Consider both fractions one at a time. Separate the trigonometric and non-trigonometric parts in each fraction before transforming the expressions.
y=- gx^2/2v_0 ^2(1+tan^2θ)+xtanθ
Practice makes perfect
The height of the firework and its horizontal displacement are related by the given equation. We will rewrite so that tangent will be the only trigonometric function in the equation.
y=- gx^2/2v_0 ^2cos^2θ+xsinθ/cosθ
We will consider each of the two fractions one at a time.
First Fraction
Consider the first part of our expression.
- gx^2/2v_0 ^2cos^2θLet's start with separating the trigonometric and non-trigonometric parts of the expression.
- gx^2/2v_0 ^2cos^2θ=- gx^2/2v_0 ^2 * 1/cos^2θ
Since the non-trigonometric part will not have any impact on the transformations of the trigonometric functions, we will just consider 1cos^2θ. Recall one of the Reciprocal Identities considering the reciprocal of cosine.
1/cosθ=secθ
Recall one of the Pythagorean Identities.
sec^2θ=1+tan^2θ
We can now use both these identities to transform our fraction.
In the obtained expression the only trigonometric function is tangent. Now we can rewrite the whole fraction using this.
- gx^2/2v_0 ^2cos^2θ=- gx^2/2v_0 ^2 * 1/cos^2θ
⇓
- gx^2/2v_0 ^2cos^2θ= - gx^2/2v_0 ^2(1+tan^2θ)
Second Fraction
Now, let's move on to the second fraction. Once again, we will separate the trigonometric and non-trigonometric parts.
xsinθ/cosθ=x * sinθ/cosθ
Recall one of the Quotient Identities considering the tangent function.
tanθ=sinθ/cosθ
Note that in our expression we have exactly sinθcosθ. Therefore, we can use the identity to transform this part of the expression into a tangent.
xsinθ/cosθ=x * sinθ/cosθ
⇕
xsinθ/cosθ=x * tanθ
Result
We have rewritten both fractions in a way that now the only trigonometric function which appears in them is tangent. Finally, we can use the obtained expressions to rewrite the whole equation.
y=- gx^2/2v_0 ^2cos^2θ+xsinθ/cosθ
⇕
y=- gx^2/2v_0 ^2(1+tan^2θ)+xtanθ
