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?
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)
To find the expression for cos theta, we can use the Pythagorean Identity in trigonometry, which states that sin^2(theta) + cos^2(theta) = 1.
Given that sin(theta) = x/3, we can substitute this value into the Pythagorean Identity and solve for cos(theta):
(sin(theta))^2 + (cos(theta))^2 = 1
(x/3)^2 + (cos(theta))^2 = 1
x^2/9 + (cos(theta))^2 = 1
(cos(theta))^2 = 1 - x^2/9
cos(theta) = sqrt(1 - x^2/9)
Therefore, the expression for cos(theta) is cos(theta) = sqrt(1 - x^2/9).
Next, let's find the expression for tan theta. We know that tan(theta) is defined as sin(theta) / cos(theta). By substituting the given value of sin(theta) = x/3, and the expression we found for cos(theta), we can calculate tan(theta):
tan(theta) = sin(theta) / cos(theta)
tan(theta) = (x/3) / sqrt(1 - x^2/9)
tan(theta) = (x/3) * (1 / sqrt(1 - x^2/9))
tan(theta) = x / (3 * sqrt(1 - x^2/9))
Therefore, the expression for tan(theta) is tan(theta) = x / (3 * sqrt(1 - x^2/9)).