I have two problems I am stuck on, if you could show me how to solve the problems it would be much appreciated.
1) Find sin 2x, cos 2x, and tan 2x from the given information.
tan x = − 1/6, cos x > 0
sin 2x =
cos 2x =
tan 2x =
2) Find sin 2x, cos 2x, and tan 2x from the given information.
cot x = 2/3, x in Quadrant I
sin 2x =
cos 2x =
tan 2x =
Sure, I can help you with that!
For problem 1:
Given that tan x = -1/6 and cos x > 0, we can find sin x using the Pythagorean identity:
sin x = √(1 - cos²x)
sin x = √(1 - (1/36))
sin x = √(35/36)
Now, we can use the double angle formulas to find sin 2x, cos 2x, and tan 2x:
sin 2x = 2sin x * cos x
cos 2x = cos²x - sin²x
tan 2x = (2 * tan x) / (1 - tan²x)
Plugging in the values we found earlier:
sin 2x = 2 * (√(35/36)) * (√(35/36)) = 2 * (√(35)/6) * (√(35)/6) = 35/18
cos 2x = (√(35/36))² - (35/36) = 35/36 - 35/36 = 0
tan 2x = (2 * (-1/6)) / (1 - (-1/6)²) = (-1/3) / (1 - 1/36) = (-1/3) / (35/36) = -6/35
So, the answers are:
sin 2x = 35/18
cos 2x = 0
tan 2x = -6/35
Now, let's move on to problem 2:
Given that cot x = 2/3 and x is in Quadrant I, we can find sin x using the Pythagorean identity:
sin x = √(1 - cos²x)
sin x = √(1 - (3/5)²)
sin x = √(1 - 9/25)
sin x = √(16/25)
sin x = 4/5
Now, we can use the double angle formulas to find sin 2x, cos 2x, and tan 2x:
sin 2x = 2sin x * cos x
cos 2x = cos²x - sin²x
tan 2x = (2 * tan x) / (1 - tan²x)
Plugging in the values we found earlier:
sin 2x = 2 * (4/5) * (√(1 - (16/25))) = 8/5 * (√(9/25)) = 8/5 * (3/5) = 24/25
cos 2x = (√(1 - (16/25)))² - (4/5)² = (9/25) - (16/25) = -7/25
tan 2x = (2 * (2/3)) / (1 - (2/3)²) = (4/3) / (1 - 4/9) = (4/3) / (5/9) = 36/15 = 12/5
So, the answers are:
sin 2x = 24/25
cos 2x = -7/25
tan 2x = 12/5
To solve both of these problems, we can use the double-angle identities for sine, cosine, and tangent, which are as follows:
sin 2x = 2sin(x)cos(x)
cos 2x = cos^2(x) - sin^2(x)
tan 2x = 2tan(x) / (1 - tan^2(x))
Let's solve each problem step by step:
Problem 1:
Given: tan x = -1/6 and cos x > 0
Step 1: Find sin x.
We know that tan x = sin x / cos x.
Since tan x = -1/6, we can set up the equation:
-1/6 = sin x / cos x
To isolate sin x, we multiply both sides by cos x:
-1/6 * cos x = sin x
Step 2: Find cos x.
Since cos x > 0, we know that x is in either Quadrant I or IV.
Since tan x is negative, we can conclude that x is in Quadrant II.
In Quadrant II, sin x is positive, so we can use the equation:
sin^2(x) + cos^2(x) = 1
Substituting the value of sin x from Step 1:
(-1/6 * cos x)^2 + cos^2(x) = 1
Simplifying the equation:
1/36 * cos^2(x) + cos^2(x) = 1
(1 + 36) / 36 * cos^2(x) = 1
37/36 * cos^2(x) = 1
cos^2(x) = 36/37
Since cos x > 0, we take the positive square root:
cos x = sqrt(36/37)
cos x = 6/7
Step 3: Find sin 2x, cos 2x, and tan 2x using the double-angle identities.
sin 2x = 2 * sin(x) * cos(x)
Substituting the values:
sin 2x = 2 * (-1/6 * cos x) * cos x
sin 2x = -2/6 * (6/7)
sin 2x = -2/7
cos 2x = cos^2(x) - sin^2(x)
Substituting the values:
cos 2x = (6/7)^2 - (-1/6)^2
cos 2x = 36/49 - 1/36
cos 2x = 1295/1764
tan 2x = 2 * tan(x) / (1 - tan^2(x))
Substituting the values:
tan 2x = 2 * (-1/6) / (1 - (-1/6)^2)
tan 2x = -2/6 / (1 - 1/36)
tan 2x = -1/3 / (35/36)
tan 2x = -1/3 * 36/35
tan 2x = -12/35
Therefore, for Problem 1:
sin 2x = -2/7
cos 2x = 1295/1764
tan 2x = -12/35
Problem 2:
Given: cot x = 2/3 and x is in Quadrant I
Step 1: Find tan x.
We know that cot x = 1/tan x.
Since cot x = 2/3, we can set up the equation:
2/3 = 1 / tan x
To isolate tan x, we take the reciprocal of both sides:
tan x = 3/2
Step 2: Find sin x.
Since x is in Quadrant I, both sine and cosine are positive.
We can use the equation:
sin^2(x) + cos^2(x) = 1
Substituting the value of tan x from Step 1:
(3/2)^2 + cos^2(x) = 1
9/4 + cos^2(x) = 1
cos^2(x) = 1 - 9/4
cos^2(x) = 4/4 - 9/4
cos^2(x) = -5/4
Since cos^2(x) cannot be negative, we made an error somewhere.
However, since x is in Quadrant I, the value of cos x must be positive.
Thus, it is not possible to have a solution for this problem.
Therefore, for Problem 2, there is no valid solution for sin 2x, cos 2x, and tan 2x.
tan is - in quadrants 2 and 4
cos is + in quadrants 1 and 4
so
we are in quadrant 4, x = 6, y = -1
hypotenuse = sqrt (37) cute
tan 2x = 2 tan x /(1-tan^2 x)
= 2 (-1/6) /(1 - 1/36)
=-(1/3) / (35/36)
= -12/35 well that was easy
now we better find out what quadrant 2x is in
x is about - 10 degrees which is 350 degrees counterclockwise from x axis
so 2 x is about 700 degrees
700 - 360 = 340 or - 20
so
we are still in quadrant 4 (but you knew that :)
now
tan 2x= -12/35
draw that in quad 4
hypotenuse = sqrt (144+1225)
you do it, I call it h
then sin 2x = -12/h
and cos 2x = 35/h