For the acute angles in a right triangle sin (2x) = cos (4x+12 degrees) What is the measure of the larger angle?

Answer 1

Let's solve for x first. We have:

sin(2x) = cos(4x + 12 degrees)

Using the double angle formula for sine, we can express sin(2x) in terms of cosine:

2sin(x)cos(x) = cos(4x + 12 degrees)

Dividing both sides by cos(4x + 12 degrees), we get:

2sin(x)cos(x) / cos(4x + 12 degrees) = 1

Using the identity sin^2(x) + cos^2(x) = 1, we can rewrite sin(x) and cos(x) in terms of one another:

2(sin(x))^2 / cos(4x + 12 degrees) = 1

Simplifying, we have:

2(sin(x))^2 = cos(4x + 12 degrees)

Now, let's use the identity cos(90 degrees - θ) = sin(θ) to rewrite the cosine term:

2(sin(x))^2 = sin(90 degrees - (4x + 12 degrees))

Using the identity sin(90 degrees - θ) = cos(θ), we get:

2(sin(x))^2 = cos(4x + 12 degrees)

Now, we have a quadratic-like equation involving sin(x) and cos(4x + 12 degrees). To solve for x, we need to find the values of x that satisfy this equation.

Squaring both sides of the equation, we get:

4(sin(x))^4 = cos^2(4x + 12 degrees)

Recall that sin^2(θ) = 1 - cos^2(θ). We can rewrite the equation as:

4(sin(x))^4 = 1 - sin^2(4x + 12 degrees)

Expanding the squared term, we have:

4(sin(x))^4 = 1 - (sin(4x + 12 degrees))^2

Rearranging the equation, we get:

(sin(4x + 12 degrees))^2 + 4(sin(x))^4 - 1 = 0

Now, let's solve this equation. Unfortunately, it is a complicated equation to solve algebraically. We can use numerical methods or technology to find the approximate solutions for x.

Once we have the values of x, we can find the acute angles by substituting them into sin(x) and cos(4x + 12 degrees). The larger acute angle is the one with the greater value.