recall that sinØ = opposite/hypotenuse
= x/3
so you have a right angled triangle with opposite = x and hypotenuse = 3
let the adjacent be a
a^2 + x^2 = 3^2
a^2 = 9 - x^2
a = ± √(x^2 - 9) , but Ø is in I, so a = √(x^2 - 9)
cosØ = √(x^2 - 9)/3
tanØ = x/a = x/√(x^2-9)
sin theta = x/3, with theta in the first quadrant, find the expression for cos theta and tan theta.
Answers will be in algebraic terms of x. Give exact answers with 0 < theta <pi/2 (0 less than or equal to theta, less than or equal to pi/2
Cos theta ?
Tan Theta?
1 answer
2 answers