draw your triangles. These are two of the simplest standard integer-sided ones. The have sides
3-4-5 and 5-12-13
So, if
sinA = 3/5, cosA = 4/5
cosB = 5/13, sinB = 12/13
Now just use your definitions and formulas:
cotB = cosB/sinB = 5/12
sin2A = 2sinAcosA = 2(3/5)(4/5) = 24/25
cos(5π/6+B) = cos(5π/6)cosB - sin(5π/6)sinB
= (-√3/2)(5/13)-(1/2)(12/13)
= -(12+5√3)/338
tan(A-π/4) = (tanA-tan π/4)/(1+tanA tan π/4)
= (3/4 - 1)/(1 + (3/4)(1))
= (1/4)/(7/4)
= 1/7
If sinA=3/5 and cosB=5/13 and if A and B are measures of two angles in Quadrant I, find the exact value of the following functions.
cotB =
sin2A=
3)cos(5pi/6 + B) =
tan(A - pi/4) =
2 answers
the last is -1/7
But I'm sure you saw my typo ...
But I'm sure you saw my typo ...