$$ 7.1 \frac{2 \cos \left(90^{\circ}+x ight)-\sin \left(360^{\circ}-x ight)}{3 \sin \left(90^{\circ}-x ight)-\cos \left(180^{\circ}-x ight)} {} 7.2 \frac{\sin \left(180^{\circ}+x ight)}{ an \left(180^{\circ}-x ight) an 45^{\circ} \cdot \cos \left(360^{\circ}-x ight)} {} 7.3 \frac{2 \cos \left(-180^{\circ}-x ight) \cdot \sin (-x)}{\sin \left(x-180^{\circ} ight) \cdot \cos \left(x-360^{\circ} ight)} {} 7.4 \frac{\cos ^{2}\left(180^{\circ}+x ight)}{\sin \left(180^{\circ}-x ight) \cdot \cos \left(x-90^{\circ} ight)}+1 {} 7.5\left[\sin \left(90^{\circ}+x ight) \sin (-x) an \left(180^{\circ}-x ight) ight]-\left[\sin \left(90^{\circ}-x ight) \cos \left(x-180^{\circ} ight) ight] $$

Question

$$ 7.1 \frac{2 \cos \left(90^{\circ}+xight)-\sin \left(360^{\circ}-xight)}{3 \sin \left(90^{\circ}-xight)-\cos \left(180^{\circ}-xight)} {} 7.2 \frac{\sin \left(180^{\circ}+xight)}{	an \left(180^{\circ}-xight) 	an 45^{\circ} \cdot \cos \left(360^{\circ}-xight)} {} 7.3 \frac{2 \cos \left(-180^{\circ}-xight) \cdot \sin (-x)}{\sin \left(x-180^{\circ}ight) \cdot \cos \left(x-360^{\circ}ight)} {} 7.4 \frac{\cos ^{2}\left(180^{\circ}+xight)}{\sin \left(180^{\circ}-xight) \cdot \cos \left(x-90^{\circ}ight)}+1 {} 7.5\left[\sin \left(90^{\circ}+xight) \sin (-x) 	an \left(180^{\circ}-xight)ight]-\left[\sin \left(90^{\circ}-xight) \cos \left(x-180^{\circ}ight)ight] $$

UpStudy AI · Accepted Answer

Alright, I need to simplify these trigonometric expressions. Let's tackle them one by one, starting with the first one.

**7.1**  
$$
\frac{2 \cos \left(90^{\circ}+x\right) - \sin \left(360^{\circ}-x\right)}{3 \sin \left(90^{\circ}-x\right) - \cos \left(180^{\circ}-x\right)}
$$

First, I recall some trigonometric identities:
- $$\cos(90^{\circ} + x) = -\sin(x)$$
- $$\sin(360^{\circ} - x) = -\sin(x)$$
- $$\sin(90^{\circ} - x) = \cos(x)$$
- $$\cos(180^{\circ} - x) = -\cos(x)$$

Applying these identities:
$$
\frac{2(-\sin(x)) - (-\sin(x))}{3\cos(x) - (-\cos(x))} = \frac{-2\sin(x) + \sin(x)}{3\cos(x) + \cos(x)} = \frac{-\sin(x)}{4\cos(x)} = -\frac{\sin(x)}{4\cos(x)} = -\frac{1}{4} \tan(x)
$$

So, the simplified form of 7.1 is $$-\frac{1}{4} \tan(x)$$.

**7.2**  
$$
\frac{\sin \left(180^{\circ}+x\right)}{\tan \left(180^{\circ}-x\right) \tan 45^{\circ} \cdot \cos \left(360^{\circ}-x\right)}
$$

Using identities:
- $$\sin(180^{\circ} + x) = -\sin(x)$$
- $$\tan(180^{\circ} - x) = -\tan(x)$$
- $$\tan(45^{\circ}) = 1$$
- $$\cos(360^{\circ} - x) = \cos(x)$$

Substituting:
$$
\frac{-\sin(x)}{(-\tan(x)) \cdot 1 \cdot \cos(x)} = \frac{-\sin(x)}{-\tan(x) \cos(x)} = \frac{\sin(x)}{\tan(x) \cos(x)} = \frac{\sin(x)}{\frac{\sin(x)}{\cos(x)} \cos(x)} = \frac{\sin(x)}{\sin(x)} = 1
$$

So, 7.2 simplifies to $$1$$.

**7.3**  
$$
\frac{2 \cos \left(-180^{\circ}-x\right) \cdot \sin (-x)}{\sin \left(x-180^{\circ}\right) \cdot \cos \left(x-360^{\circ}\right)}
$$

Using identities:
- $$\cos(-180^{\circ} - x) = \cos(180^{\circ} + x) = -\cos(x)$$
- $$\sin(-x) = -\sin(x)$$
- $$\sin(x - 180^{\circ}) = -\sin(x)$$
- $$\cos(x - 360^{\circ}) = \cos(x)$$

Substituting:
$$
\frac{2(-\cos(x)) \cdot (-\sin(x))}{(-\sin(x)) \cdot \cos(x)} = \frac{2\cos(x)\sin(x)}{-\sin(x)\cos(x)} = \frac{2}{-1} = -2
$$

So, 7.3 simplifies to $$-2$$.

**7.4**  
$$
\frac{\cos ^{2}\left(180^{\circ}+x\right)}{\sin \left(180^{\circ}-x\right) \cdot \cos \left(x-90^{\circ}\right)} + 1
$$

Using identities:
- $$\cos(180^{\circ} + x) = -\cos(x)$$
- $$\cos^2(180^{\circ} + x) = \cos^2(x)$$
- $$\sin(180^{\circ} - x) = \sin(x)$$
- $$\cos(x - 90^{\circ}) = \sin(x)$$

Substituting:
$$
\frac{\cos^2(x)}{\sin(x) \cdot \sin(x)} + 1 = \frac{\cos^2(x)}{\sin^2(x)} + 1 = \cot^2(x) + 1
$$

Recall the identity $$\cot^2(x) + 1 = \csc^2(x)$$, so:
$$
\csc^2(x)
$$

Therefore, 7.4 simplifies to $$\csc^2(x)$$.

**7.5**  
$$
\left[\sin \left(90^{\circ}+x\right) \sin (-x) \tan \left(180^{\circ}-x\right)\right] - \left[\sin \left(90^{\circ}-x\right) \cos \left(x-180^{\circ}\right)\right]
$$

Using identities:
- $$\sin(90^{\circ} + x) = \cos(x)$$
- $$\sin(-x) = -\sin(x)$$
- $$\tan(180^{\circ} - x) = -\tan(x)$$
- $$\sin(90^{\circ} - x) = \cos(x)$$
- $$\cos(x - 180^{\circ}) = -\cos(x)$$

Substituting:
$$
\left[\cos(x) \cdot (-\sin(x)) \cdot (-\tan(x))\right] - \left[\cos(x) \cdot (-\cos(x))\right] = \left[\cos(x) \cdot \sin(x) \cdot \tan(x)\right] - \left[-\cos^2(x)\right] = \cos(x) \sin(x) \tan(x) + \cos^2(x)
$$

Recall that $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$, so:
$$
\cos(x) \sin(x) \cdot \frac{\sin(x)}{\cos(x)} + \cos^2(x) = \sin^2(x) + \cos^2(x) = 1
$$

So, 7.5 simplifies to $$1$$.

7.1: $$-\frac{1}{4} \tan(x)$$ 7.2: $$1$$ 7.3: $$-2$$ 7.4: $$\csc^2(x)$$ 7.5: $$1$$

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