Question

4cos^2(45-4a)-2sin8a=2

Answer 1

2cos²(45+4a) + sin8a = 1

cos²x = (1+cos2x)/2   - формула

2cos²(45+4a) +sin8a = 1

2(1+cos2(45+4a))/2 +sin8a = 1

1+cos2(45+4a) + sin8a   = 1

1+ cos(90+8a)+ sin8a = 1

cos(90+a) = -sina   формула

1+ cos(90+8a) + sin8a = 1

1-sin8a +sin8a  = 1

1 = 1

Answer 2

To solve the equation 4cos^2(45-4a) - 2sin8a = 2, we'll start by applying trigonometric identities to simplify the equation.

First, we'll use the identity cos^2(x) + sin^2(x) = 1:

cos^2(x) = 1 - sin^2(x)

Next, we'll use the double-angle formula for sine:

sin(2x) = 2sin(x)cos(x)

Now, let's apply these identities to the given equation:

4cos^2(45-4a) - 2sin(8a) = 2

Using the identity for cos^2(45-4a):

4(1 - sin^2(45-4a)) - 2sin(8a) = 2

Now, we need to convert sin(8a) to a function of sin(2a) using the double-angle formula for sine:

sin(8a) = 2sin(4a)cos(4a)

But we also have the double-angle formula for sine:

sin(4a) = 2sin(2a)cos(2a)

So, we can rewrite sin(8a) as:

sin(8a) = 2(2sin(2a)cos(2a))cos(4a)

Now, substitute sin(8a) back into the equation:

4(1 - sin^2(45-4a)) - 2(2sin(2a)cos(2a)cos(4a)) = 2

Next, we need to use another identity for cos(2x):

cos(2x) = 2cos^2(x) - 1

Now, let's substitute cos(2x) = 2cos^2(2a) - 1:

4(1 - sin^2(45-4a)) - 2(2sin(2a)(2cos^2(2a) - 1)) = 2

Simplify further:

4(1 - sin^2(45-4a)) - 4sin(2a)(2cos^2(2a) - 1) = 2

Now, recall that sin(45-θ) = sin(θ) and cos(45-θ) = sin(θ). So, sin^2(45-4a) = sin^2(4a) and cos^2(2a) = 1 - sin^2(2a):

4(1 - sin^2(4a)) - 4sin(2a)(2(1 - sin^2(2a)) - 1) = 2

Simplify further:

4(1 - sin^2(4a)) - 4sin(2a)(2 - 2sin^2(2a) - 1) = 2

Now, distribute the -4sin(2a) term:

4(1 - sin^2(4a)) - 4sin(2a)(1 - 2sin^2(2a)) = 2

Expand further:

4 - 4sin^2(4a) - 4sin(2a) + 8sin(2a)sin^2(2a) = 2

Combine like terms:

-4sin^2(4a) + 8sin(2a)sin^2(2a) - 4sin(2a) + 2 = 0

Now, the equation is in terms of sin(2a) and sin(4a). Let's use a substitution to simplify it further. Let's set sin(2a) = x, and sin(4a) = x^2:

-4x^2 + 8x^3 - 4x + 2 = 0

Now, we have a cubic equation in terms of x. Unfortunately, cubic equations do not have a general algebraic solution like quadratic equations. We can use numerical methods or calculators to find the approximate solutions for x (sin(2a)).

Once we find the values of x, we can use inverse trigonometric functions to find the values of a. However, do note that cubic equations can have one or more real roots, and it might not always be possible to find exact solutions for a.

If you have specific values or a range of a you are interested in, I can try to provide numerical approximations for sin(2a) and sin(4a) for those values.