Question

Cos3П/8+cos5П/8 пж нада решить

Answer 1

Ответ:

$cos\frac{3\pi}{8}+cos\frac{5\pi}{8}=2cos\frac{\frac{5\pi}{8}+\frac{3\pi}{8}}{2}cos\frac{\frac{5\pi}{8}-\frac{3\pi}{8}}{2}=2cos\frac{\frac{8\pi}{8}}{2}cos\frac{\frac{2\pi}{8}}{2}=\\\\=2cos\frac{\pi}{2}cos\frac{\pi}{8}=2*0*cos\frac{\pi}{8}=0$

Пошаговое объяснение:

При решении пользуемся формулой суммы косинусов:

$cosa+cos\beta=2cos\frac{a+\beta}{2}cos\frac{a-\beta}{2}$

Answer 2

To solve the trigonometric expression, we need to use some trigonometric identities. First, let's recall the following identities:

1. Cosine of the sum of two angles: cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

2. Cosine of the difference of two angles: cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

Now, let's focus on the given expression: cos(3π/8) + cos(5π/8)

We can use the double-angle formula to rewrite the cosine expressions as follows:

1. cos(2θ) = 2cos²(θ) - 1
2. cos(2θ) = 1 - 2sin²(θ)

Using these formulas, we can manipulate the given expression:

cos(3π/8) + cos(5π/8)

First, we'll convert π/8 to 2π/16 to match the double-angle formula.

cos(6π/16) + cos(10π/16)

Now, let's use the double-angle formulas:

cos(6π/16) = 2cos²(3π/16) - 1 cos(10π/16) = 2cos²(5π/16) - 1

So the expression becomes:

2cos²(3π/16) - 1 + 2cos²(5π/16) - 1

Now, we can use the Pythagorean identity to relate the square of sine and cosine:

sin²(θ) + cos²(θ) = 1

Therefore:

cos²(θ) = 1 - sin²(θ)

Now, let's apply this identity to the expression:

2(1 - sin²(3π/16)) - 1 + 2(1 - sin²(5π/16)) - 1

Simplify further:

2 - 2sin²(3π/16) - 1 + 2 - 2sin²(5π/16) - 1

Combine like terms:

4 - 2sin²(3π/16) - 2sin²(5π/16)

Now, to proceed, we need to know the values of sin(3π/16) and sin(5π/16), which are not immediately obvious. At this point, we could use numerical methods or a calculator to find their approximate values. However, I can't provide the final numerical result without more information.

If you need the exact value of the expression, it would be best to leave it in this form:

4 - 2sin²(3π/16) - 2sin²(5π/16)

But if you have specific values for sin(3π/16) and sin(5π/16), you can plug them in and calculate the final result.