If cos A= -6/7, in quadrant 2, determine the exact value of tan 2A

Question

I don't understand how to do this?

Answer 1

Several ways to do this.
This might be the easiest to understand ...

Make a sketch of a right-angled triangle in quad II
since cosA = x/r = -6/7
x = -6 and r = 7 (remember r is always positive)

by Pythagoras:
x^2 + y^2 = r^2
36 + y^2 = 49
y^2 = 13
y = ±√13 , but we are in quad II, so y = +√13

so now we can find sinA = √13/7

We need several identities:
tan 2A = sin 2A/cos 2A
and sin 2A = 2sinAcosA, cos 2A = cos^2 A - sin^2 A

so tan 2A = (2sinAcosA)/(cos^2 A - sin^2 A)
= 2(√13/7)(-6/7) / (36/49 - 13/49)
= (-12√13/49) / (23/49)
= -12√13/23

Another way:
recall tan 2A = 2tanA/(1 - tan^2 A)
and from above: tanA = sinA/cosA = (√13/7) / (-6/7)
= -√13/6

so tan 2A = (-2√13/6) / (1 - 13/36)
= (-2√13/6) / (23/36)
= (-2√13/6)(36/23)
= -12√13/23

The second way looks actually easier, but involves a more complicated formula