Question

4sin20°•cos20°•сos80°​

Answer 1

Ответ:

$\frac{ \sqrt{3} }{2} - \sin(40°)$

Объяснение:

Условие = 2sin(40°)×cos(80°) = 2 × 1/2 × ( sin (120°) + sin(-40°)) = √3/2 - sin(40°).

Answer 2

To calculate the value of the expression $4\sin(20^\circ)\cos(20^\circ)\cos(80^\circ)$, you can use trigonometric identities.

1. First, you can apply the double-angle formula for the sine function:

   $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$

   In your expression, $\theta = 20^\circ$, so:

   $\sin(40^\circ) = 2\sin(20^\circ)\cos(20^\circ)$

2. Next, use the sum-to-product formula for sine:

   $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$

   In your case, $A = 40^\circ$ and $B = 80^\circ$, so:

   $\sin(40^\circ + 80^\circ) = \sin(40^\circ)\cos(80^\circ) + \cos(40^\circ)\sin(80^\circ)$

3. Now, since you have $\sin(40^\circ) = 2\sin(20^\circ)\cos(20^\circ)$ from step 1, you can substitute this into the equation from step 2:

   $2\sin(20^\circ)\cos(20^\circ)\cos(80^\circ) = \sin(40^\circ)\cos(80^\circ) + \cos(40^\circ)\sin(80^\circ)$

4. You can now calculate the values of $\sin(40^\circ)$ and $\cos(80^\circ)$ and $\cos(40^\circ)$ and $\sin(80^\circ)$ using the trigonometric identities:

   $\sin(40^\circ) = \sin(2\cdot20^\circ) = 2\sin(20^\circ)\cos(20^\circ)$ (from the double-angle formula for sine)

   $\cos(80^\circ) = \cos(2\cdot40^\circ) = 2\cos^2(40^\circ) - 1$

   $\cos(40^\circ) = \cos(2\cdot20^\circ) = 2\cos^2(20^\circ) - 1$

   $\sin(80^\circ) = \sin(2\cdot40^\circ) = 2\sin(40^\circ)\cos(40^\circ)$ (from the double-angle formula for sine)

5. Substitute these values into the equation:

   $2\sin(20^\circ)\cos(20^\circ)\cos(80^\circ) = (2\sin(20^\circ)\cos(20^\circ))(2\cos^2(40^\circ) - 1) + (2\cos^2(20^\circ) - 1)(2\sin(40^\circ)\cos(40^\circ))$

6. Calculate each of these components separately:

   $2\sin(20^\circ)\cos(20^\circ) = \sin(40^\circ)$

   $2\sin(40^\circ)\cos(40^\circ) = \sin(80^\circ)$

   $\cos^2(40^\circ)$ and $\cos^2(20^\circ)$ can be calculated using the Pythagorean identity for cosine: $\cos^2(\theta) + \sin^2(\theta) = 1$