about multiple angles of trigonometry

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August 2nd 2008, 01:45 AM
wintersoltice

about multiple angles of trigonometry

my answer is different from the provided answer.
can someone please help me to check where i gone wrong??

the question is:

given that sin X= - 5/13, cos Y = - 4/5 and that X and Y are in the same quadrant, find the value of sin(X/2).

my solution :
cos X = 2 cos^2 (X/2) -1
cos (X/2) = square root [(1/13)/2]
cos (X/2)= 1/(square root 26)
cos (X/2)= (square root 26)/26

then, sin X = 2(sin X/2)(cos X/2)
-5/13=2(sin X/2)[(square root 26)/26]
-5/13 x [26/(square root 26)] = 2(sin X/2)
[-10/(square root 26)] x 1/2 = sin X/2
sin (X/2) = -5/(square root 26)
rationalise
sin (X/2) = [(-5square root 26)/26]

but the provided answer is (5square root 26)/26 .....

which means mine is negative but the provided answer is positive...

can anyone tell me where i gone wrong???
August 2nd 2008, 01:54 AM
kalagota

Quote:

Originally Posted by wintersoltice

my answer is different from the provided answer.
can someone please help me to check where i gone wrong??

the question is:

given that sin X= - 5/13, cos Y = - 4/5 and that X and Y are in the same quadrant, find the value of sin(X/2).

my solution :
cos X = 2 cos^2 (X/2) -1
cos (X/2) = square root [(1/13)/2]
cos (X/2)= 1/(square root 26)
cos (X/2)= (square root 26)/26

then, sin X = 2(sin X/2)(cos X/2)
-5/13=2(sin X/2)[(square root 26)/26]
-5/13 x [26/(square root 26)] = 2(sin X/2)
[-10/(square root 26)] x 1/2 = sin X/2
sin (X/2) = -5/(square root 26)
rationalise
sin (X/2) = [(-5square root 26)/26]

but the provided answer is (5square root 26)/26 .....

which means mine is negative but the provided answer is positive...

can anyone tell me where i gone wrong???

it was said that X and Y are in the same quadrant. sin X is negative, so X must be in III or IV quadrant.. cos Y is also negative, thus, Y is in II or III..

therefore, X and Y must be in III..
note that if you divide an angle in the III quadrant, the resulting angle will be on the II quadrant..

thus, your cos (X/2) must be negative..
August 2nd 2008, 04:46 AM
wintersoltice

does that mean my answer is correct???
August 2nd 2008, 06:58 PM
kalagota

Quote:

Originally Posted by wintersoltice

does that mean my answer is correct???

no.. the answer is the positive one..

look at your calculations.

sin x = 2 sin (x/2) cos (x/2)

sin x is negative, cos (x/2) is also negative, then sin (x/2) must be positive..

besides, since x/2 is in the second quadrant, it should be logical that sin (x/2) must be positive..