# If 4sinθ + 6cosθ = sqrt(3), then 4cosθ - 6sinθ is equal to a.+ sqrt(6), - sqrt(6) b.+ sqrt(7), - sqrt(7) c. 5 d. 8

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### 1 Answer

You need to raise to square` 4sin theta + 6cos theta = sqrt 3` , such that:

`16sin^2 theta + 48sin theta*cos theta + 36 cos^2 theta = 3`

You need to raise to square the relation (`4cos theta - 6sin theta` ) = `x` to square, such that:

`16cos^2 theta - 48sin theta*cos theta + 36 sin^2 theta = x^2`

You need to perform the addition of the relations `(4sin theta + 6cos theta)^2+ (4cos theta - 6sin theta)^2` , such that:

`16sin^2 theta + 48sin theta*cos theta + 36 cos^2 theta + 16cos^2 theta - 48sin theta*cos theta + 36 sin^2 theta = 3 + x^2`

You need to group theterms such that:

`(16sin^2 theta + 16cos^2 theta) + (36 cos^2 theta + 36 sin^2 theta) + 48sin theta*cos theta - 48sin theta*cos theta = 3 + x^2`

Reducing duplicate members yields:

`16(sin^2 theta + cos^2 theta) + 36(sin^2 theta + cos^2 theta) = 3 + x^2`

Since `sin^2 theta + cos^2 theta = 1` , yields:

`16 + 36 = 3 + x^2 => x^2 = 52 - 3 => x^2 = 49 => x_(1,2) = +-sqrt49 => x_(1,2) = +-7`

**Hence, evaluating the expression `4cos theta - 6sin theta,` under the given conditions, yields **`4cos theta - 6sin theta = +-7` , **hence, the answer b. would be good if you would remove the radicals.**