Question
𝑐𝑜𝑠 (𝑠𝑖𝑛^−1 ((√3)/2)− 𝑐𝑜𝑡^−1(√3))
Answers
oobleck
These are easy to figure if you draw the right triangles involved. If you label the two sides specified by the trig function, you can then label the 3rd side and see the rest of the trig functions.
You have cos(x-y) = cosx cosy + sinx siny
sinx = √3/2
so, cosx = 1/2
coty = √3
so, siny = 1/2
cosy = √3/2
cos(x-y) = (1/2)(√3/2) + (√3/2)(1/2) = √3/2
check:
x = π/3
y = π/6
cos(π/3 - π/6) = cos(π/6) = √3/2
You have cos(x-y) = cosx cosy + sinx siny
sinx = √3/2
so, cosx = 1/2
coty = √3
so, siny = 1/2
cosy = √3/2
cos(x-y) = (1/2)(√3/2) + (√3/2)(1/2) = √3/2
check:
x = π/3
y = π/6
cos(π/3 - π/6) = cos(π/6) = √3/2