tanx= -12/5 x in quadrant 2
find;
sin2x=
cos2x=
tan2x=
draw a sketch of x y axis system
x = -5
y = 12
hypotenuse = 13
sin x = 12/13
cos x = -5/13
tan x = -12/5
now use 2x formulas
for example
sin 2x = 2 sin x cos x = 24/13 * -5/13
= -120/169
To find sin(2x), cos(2x), and tan(2x), we'll need to use the double-angle identities.
First, let's find sin(x) and cos(x) using the given information.
Given: tan(x) = -12/5, x is in quadrant 2.
In quadrant 2, the sine is positive and the cosine is negative.
Using the definition of tangent:
tan(x) = sin(x) / cos(x)
-12/5 = sin(x) / (-cos(x))
Since sin(x) is positive in quadrant 2, and cos(x) is negative, we can rewrite the equation as:
-12/5 = -sin(x) / cos(x)
By cross-multiplying, we get:
5(-sin(x)) = (-12)(cos(x))
-5sin(x) = -12cos(x)
Divide both sides by -5 to isolate sin(x):
sin(x) = 12/5cos(x)
Now, using the Pythagorean identity (sin^2(x) + cos^2(x) = 1), we can solve for cos(x):
sin^2(x) + cos^2(x) = 1
(12/5cos(x))^2 + cos^2(x) = 1
144/25cos^2(x) + cos^2(x) = 1
(144/25 + 1)cos^2(x) = 1
(144/25 + 25/25)cos^2(x) = 1
169/25cos^2(x) = 1
cos^2(x) = 25/169
cos(x) = ±sqrt(25/169)
cos(x) = ±5/13
Since cos(x) is negative in quadrant 2, we have:
cos(x) = -5/13, and sin(x) = 12/5( -5/13) = -12/13.
Now, let's find sin(2x), cos(2x), and tan(2x).
Using the double-angle formulas:
sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x)
tan(2x) = sin(2x) / cos(2x)
Plugging in the values we found:
sin(2x) = 2(-12/13)(-5/13) = 120/169
cos(2x) = (-5/13)^2 - (-12/13)^2 = -39/169
tan(2x) = (120/169) / (-39/169) = -120/39 = -40/13
Therefore, sin(2x) = 120/169, cos(2x) = -39/169, and tan(2x) = -40/13.
To find the values of sin(2x), cos(2x), and tan(2x), we can use the double-angle identities. However, before we can do that, let's find the value of x.
Given that tan(x) = -12/5 and x is in quadrant 2, we know that the tangent of an angle in quadrant 2 is negative. Therefore, tan(x) = -12/5 implies that the magnitude of tan(x) is 12/5.
Using the identity tan(x) = sin(x)/cos(x), we can find sin(x) and cos(x).
tan(x) = sin(x)/cos(x)
(-12/5) = sin(x)/cos(x)
Since x is in quadrant 2, both sin(x) and cos(x) are positive.
Now, let's solve for sin(x) using the equation (-12/5) = sin(x)/cos(x):
sin(x) = (-12/5) * cos(x)
Next, we can use the Pythagorean identity to solve for cos(x):
cos^2(x) + sin^2(x) = 1
(cos(x))^2 + ((-12/5) * cos(x))^2 = 1
Now, let's solve for cos(x) using the above equation:
25(cos(x))^2 + 144(cos(x))^2 = 25
169(cos(x))^2 = 25
(cos(x))^2 = 25/169
Taking the square root of both sides, we get:
cos(x) = ±(5/13)
Since x is in quadrant 2 and cos(x) is positive, cos(x) = 5/13.
Now, substitute the values of sin(x) = (-12/5) * cos(x) and cos(x) = 5/13 into the double-angle identities:
sin(2x) = 2 * sin(x) * cos(x)
cos(2x) = cos^2(x) - sin^2(x)
tan(2x) = (2 * tan(x)) / (1 - tan^2(x))
Calculating the above equations, we get:
sin(2x) = 2 * ((-12/5) * (5/13)) = -24/65
cos(2x) = (5/13)^2 - ((-12/5) * (5/13))^2 = 169/169 - 144/169 = 25/169
tan(2x) = (2 * (-12/5)) / (1 - (-12/5)^2) = -24/65
Therefore, sin(2x) = -24/65, cos(2x) = 25/169, and tan(2x) = -24/65.