If cos A= -6/7, in quadrant 2, determine the exact value of tan 2A
I don't understand how to do this?
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
To find the exact value of tan 2A, we first need to determine the value of A. We are given that cos A = -6/7 and that A is in quadrant 2.
In quadrant 2, cosine is negative, so the cosine ratio is negative. Since cos A = -6/7, we can find the value of sin A using the Pythagorean identity:
sin^2 A + cos^2 A = 1
Substituting the value of cos A, we get:
sin^2 A + (-6/7)^2 = 1
sin^2 A + 36/49 = 1
sin^2 A = 1 - 36/49
sin^2 A = 49/49 - 36/49
sin^2 A = 13/49
sin A = ±√(13/49)
sin A = ±(√13)/7
Since A is in quadrant 2, sin A is positive. Therefore, sin A = (√13)/7.
Now that we have the values of sin A and cos A, we can find the value of tan A:
tan A = sin A / cos A
tan A = (√13)/7 / (-6/7)
tan A = -√13/6
To find tan 2A, we use the double-angle formula for tangent:
tan 2A = 2tan A / (1 - tan^2 A)
Substituting the value of tan A, we get:
tan 2A = 2(-√13/6) / (1 - (-√13/6)^2)
tan 2A = -2(√13/6) / (1 - 13/36)
tan 2A = -2(√13/6) / (36/36 - 13/36)
tan 2A = -2(√13/6) / (23/36)
tan 2A = -2√13/23
Therefore, the exact value of tan 2A is -2√13/23.
To find the exact value of tan 2A, we'll use the double angle formula for tangent.
The double angle formula for tangent states that:
tan 2A = (2 * tan A) / (1 - tan^2 A)
To find the exact value of tan 2A, we need to determine the values of sin A and cos A based on the given information.
Given that cos A = -6/7 and A is in quadrant 2, we can determine sin A using the Pythagorean identity:
sin^2 A + cos^2 A = 1
Plugging in our given value of cos A:
sin^2 A + (-6/7)^2 = 1
Simplifying:
sin^2 A + 36/49 = 1
sin^2 A = 1 - 36/49
sin^2 A = 13/49
Since A is in quadrant 2, sin A is positive. Taking the square root of both sides:
sin A = √(13/49)
sin A = √13 / 7
Now that we have the values of sin A and cos A, we can find tan A:
tan A = sin A / cos A
tan A = (√13 / 7) / (-6/7)
tan A = -√13 / 6
Next, let's use the double angle formula for tangent:
tan 2A = (2 * tan A) / (1 - tan^2 A)
tan 2A = (2 * (-√13 / 6)) / (1 - (-√13 / 6)^2)
Simplifying further:
tan 2A = (-2√13 / 6) / (1 - 13/36)
tan 2A = (-2√13 / 6) / (36/36 - 13/36)
tan 2A = (-2√13 / 6) / (23/36)
To simplify this expression even further, we can multiply both the numerator and denominator by the reciprocal of 23/36, which is 36/23:
tan 2A = (-2√13 / 6) * (36/23)
tan 2A = (-2 * 36√13) / (6 * 23)
tan 2A = (-72√13) / 138
Finally, we can simplify the expression:
tan 2A = -36√13 / 69
Therefore, the exact value of tan 2A, given cos A = -6/7 in quadrant 2, is -36√13 / 69.