Trigonometry

Chucky

Find all angles between 0 and 360 that satisfy sin 2θ.


How do I do this? It's 23:39 on a Friday night and I'm trying to do maths... ...I need a social life (But so do you if you answer before the morning)

Chucky

Is it just like this (I'm rusty on Trigonometry):


sin2θ = 2 Sinθ Cosθ = -0.7

Sinθ Cosθ = -0.35

Sinθ = -0.35 OR Cosθ = -0.35


...and work out he angles from there yeh?

Nietzschean

Since sin is continious between 0 and 360...

0<= θ<=180
All angles in that range will have 2θ within the 0 to 360 range, which are all valid for 2θ.

Though that question seems odd, is there some part missing?

Chucky

Oh bugger yes, it should be:

Find all angles between 0 and 360 that satisfy sin 2θ = -0.7

Nietzschean

ah well then...
Sin is negitive in 2 quadrants, the 3rd and 4th...
so getting asin(.7) = 44.427
in the first and 3rd quadrents this becomes :
180 + 44.427 = 224.427
270 + 44.427 = 314.427

So these are our 2 values for 2θ....
therefore θ = 157.2135 or 112.2135

Crumbs

Nietzschean wrote:

180 + 44.427 = 224.427
270 + 44.427 = 314.427

Should that 2nd value be 360 - 44.427 = 315.573 ?

And you can get further values of 2θ by adding 360 to each angle, such as:
224.427 + 360 = 584.427
315.573 + 360 = 675.573

so that when you calculate θ, your answers are still <360, ie. 112.214, 157.787, 292.214 and 337.787

Nietzschean

Crumbs wrote:

Should that 2nd value be 360 - 44.427 = 315.573 ?

Quite correct, my apoligies, tis a bit early in the day.....

Chucky

Is it that easy? That's what I ended up doing last night with it but I wasn't sure. It seemed as if it should be harder.

Nietzschean

most questions of that style are trick, people assume them to be much harder than they are...

Chucky

Sorry, I have another Trigonometry question:

Given that:

8sinθ / (5sinθ - cosθ) = 1

Find the values of the angles which lie between 0 and 360

breadmonkey

Come on man, do your own homework.

Chucky

It's not homework at all. I am simply revising maths for myself and ask for assistence considering I never covered a question of this nature in the Leaving Cert years.

David19

Chucky - Can you simplify it down? Multiply across by 5sinθ - cosθ.

Chucky

Yeh I can simplify it down but where do you go from there? That's where the bit of inventiveness is required and i managed to crack it today.


Once you get it down to 3Sinθ = - Cosθ you then divide across by Cosθ to get:

3Sinθ / Cosθ = -1

That is equal to:

3Tanθ = -1


From there it is easy to get the angles.

David19

Ok, fair play. These things just take lots of practice.