Find sin 2x, cos 2x, and tan 2x from the given information.
[1]. sin x = 8/17, x in Quadrant I
1). sin 2x =________.
2). cos 2x =________.
3). tan 2x =________.
[2]. sin x = -5/13, x in Quadrant III
1). sin 2x =________.
2). cos 2x =________.
3). tan 2x =________.
[Note: I'm having a lot of trouble with these. From what I'm told the answers are suppose to come in fraction forms]
oops
sinx = 8/17
so, cosx = 15/17
sin2x = 2 sinx cosx = 2 * 8/17 * 15/17 = 240/289
To find sin 2x, cos 2x, and tan 2x, we can use the double-angle identities. These identities can be derived from the sum and difference identities. The double-angle identities for sine, cosine, and tangent are as follows:
1. sin 2x = 2sin x * cos x
2. cos 2x = cos^2 x - sin^2 x
3. tan 2x = (2tan x) / (1 - tan^2 x)
Let's solve the first problem:
[1]. sin x = 8/17, x in Quadrant I
Since sin x = 8/17, we can let the adjacent side be 8 and the hypotenuse be 17. Using the Pythagorean theorem, we can find the opposite side:
opposite side = sqrt(hypotenuse^2 - adjacent side^2)
= sqrt(17^2 - 8^2)
= sqrt(289 - 64)
= sqrt(225)
= 15
So, in Quadrant I, using a right-angled triangle with opposite side = 15 and adjacent side = 8, we have sin x = 8/17.
Now, let's find sin 2x:
sin 2x = 2sin x * cos x
First, let's find cos x:
Using the Pythagorean identity: sin^2 x + cos^2 x = 1
cos^2 x = 1 - sin^2 x
cos^2 x = 1 - (8/17)^2
cos^2 x = 1 - 64/289
cos^2 x = (289 - 64) / 289
cos^2 x = 225 / 289
cos x = sqrt(225 / 289)
cos x = 15 / 17
Now we can find sin 2x:
sin 2x = 2sin x * cos x
sin 2x = 2 * (8/17) * (15/17)
sin 2x = (2 * 8 * 15) / (17 * 17)
sin 2x = 240 / 289
Therefore, in the given problem:
1. sin 2x = 240/289
2. cos 2x = ...
3. tan 2x = ...
Now, let's solve the second problem:
[2]. sin x = -5/13, x in Quadrant III
Since sin x = -5/13, we can let the opposite side be -5 and the hypotenuse be 13. Using the Pythagorean theorem, we can find the adjacent side:
adjacent side = sqrt(hypotenuse^2 - opposite side^2)
= sqrt(13^2 - (-5)^2)
= sqrt(169 - 25)
= sqrt(144)
= 12
So, in Quadrant III, using a right-angled triangle with opposite side = -5 and adjacent side = -12, we have sin x = -5/13.
Now, let's find sin 2x:
sin 2x = 2sin x * cos x
First, let's find cos x:
Using the Pythagorean identity: sin^2 x + cos^2 x = 1
cos^2 x = 1 - sin^2 x
cos^2 x = 1 - (-5/13)^2
cos^2 x = 1 - 25/169
cos^2 x = (169 - 25) / 169
cos^2 x = 144 / 169
cos x = sqrt(144 / 169)
cos x = 12 / 13
Now we can find sin 2x:
sin 2x = 2sin x * cos x
sin 2x = 2 * (-5/13) * (12/13)
sin 2x = (-10 * 12) / (13 * 13)
sin 2x = -120 / 169
Therefore, in the given problem:
1. sin 2x = -120/169
2. cos 2x = ...
3. tan 2x = ...
Please let me know if you need help with finding cos 2x and tan 2x as well.
sinx = 8/17
so, cosx = 15/7
sin2x = 2 sinx cosx = 2 * 8/7 * 15/7 = 240/49
do the others similarly.
I will do the 2nd one, which is the harder of the two.
Follow my steps to do the first one.
sinx = -5/13 ,and the angle x is in III
recall that sinØ = y/r
so y = -5, r = 13
sketch a right-angled triangle in III with hypotenuse 13 and y = -5
x^2 + y^2 = r^2
x^2 + 25 =169
x^2 = 144
x = ± 12 , but we are in III, so x = -12
giving us cosx = -12/13
recall that sin 2x = 2sinx cosx
= 2(-5/13)(-12/13)
= 120/169
recall cos 2x = cos^2 x - sin^2 x
= 144/169 - 25/169
= 119/169
recall tan 2x = sin2x/cos2x
= (120/169) / (119/169)
= 120/119