Hello, I need a bit of help with the following problem:
If sinA=3/5 when pi/2 < A < pi and cosB=5/13 when 3pi/2 < B < 2pi, find the exact value of the following equations:
cotB =
sin2A =
tan(A-pi/4)=
As you would do normally for these kind of problems, sketch a right-angled triangle
since sinA = 3/5 = opposite/hypotenuse, you should recognize the standard 3-4-5 Pythagorean triangle
cosA = -4/5 , since A is in quad II
then sin 2A = 2sinAcosA = .... you have both
for cosB = 5/13, you have the other common right-angled triangle with sides 5-12-13
sinB = -12/15, since B is in quad IV
cotB = cosB/sinB = (-12/15) / (5/13) = -12/5
tan(A-pi/4) <--- several ways to do this, how about this
= (tanA - tanπ/4)/(1 + tanAtanπ/4) , we know tanπ/4 = 1
= (tanA - 1)/(1 + tan A)
but tanA= -3/4
= (-3/4-1)/(1 - 3/4)
= (-7/4) / (1/4)
= -7
check my arithmetic