i think you can build 2 triangle from cos(a) and sin(b)
#exact value?
mild pivot
empty raven
To find the missing functions, use the main trigonometric identity. However, recall which sign the functions have at which quarters.
Well, for the second one, you need cosine.
So, in general, we have:
cos(t)^2 + sin(t)^2 = 1
For π/2 < a < π we have sin(a) > 0, so:
sin(a) = √(1 - cos(a)^2)
For π < b < 3π/2 we have cos(b) < 0, so:
cos(b) = -√(1 - sin(b)^2)
Ao, now you can find the values of sin(a) and cos(b) and calculate sin(a + b).
empty raven
If cos(x)^2 + sin(x)^2 = 1, then cos(x) = ±√(1 - sin(x)^2), where the sign is determined by the location of the angle, an analogously for sin(x).
mild pivot
yeah you can either use the identity or construct a right triangle
the 2nd way will give you all 6 trig functions
you know cosa = A/H right?
so make -4 = A, 5 = H
you do same thing for sinb as well
then solve for remaining side
(just treat them as acute angle btw it makes things easier)
empty raven
Hm... Well, that works, but I don't really like to think of trigonometric functions as geometric ratios in algebra, but rather as, well, functions 😄
With some properties of symmetry, periodicity and such, of course.
verbal harnessBOT
@spring quartz