sin, cos, tan help?

Question

If x = 9tan(theta) and theta is in the first quadrant, find the expression for sin(theta) and cos(theta) in terms of x.

can someone show me how to do this step by step?

Answer 1

Hypotenuse = h
h ² = x ² + 9 ²
tan Ө = x / 9
sin Ө = x / √( x ² + 9 ² )
cos Ө = 9 / √( x ² + 9 ² )

Answer 2

sin(theta) = x / SQRT(x^2 + 9^2)
cos(theta) = 9 / SQRT(x^2 + 9^2)

Draw a right triangle in quadrant one, with theta as the angle between the x axis and the hypotenuse of the triangle. We need to find the lengths of all three sides of the triangle.

x = 9 tan(theta), tan(theta) = x/9, says the length of the side on the x axis is 9, and the vertical line part of the triangle has length x.

The hypotenuse must have length SQRT(x^2 + 9^2).

The answers for sin and cosine now follow because we have all 3 sides of the triangle.

Answer 3

x = 9tan(theta) --- Using A for Theta
=> Tan A = x/9

=> Tan^2 A = x^2 / 81

add 1 to both the sides you get-

=> 1 + tan^2 A = 1 + x^2 / 81
=> sec^2 A = (x^2 + 81) ÷ 81
=> 1 / cos^2 A = (x^2 + 81) ÷ 81
=> cos^2 A = 81 ÷ (x^2 + 81)
=> cosA = sqrt of [ 81 ÷ (x^2 + 81) ]
=> cos A = 9 ÷ sqrt (x^2 + 81) ............ Answer
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Again as cos^2 A = 81 ÷ (x^2 + 81)

=> 1 - (cos^2 A) = 1 - [81 ÷ (x^2 + 81)]
=> sin^2 A = (x^2) / (x^2 + 81)
=> sin A = x / sqrt of (x^2 + 81) ....................... Answer

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