Huge help in Trigonometry needed!

[Hiep HO]

Hi Guys!

I was about to finish of a worksheet that was handed to me
on Friday as homework...
Until I came across these 2 questions that baffled ma mind...
Would appreciate it if you kind chaps could help me out?


=========Typed exactly from my sheet=========

1. Solve the following equations for 0 ≤ θ ≤ 2π giving your answer in terms of π.
You should not need your calculator.

(i) sinθ = √3/2 (ii) cosθ = -√3/2 (ii) tanθ = √3


2. Solve the following equations for 0 ≤ θ ≤ 2π giving your answers to 1 d.p. where necessary.

(i) cos θ = 0.4
(ii) tan θ = -1.2
(iii) sin² θ = 1

--------------------------------------
==KEY==
π - Pi
θ - Theta
--------------------------------------

Thanks to everyone who particated I stay alive for 1 more day
Well appreciated!


I don't know how but I'll return the favour to you guys!!!

 

[xGodZzZ]

1 (i) theta = 60 degrees, which is pi/3

i think. I dont feel like doing the others.

 

[malyce]

Hi Guys!

I was about to finish of a worksheet that was handed to me
on Friday as homework...
Until I came across these 2 questions that baffled ma mind...
Would appreciate it if you kind chaps could help me out?


=========Typed exactly from my sheet=========

1. Solve the following equations for 0 ≤ θ ≤ 2π giving your answer in terms of π.
You should not need your calculator.

(i) sinθ = √3/2 (ii) cosθ = -√3/2 (ii) tanθ = √3


2. Solve the following equations for 0 ≤ θ ≤ 2π giving your answers to 1 d.p. where necessary.

(i) cos θ = 0.4
(ii) tan θ = -1.2
(iii) sin² θ = 1

--------------------------------------…
==KEY==
π - Pi
θ - Theta
--------------------------------------…

Thanks to everyone who particated I stay alive for 1 more day
Well appreciated!


I don't know how but I'll return the favour to you guys!!!

Are you trying to find what theta is? I'm not quite sure I understand but I know I can help. Trig is what us engineers do

If that's the case, to solve for theta it's pretty much simple alegbra.

1. )
i) sin&#952; = &#8730;3/2. to find what &#952; is we must divide by a sin. How do you divide by a sin? Easy. Take the arcsin (sin^-1) of the fraction on the other side. So &#952; = sin^1(&#8730;3/2). To then solve for n, we know that &#952; is <= 2n, so &#952;/2 = n. n could also be larger than that. This is how I interpreted the problem: solve for theta, use theta to solve for how this relates to 2pi.
To do this, you must use the unit circle.
ii) Same situation. &#952; = cos^-1(-&#8730;3/2). Use that theta to solve for how this relates to 2pi (I'm assuming if you're in trig you can do algebra ).
iii) same situation again. How about you solve this one for practice?
If you have troubles with this, skip down to my note in pink for part 2. The orange text will give you a much easier method of finding relations to 2 pi than using a unit circle.

Hint: I am sure these will turn out to be theta = some factor of pi. Not just because it says it at the end of your homework, but because &#8730;3/2 is a trig identity for pi, same with the other values. Remember your circle that goes from zero to 2pi in increments of 1/4?

Like I said, I am unsure if these are the correct answers for n, but I do know they are correct for theta.

2) Basically the same situation as number 1. Use arcsins, arccosins, and arctangents to solve for theta (sin^-1 is arcsin). Round to 1 decimal place for the 2pi value (it might be some crazy fraction). In order to relate theta to 2pi you must now realize that 360 degrees = 2pi, so dividing 360 degrees by theta will give you a fraction of 360/theta. This is your ratio of what 2pi is in relation to theta. So a theta of 90 degrees would be 1/4*(2pi) or 1/4. You could do this above as well, but I wanted you to learn to use the unit circle


Edit #3. I finally realized that wasn't actually an n, it was pi. So from 0 &#8804; &#952; &#8804; 2pi. Solve for theta, and then find out how that theta relates to two pi (you MUST use the unit circle to solve for the relationship between your theta and pi. This is the unit circle.

 

[Zstunner]

I might be able to help you out shortly if you still need it. Finishing up some stakes currently.

 

[Koot]

If theta were 60 degrees, expressed in pi it would be pi/3. Pi is 180 degrees as 1 pi rad would take up half the circle.

For other degrees just take theta * (pi/180).

When you have one answer, use reference angles to find the other answer that is also in the domain of 0 to 2pi.

Reply if you still have questions.

 

[Hiep HO]

Thanks for the huge favour guys, the world needs more people like you guys!

Anyone got any ideas for queston 2??

 

[malyce]

Thanks for the huge favour guys, the world needs more people like you guys!

Anyone got any ideas for queston 2??

It's the same as question number one. We know that 2*pi = 360 degree, this is important.

2*pi = 360

First you must solve for theta, like you did in question 1. This should give you some degree value. You then take that degree value (lets say number 1 gives you 10.7 degrees (this is not correct, I don't have my calc with me).

Take 360/10.7 to find out how your theta value compares to 2*pi. The answer you get as a result of 360/10.7 is your answer as to how this compares to 2*pi. It's that simple.

Once again, all you need to do is 360/(your theta value).

For iii, because it is a sin squared, you must first take the square root of both sides before you take the arcsin of both sides!