Question

(1+cos2x-sin2x)/(cos x+cos(pi/2+x)) если cos x=-1/2

Answer 1

Ответ:

-1

Пошаговое объяснение:

(cos²x+sin²x+cos²x-sin²x-2sinxcosx)/(cosx-sinx)=

(2cos²x-2sinxcosx)/(cosx-sinx)=(2cosx(cosx-sinx))/(cosx-sinx)=

2cosx=2·(-1/2)=-1

Answer 2

To find the value of the expression $\frac{{1 + \cos(2x) - \sin(2x)}}{{\cos(x) + \cos\left(\frac{\pi}{2} + x\right)}}$ when $\cos(x) = -\frac{1}{2}$, we first need to compute the values of $\cos(2x)$ and $\sin(2x)$.

Using trigonometric identities, we have:

$$\begin{align*} \cos(2x) &= 2\cos^2(x) - 1 \\ \sin(2x) &= 2\sin(x)\cos(x) \end{align*}$$

Given that $\cos(x) = -\frac{1}{2}$, we can find $\sin(x)$ using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$.

$$\sin(x) = \sqrt{1 - \cos^2(x)} = \sqrt{1 - \left(-\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}$$

Now, we can calculate $\cos(2x)$ and $\sin(2x)$:

$$\begin{align*} \cos(2x) &= 2\left(-\frac{1}{2}\right)^2 - 1 = 2\left(\frac{1}{4}\right) - 1 = -\frac{1}{2} - 1 = -\frac{3}{2} \\ \sin(2x) &= 2\sin(x)\cos(x) = 2\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{1}{2}\right) = -\sqrt{3} \end{align*}$$

Now, let's substitute these values back into the original expression:

$$\frac{{1 + \cos(2x) - \sin(2x)}}{{\cos(x) + \cos\left(\frac{\pi}{2} + x\right)}} = \frac{{1 - \frac{3}{2} + \sqrt{3}}}{{-\frac{1}{2} + \cos\left(\frac{\pi}{2} - x\right)}}$$

Since $\cos(\frac{\pi}{2} + x) = \sin(x)$, and we know $\sin(x) = \frac{\sqrt{3}}{2}$, we can simplify further:

$$\frac{{1 - \frac{3}{2} + \sqrt{3}}}{{-\frac{1}{2} + \cos\left(\frac{\pi}{2} - x\right)}} = \frac{{1 - \frac{3}{2} + \sqrt{3}}}{{-\frac{1}{2} + \frac{\sqrt{3}}{2}}}$$