Integrating trig

[woeful schooner]

$$\int_{0}^{\frac{\pi}{12}}\sin(3x)dx$$

[wheat spade]

Rizz Cooker

[carmine aspen]

$$[-\frac{1}{3}cos(3x)]_{0}^{\frac{\pi}{12}}$$

[woeful schooner]

yeah issue is the answer i got to

[wheat spade]

AmbivalentChamp

[carmine aspen]

hm?

[woeful schooner]

the answer is somehow 1/3 - root2/6

[carmine aspen]

yeah?

[woeful schooner]

its meant to be in exact form

issue is im not getting to that

[carmine aspen]

-⅓cos(3(π/12)) - - ⅓cos(0)?

[woeful schooner]

yeah but how do i work towards exact values

the answer has to specifically be in the form 1/3 - root2/6

[carmine aspen]

cos(3(π/12) = root2/2 (cos(π/4))

×-⅓ = -root2/6

Cos(0) = 1

so - - ⅓ × 1 = ⅓

so ⅓ - root2/6

[woeful schooner]

now im more confused

its so joever for me

where did that pi/4 come from

[carmine aspen]

3 × π/12

= 3π/12

[woeful schooner]

oh yeah

[carmine aspen]

= ¼π

[woeful schooner]

what about the root2/2

[carmine aspen]

-⅓(cos(π/4)) = -⅓(root2/2)

cos(π/4) = cos(45) = 1/root2 = root2/2

[woeful schooner]

omds i found my problem

i was in degrees on my calculator

ffs

thx for the help though