Hi can anyone help me work out this question.
If tanx = -(4/3) and 180<x<360 find the value of
(5sinx+6cosx)/(7cosx-3sinx)
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Hi can anyone help me work out this question.
If tanx = -(4/3) and 180<x<360 find the value of
(5sinx+6cosx)/(7cosx-3sinx)
It's all about deriving the values of sin and cos from that of tan.Quote:
Originally Posted by fiery123
tan2x = sin2x / cos2x = (1 - cos2x) / cos2x = 1/cos2x - 1
Rearranging:
1 + tan2x = 1/cos2x
1 / (1 + tan2x) = cos2x
1 / Sqrt(1 + tan2x) = cos x
Likewise you can work it out for the sine:
tan2x = sin2x / cos2x = sin2x / (1 - sin2x)
...
tan x / Sqrt(1 + tan2x) = sin x
Now you can substitute the value. Taking the square root as positive:
sin x = (-4/3) / Sqrt(1 + (-4/3)2) = -4/5
cos x = 1 / Sqrt(1 + (-4/3)2) = 3/5
The square root can be positive or negative, but if x lies between 180 and 360, then the sine must be negative. Therefore, because the tangent is negative, the cosine must be positive.
Finally:
(5 sin x + 6 cos x) / (7 cos x - 3 sin x) = -2 / 33
Right thanks a lot, I understand now.
You're welcome.Quote:
Originally Posted by fiery123
Keep this equality:
sin2x + cos2x = 1
as an ace in your sleeve. It may come to the rescue more often than not.