Question

Rewrite each expression as an algebraic expression involving no trigonometric functions.

sin(tan (inverse) 5x)

Answer 1

Let y = sin(tan(inverse)5x)

By definition, tan(inverse)5x = θ where tanθ = 5x.

So, y = sinθ, where sinθ = 5x.

Thus, the expression sin(tan (inverse) 5x) can be rewritten as y = 5x.

Answer 2

`\sin{(\tan^(-1){(5x)})}=(5x)/(√(25x^2+1))=(5x√(25x^2+1))/(25x^2+1)`

Explanation:

You want sin(α) where tan(α) = 5x.

Trig relations

We can use the trig relations ...

• sin(α) = 1/csc(α)
• csc(α)² = 1 + cot(α)²
• cot(α) = 1/tan(α)

Given tan(α) = 5x, ...

cot(α) = 1/(5x)

csc(α)² = 1 + cot(α)² = 1 +1/(5x)² = (25x² +1)/(25x²)

csc(α) = √(25x² +1)/5x

sin(α) = 5x/√(25x² +1)

`\boxed{\sin{(\tan^(-1){(5x)})}=(5x)/(√(25x^2+1))=(5x√(25x^2+1))/(25x^2+1)}`

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Additional comment

Sometimes we like the denominator to be "rational." The second form of the answer has the radical in the numerator for that purpose.

In the attached figure, the sides are labeled so that ...

Tan = Opposite/Adjacent = 5x/1 = 5x

By the Pythagorean theorem, the hypotenuse is the root of the sum of the squares of the sides. The desired sine ratio is ...

Sin = Opposite/Hypotenuse = 5x/√(25x² +1)