tanx = 5/12 and sinx<0
Find sin2x and cos2x
How to get the answers of sin2x = 120/169 and cos 2x = 119/169?
tanx = 5/12
r^2 = x^2 + y^2
= 12^2 + 5^2
r = 13
tanx >0 and sinx<0, so x must be in quad. III and
and sinx = -5/13 , cosx = -12/13
sin 2x = 2sinxcosx = 2(-5/13)(-12/13)
= 120/169
cos 2x = cos^2 x - sin^2 x
= 144/169-25/169
=119/169
do the others the same way, or the way
drwls showed you in the other post.
tan x = 13 cot x
tan x = 13 cot x
tan x = 13/tan x
tan^2 x = 13
tan x = ±√13
x = 74.5º,105.5º, 254.5º, 285.5º
To find the values of sin2x and cos2x, we can use trigonometric identities and the given information.
First, let's recall the double angle identities for sine and cosine:
sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x)
Given that tan(x) = 5/12 and sin(x) < 0, we can determine the values of sin(x), cos(x), and tan(x).
We know that tan(x) = sin(x)/cos(x). So, we can equate this to the given value:
sin(x)/cos(x) = 5/12
To find sin(x) and cos(x), we can use the Pythagorean identity:
sin^2(x) + cos^2(x) = 1
We know sin(x) < 0, so we can assume sin(x) = -a, where a is a positive value. Therefore:
(-a)^2 + cos^2(x) = 1
a^2 + cos^2(x) = 1
Now, we can substitute the value of cos(x) from tan(x) = sin(x)/cos(x):
a^2 + (5/12)^2 = 1
Simplifying this equation, we get:
a^2 + 25/144 = 1
a^2 = 1 - 25/144
a^2 = 119/144
Taking the square root of both sides, we get:
a = √(119/144)
a = √119 / 12
We now have the values of sin(x) = -a and cos(x) = 5/12.
Next, let's find sin(2x) using the double-angle identity:
sin(2x) = 2sin(x)cos(x)
= 2(-a)(5/12)
= -10a/12
= -5a/6
Substituting the value of a, we get:
sin(2x) = -5(√119/12) / 6
= -√119 / 12
Finally, let's find cos(2x) using the double-angle identity:
cos(2x) = cos^2(x) - sin^2(x)
= (5/12)^2 - (-a)^2
= 25/144 - 119/144
= -94/144
Simplifying this fraction, we get:
cos(2x) = -47/72
Therefore, the values of sin2x and cos2x are sin2x = -√119 / 12 and cos2x = -47/72.
Note: The values of sin2x = 120/169 and cos2x = 119/169 provided in the question are incorrect.