Manipulating Trig Functions in terms of "a" and "b"

[coooool222]

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) ???

 

[BigBeachBanana]

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) ???

Hint: [imath]\sin(\theta)[/imath] has a period of [imath]2\pi[/imath]

 

[coooool222]

Hint: [imath]\sin(\theta)[/imath] has a period of [imath]2\pi[/imath]

ah i see so it's still a/b since sin(theta+4pi) = sin(theta)

 

[Steven G]

ah i see so it's still a/b since sin(theta+4pi) = sin(theta)

...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 angle

 

[khansaheb]

c)tan^2(theta) = a^2/b^2-a^2

That should be written as

tan^2(theta) = a^2/(b^2-a^2)

A bit of nit-picking, but an important nit to pick.

 

[khansaheb]

d) sin(theta + 4pi) = i have no idea for this one, this is a shift to the left so I'm guessing?

One way to calculate:

Sin(A+B) = Sin(A) * Cos(B) + Cos(A) * Sin(B)

Sin(4pi + x) = Sin(4pi) * Cos(x) + Cos(4pi) * Sin(x) ..... continue....