Question

How do you find the sin of theta if sin of theta is radical 5 times cos of theta and 0<theta<pi/2?

How do you find the `sinθ` if `sinθ=sqrt(5cosθ)` and 0<θ<`pi/2`?

Answer 1

No solution.

Explanation:

`"Given : " sin theta = sqrt(5 cos theta), theta in 0 < pi/2`

`sin^2 theta = 5 cos theta, color(crimson)(" squaring both sides"`

`1 - cos^2 theta = 5 cos theta, color(crimson)(" as " sin^2 theta + cos^2 theta = 1`

`cos^2 theta + 5cos theta - 1 = 0`

`Let cos theta = x`

`x^2 + 5 x - 1 = 0`

`x = (-5 +- sqrt(5^2 -4 * 1 * 1 )) / (2*1)`

`x = (-5 +-sqrt21) / 2`

`x =(-5 +- 4.9) /2`

`x = -0.05, -4.95`

`cos theta = -0.05, cancel(-4.05), " as cos can have a value between -1 and 1"`

`cos theta = -0.05`

`"Since " theta in 0 < pi/2, " where cos is always positive, we cannot have a solution for the given sum."`