If x = 9tanθ, use trigonometric substitution to write square root 81 + x^2 as a trigonometric function of θ, where -pi/2 < θ < pi/2
√(81+x^2)
= √(81+81tan^2θ)
= √(81(1+tan^2θ))
= √(81sec^2θ)
= 9secθ
Looks like you need to brush up on your basic trig identities.
To write the expression √(81 + x^2) as a trigonometric function of θ, we will use the trigonometric substitution x = 9tanθ.
First, we'll substitute x in the expression:
√(81 + x^2) → √(81 + (9tanθ)^2)
Next, we'll simplify the expression:
√(81 + (9tanθ)^2) → √(81 + 81tan^2θ)
Since tan^2θ + 1 = sec^2θ, we can write it as:
√(81 + 81tan^2θ) → √(81(1 + tan^2θ))
Now, we can factor out 81:
√(81(1 + tan^2θ)) → √(81) * √(1 + tan^2θ) → 9 * √(1 + tan^2θ)
Finally, we can rewrite tan^2θ using the Pythagorean identity tan^2θ + 1 = sec^2θ:
9 * √(1 + tan^2θ) → 9 * √(1 + sec^2θ - 1) → 9 * √sec^2θ → 9secθ
Therefore, √(81 + x^2), where -π/2 < θ < π/2, is equal to 9secθ.
To write √(81 + x^2) as a trigonometric function of θ, we can use the substitution x = 9tanθ.
First, let's substitute x into the expression:
√(81 + x^2) = √(81 + (9tanθ)^2)
= √(81 + 81tan^2θ)
= √(81(1 + tan^2θ))
= √(81sec^2θ) (using the identity 1 + tan^2θ = sec^2θ)
Next, we can simplify by factoring out 81 and taking the square root of sec^2θ:
√(81sec^2θ) = √(9^2 * sec^2θ)
= 9√sec^2θ
Now, remember that we are given that -π/2 < θ < π/2, which means θ is within the range of the principal branch of the secant function.
In the principal branch, the square root of sec^2θ is equal to the absolute value of secθ. So we have:
√(81 + x^2) = 9 |secθ|
Therefore, √(81 + x^2) can be written as the trigonometric function of θ: 9|secθ|.