Evaluating Tangent Values from Trigonometric Equations

When dealing with trigonometric equations, it is often important to find the value of a specific function, such as the tangent. In this article, we will explore an example where we need to find the value of tan theta given the condition 8sinθ 4cosθ.

Step-by-Step Solution

Given: 8sinθ 4cosθ

We start by taking the logarithm of both sides, but for simplicity, we can directly manipulate the equation to find tan theta.

Using Basic Trigonometric Identities

First, let's express the equation in terms of sine and cosine:

[ 8sintheta 4costheta ]

Next, we divide both sides by 2cosθ:

[ frac{8sintheta}{8costheta} frac{4costheta}{8costheta} ]

This simplifies to:

[ frac{sintheta}{costheta} frac{1}{2} ]

By the quotient identity for tangent, we have:

[ tantheta frac{1}{2} ]

Thus, the value of tan θ is ?.

Alternative Method: Using Tangent and Secant Identities

Alternatively, we can use the identities involving tangent and secant. Start with:

[ 8sintheta 4costheta ]

Divide both sides by cos θ:

[ 8tantheta 4sectheta ]

Rearrange terms:

[ 8tantheta - 1 4sectheta ]

Square both sides:

[ (8tantheta - 1)^2 (4sectheta)^2 ]

Expand and simplify:

[ 64tan^2theta - 16tantheta 1 16sec^2theta ]

Since sec2θ 1 tan2θ, substitute:

[ 64tan^2theta - 16tantheta 1 16(1 tan^2theta) ]

This simplifies to:

[ 64tan^2theta - 16tantheta 1 16 16tan^2theta ]

Combine like terms:

[ 48tan^2theta - 16tantheta - 15 0 ]

This is a quadratic equation in tanθ. Solve using the quadratic formula:

[ tantheta frac{-b pm sqrt{b^2 - 4ac}}{2a} ]

Here, a 48, b -16, and c -15. Thus:

[ tantheta frac{16 pm sqrt{(-16)^2 - 4(48)(-15)}}{2(48)} ]

Calculate the discriminant:

[ sqrt{256 2880} sqrt{3136} 56 ]

Substitute back:

[ tantheta frac{16 pm 56}{96} ]

This results in:

[ tantheta frac{3}{4} quad text{or} quad tantheta -frac{5}{12} ]

Both values satisfy the original equation:

[ 8sintheta 4costheta ]

Using Right Triangle Trigonometry

A third approach involves the right triangle representation of sine and cosine. Let θ x:

[ 8sin x 4cos x ]

Divide both sides by cos x:

[ 8tan x - 1 4sec x ]

Square both sides:

[ (8tan x - 1)^2 (4sec x)^2 ]

Expand and simplify:

[ 64tan^2 x - 16tan x 1 16sec^2 x ]

Substitute sec2x 1 tan2x:

[ 64tan^2 x - 16tan x 1 16(1 tan^2 x) ]

Combine like terms:

[ 48tan^2 x - 16tan x - 15 0 ]

Solve the quadratic equation:

[ tan x frac{16 pm 56}{96} ]

Thus, the tangent values are:

[ tan x frac{3}{4} quad text{or} quad tan x -frac{5}{12} ]

Both solutions are valid and satisfy the original equation.

Conclusion

In this article, we demonstrated three different methods to find the value of tanθ given the equation 8sinθ 4cosθ. The methods utilized basic trigonometric identities, the tangent and secant relationship, and right triangle trigonometry. The final values obtained were tanθ ? through the first method, and tanθ 3/4 and -5/12 through the second and third methods, with both solutions being valid.