that once I get back.

 

[sweiss]

Excellent work so far.

sin x = 9/sqrt(85) ...............................correct

Now then

cos (x) = ? .......................................(remember x is in 2nd. quadrant)

sin(2x) = 2 * sin(x) * cos(x) = ?

Thanks!
cos (x) = -2sqrt(85)/85
sin(2x) = 2 * sin (x) * cos (x) =
2(9sqrt85/85)(-2sqrt85/85) = -0.424 (rounded)

 

[Deleted member 4993]

Thanks!
cos (x) = -2sqrt(85)/85
sin(2x) = 2 * sin (x) * cos (x) =
2(9sqrt85/85)(-2sqrt85/85) = -0.424 (rounded)

In trigonometry - most of the time exact answers are expected - simplified but leave the answer in simplified fraction form. In this case your answer should be:

sin(2x) = - 36/85 ..............................(edited)

 

[sweiss]

In trigonometry - most of the time exact answers are expected - simplified but leave the answer in simplified fraction form. In this case your answer should be:

sin(2x) = 36/85

That makes sense! Would sin(2x) be positive since it is quadrant 2? The reason this is confusing me is that when I compute sin(2x)=2 * sin(x) * cos(x) I get a negative product since sine is positive and cosine is negative...

Thank you so much for the time you have taken so far!

 

[sweiss]

That makes sense! Would sin(2x) be positive since it is quadrant 2? The reason this is confusing me is that when I compute sin(2x)=2 * sin(x) * cos(x) I get a negative product since sine is positive and cosine is negative...

Thank you so much for the time you have taken so far!

I certainly don't mean to bother, but I have to submit some of my work today, and was hoping to understand how the answer comes out positive within this problem. I understand if you can't get back with me, but just wanted to refresh my reply in case it got lost in the mass of messages! Thanks again for all your assistance thus far.

 

[JeffM]

Did you use the arcsine function to evaluate 2x and then calculate x?

By the way, I edited my final response in your other thread. It is an important edit.

BIG EDIT

My clue above about calculating x may be too obscure. Just because x is in the second quadrant does not entail that 2x is also.

 

[sweiss]

What makes you say that your answer is in the second quadrant? Did you use the arcsine function to evaluate 2x and then calculate x?

By the way, I edited my final response in your other thread. It is an important edit.

I didn't use arcsine at any point. I found sin(x)=(9sqrt85)/85 and cos(x)=(-2sqrt85)/85 by using cot(x) provided in the original problem and placing them in Quadrant II. I then used the addition formula to get sin2(x)= 2 * sin(x) * cos(x) which got me a fractional answer of -36/85. The previous person that I was communication with arrived at the same answer but theirs was positive. I'm trying to see where I went wrong... Not sure if that all makes sense though.

I just look back at your edit on the other thread. Thats a very helpful point that I hadn't thought of. tan^2x definitely makes sense for that problem.

 

[JeffM]

I didn't use arcsine at any point. I found sin(x)=(9sqrt85)/85 and cos(x)=(-2sqrt85)/85 by using cot(x) provided in the original problem and placing them in Quadrant II. I then used the addition formula to get sin2(x)= 2 * sin(x) * cos(x) which got me a fractional answer of -36/85. The previous person that I was communication with arrived at the same answer but theirs was positive. I'm trying to see where I went wrong... Not sure if that all makes sense though.

I just look back at your edit on the other thread. Thats a very helpful point that I hadn't thought of. tan^2x definitely makes sense for that problem.

unlike the other problem, I have not worked through this problem. But looking at the original statement, I see it places x in the second quadrant. So what quadrant or quadrants may 2x be in. That will tell you the sign of sin(2x). You are correct: you do not need the arcsine function. Sorry. I was trying to hurry in response to your request and did not think hard enough on my first answer.

 

[sweiss]

unlike the other problem, I have not worked through this problem. But looking at the original statement, I see it places x in the second quadrant. So what quadrant or quadrants may 2x be in. That will tell you the sign of sin(2x). You are correct: you do not need the arcsine function. Sorry. I was trying to hurry in response to your request and did not think hard enough on my first answer.

Hmmm, I'm honestly drawing a blank as to how I would determine what quadrant(s) 2x might be in. My instinct would be Q I& II. So that would make my final answer of sin2x positive...

 

[JeffM]

In the givens of the problem

[MATH]\dfrac{\pi}{2} < x < \pi \implies \pi < 2x < 2\pi[/MATH]
Thus, 2x is in either the third or fourth quadrant. Which means the sign of the sine is negative.

Furthermore, x is in the second quadrant so the sign of sin(x) is positive and the sign of cos(x) is negative so their product is indeed negative.

We tutors sometimes make mistakes too. Feel free to challenge us (politely of course). My initial reaction was to actually find the approximate value of x using the arcsine function, which is totally unnecessary. Simple multiplication of the original inequality is all that is needed.

 

[sweiss]

In the givens of the problem

[MATH]\dfrac{\pi}{2} < x < \pi \implies \pi < 2x < 2\pi[/MATH]
Thus, 2x is in either the third or fourth quadrant. Which means the sign of the sine is negative.

Furthermore, x is in the second quadrant so the sign of sin(x) is positive and the sign of cos(x) is negative so their product is indeed negative.

We tutors sometimes make mistakes too. Feel free to challenge us (politely of course). My initial reaction was to actually find the approximate value of x using the arcsine function, which is totally unnecessary. Simple multiplication of the original inequality is all that is needed.

Got it! I totally forgot about π < 2x < 2π within the equation. That really makes more sense now. Thanks much!

 

[Deleted member 4993]

I certainly don't mean to bother, but I have to submit some of my work today, and was hoping to understand how the answer comes out positive within this problem. I understand if you can't get back with me, but just wanted to refresh my reply in case it got lost in the mass of messages! Thanks again for all your assistance thus far.

You are correct! the value of sin(2x) is negative in this case (sin(x) is positive and cos(x) is negative). Sorry about the Khan-fusion.I corrected my response(#8).

I am going to the corner right now......

 

[sweiss]

You are correct! the value of sin(2x) is negative in this case (sin(x) is positive and cos(x) is negative). Sorry about the Khan-fusion.I corrected my response(#8).

I am going to the corner right now......

Lol! No problems...thanks so much for all your assistance, I appreciate it!