help with trigonometry equations

[fiery123]

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)

[krtxmrtz]

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.

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

[fiery123]

Right thanks a lot, I understand now.

[krtxmrtz]

Right thanks a lot, I understand now.

You're welcome.

Keep this equality:

sin²x + cos²x = 1

as an ace in your sleeve. It may come to the rescue more often than not.