Question

8*sin^2*25p/24*cos^2*23p/24-1

Answer 1

$8sin^2 (\frac{25\pi }{24})*cos^2(\frac{23\pi}{24})-1 =8*[\frac{1}{2}*sin(\frac{25\pi}{24}-\frac{23\pi}{24})+sin(\frac{25\pi}{24}+\frac{23\pi}{24})]^2\\ \\ -1=8*(\frac{1}{2}*sin\frac{\pi}{12}+sin2\pi)^2-1=8*(\frac{1}{2}*sin\frac{\pi}{12}+0)^2-1=\\ \\ =8*\frac{1}{4}*sin^2\frac{\pi}{12}-1=2*\frac{1-cos\frac{\pi}{6}}{2}-1=1-cos\frac{\pi}{6}-1=-cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}$

P.S. Формулы:

$sina*cosb= \frac{1}{2} [sin(a-b)+sin(a+b)] \\ \\ sin^2 a= \frac{1-cos(2a)}{2}$

Answer 2

It appears you have an expression involving trigonometric functions. If I understand your expression correctly, you have:

8 * sin^2(25π/24) * cos^2(23π/24) - 1

Let's break it down step by step:

1. sin^2(25π/24) is the square of the sine of 25π/24 radians.
2. cos^2(23π/24) is the square of the cosine of 23π/24 radians.

To evaluate this expression, we need to find the values of sin(25π/24) and cos(23π/24) and then perform the calculations:

sin(25π/24) and cos(23π/24) are related to the sine and cosine of the standard angles within the unit circle. You can use the properties of periodicity to simplify these values. The period of sine and cosine functions is 2π, meaning that sin(x) = sin(x ± 2π) and cos(x) = cos(x ± 2π).

Let's find equivalent angles within a single period (0 to 2π):

• 25π/24 = 24π/24 + π/24 = π + π/24
• 23π/24 = 24π/24 - π/24 = π - π/24

Now, we can evaluate sin(25π/24) and cos(23π/24):

• sin(25π/24) = sin(π + π/24) = sin(π/24)
• cos(23π/24) = cos(π - π/24) = cos(π/24)

To evaluate sin(π/24) and cos(π/24), you may need to use trigonometric identities or a calculator. Once you have those values, you can proceed with the calculation:

8 * sin^2(π/24) * cos^2(π/24) - 1

Now, calculate sin(π/24) and cos(π/24), square them, and then perform the multiplication and subtraction to find the result.