Asked by Amber
                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]
            
        [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]
Answers
                    Answered by
            Steve
            
    sinx = 8/17
so, cosx = 15/7
sin2x = 2 sinx cosx = 2 * 8/7 * 15/7 = 240/49
do the others similarly.
    
so, cosx = 15/7
sin2x = 2 sinx cosx = 2 * 8/7 * 15/7 = 240/49
do the others similarly.
                    Answered by
            Steve
            
    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
                    Answered by
            Reiny
            
    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
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.