Originally Posted by Hiep HO
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 ≤ * ≤ 2* giving your answer in terms of *.
You should not need your calculator.
(i) sin* = √3/2 (ii) cos* = -√3/2 (ii) tan* = √3
2. Solve the following equations for 0 ≤ * ≤ 2* giving your answers to 1 d.p. where necessary.
(i) cos * = 0.4
(ii) tan * = -1.2
(iii) sin² * = 1
* - Pi
* - Theta
Thanks to everyone who particated I stay alive for 1 more day
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.
i) sin* = √3/2. to find what * 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 * = sin^1(√3/2). To then solve for n, we know that * is <= 2n, so */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. * = cos^-1(-√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 :P).
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 √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 ≤ * ≤ 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.