coooool222
Junior Member
- Joined
- Jun 1, 2020
- Messages
- 93
Hint: [imath]\sin(\theta)[/imath] has a period of [imath]2\pi[/imath]View attachment 36520
I know how to do a) b) and c) but i have no clue how to do sin(theta + 4pi)
a) csc (theta) = b/a
b) cos^2(theta)-1 = (b^2-a^2-b^2)/b^2 = -a^2/b^2
c)tan^2(theta) = a^2/b^2-a^2
d) sin(theta + 4pi) = i have no idea for this one, this is a shift to the left so I'm guessing?
sin (4pi*b/a) ???
ah i see so it's still a/b since sin(theta+4pi) = sin(theta)Hint: [imath]\sin(\theta)[/imath] has a period of [imath]2\pi[/imath]
...and yes, it is a shift to the left but as you noticed shifting to the left 4pi doesn't change the sin of any angleah i see so it's still a/b since sin(theta+4pi) = sin(theta)
That should be written asc)tan^2(theta) = a^2/b^2-a^2
One way to calculate:d) sin(theta + 4pi) = i have no idea for this one, this is a shift to the left so I'm guessing?