Suppose that sin(t) = x/(x + 3)
Use this information and a reference triangle to eliminate the trigonometric function from the following expression.
inverse tangent (x/ sqrt(6x + 9))
To eliminate the trigonometric function inverse tangent (arctan) from the expression, we can use the given information sin(t) = x/(x + 3) and a reference triangle.
First, let's label the sides of the reference triangle corresponding to the angle t:
Opposite side = x
Adjacent side = x + 3
Using the Pythagorean theorem, we can find the hypotenuse:
Hypotenuse^2 = (Opposite side)^2 + (Adjacent side)^2
Hypotenuse^2 = x^2 + (x + 3)^2
Hypotenuse^2 = x^2 + (x^2 + 6x + 9)
Hypotenuse^2 = 2x^2 + 6x + 9
Hypotenuse = sqrt(2x^2 + 6x + 9)
Now, we can rewrite the expression inverse tangent (x/ sqrt(6x + 9)) using the reference triangle:
inverse tangent (x/ sqrt(6x + 9)) = inverse tangent (Opposite side / Hypotenuse)
inverse tangent (x/ sqrt(6x + 9)) = inverse tangent (x / sqrt(2x^2 + 6x + 9))
Therefore, the trigonometric function inverse tangent can be eliminated from the expression by substituting x with Opposite side and sqrt(6x + 9) with Hypotenuse using the reference triangle.
To eliminate the trigonometric function inverse tangent (arctan) and express it in terms of the given equation sin(t) = x/(x + 3), we need to construct a reference triangle and use its properties.
Let's start by rewriting the given equation: sin(t) = x/(x + 3).
Consider a right triangle with an angle t (in the reference triangle) and opposite side x, and adjacent side (x + 3). The hypotenuse can be found using the Pythagorean theorem.
From the reference triangle, we can identify the following trigonometric relationships:
- Sine (sin) of angle t is equal to the ratio of the opposite side to the hypotenuse: sin(t) = x/hypotenuse.
- Tangent (tan) of angle t is equal to the ratio of the opposite side to the adjacent side: tan(t) = x/(x + 3).
To find a relationship with inverse tangent (arctan), we need to express tan(t) in terms of x:
tan(t) = x/(x + 3).
Now, we can use the definition of the inverse tangent function to eliminate the inverse tangent in the given expression.
Given expression: inverse tangent (x/ sqrt(6x + 9))
Let's rewrite the given expression using the relationship between tangents from above:
inverse tangent (x/(x + 3)) = inverse tangent (tan(t)).
By the properties of inverse functions, inverse tangent (tan(t)) is equal to t:
inverse tangent (tan(t)) = t.
So, the expression can be simplified to:
inverse tangent (x/ sqrt(6x + 9)) = t.
Hence, the trigonometric function inverse tangent (arctan) has been eliminated from the expression and expressed in terms of the given equation sin(t) = x/(x + 3).
opposite leg is x
hypotenuse is x+3
so adjacent leg is √((x+3)^2 - x^2) = √(6x+9)
so tan(t) = opposite/adjacent = x/√(6x+9)
tan a = (x/ sqrt(6x + 9))
draw right triangle , legs x and sqrt(6x+9)
hypotenuse = sqrt (x^2 + 6 x + 9) = sqrt (x+3)^2 = x+3
then sin a = x/(x+3)