Question

Cos20/ cos²40-sin²20​

Answer 1

Ответ:

Я думаю, что так

Пошаговое объяснение:

Answer 2

It seems like you're asking for the value of the expression:

$\frac{\cos(20^\circ)}{\cos^2(40^\circ) - \sin^2(20^\circ)}$

Let's break this down step by step:

1. Use the trigonometric identity $\cos(2\theta) = 2\cos^2(\theta) - 1$ to express $\cos^2(40^\circ)$:

   $\cos(2 \cdot 20^\circ) = 2\cos^2(20^\circ) - 1$ $\cos^2(20^\circ) = \frac{\cos(2 \cdot 20^\circ) + 1}{2}$

2. Use the trigonometric identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ to express $\sin^2(20^\circ)$:

   $\sin(2 \cdot 20^\circ) = 2\sin(20^\circ)\cos(20^\circ)$ $\sin^2(20^\circ) = \frac{\sin(2 \cdot 20^\circ)}{2\cos(20^\circ)}$

Now let's substitute these expressions into the original expression:

$\frac{\cos(20^\circ)}{\cos^2(40^\circ) - \sin^2(20^\circ)} = \frac{\cos(20^\circ)}{\frac{\cos(2 \cdot 20^\circ) + 1}{2} - \frac{\sin(2 \cdot 20^\circ)}{2\cos(20^\circ)}}$

Now you can plug in the values of the trigonometric functions:

$\frac{\cos(20^\circ)}{\frac{\cos(2 \cdot 20^\circ) + 1}{2} - \frac{\sin(2 \cdot 20^\circ)}{2\cos(20^\circ)}} = \frac{\cos(20^\circ)}{\frac{2\cos^2(20^\circ) - 1 + 1}{2} - \frac{2\sin(20^\circ)\cos(20^\circ)}{2\cos(20^\circ)}}$

Simplify further:

$\frac{\cos(20^\circ)}{\cos^2(20^\circ) - \sin(20^\circ)\cos(20^\circ)}$

Now you can factor out $\cos(20^\circ)$ from the denominator:

$\frac{\cos(20^\circ)}{\cos(20^\circ)(\cos(20^\circ) - \sin(20^\circ))}$

Since $\cos(20^\circ) \neq 0$, you can cancel out the $\cos(20^\circ)$ terms:

$\frac{1}{\cos(20^\circ) - \sin(20^\circ)}$