sinx = 8/17
so, cosx = 15/7
sin2x = 2 sinx cosx = 2 * 8/7 * 15/7 = 240/49
do the others similarly.
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]
3 answers
oops
sinx = 8/17
so, cosx = 15/17
sin2x = 2 sinx cosx = 2 * 8/17 * 15/17 = 240/289
sinx = 8/17
so, cosx = 15/17
sin2x = 2 sinx cosx = 2 * 8/17 * 15/17 = 240/289
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
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
1 answer