$$ \sin (90-x) \cdot \cos \left(360^{\circ} x ight)+ an x \cos (-x) \sin (150- $$ Shat that $$ \sin (360-x) \cdot an ^{2} x $$ $$ \sin \left(x-180^{\circ} ight) \cos (90+x)=\frac{1}{\cos x} $$

Question

$$ \sin (90-x) \cdot \cos \left(360^{\circ} xight)+	an x \cos (-x) \sin (150- $$ Shat that $$ \sin (360-x) \cdot 	an ^{2} x $$ $$ \sin \left(x-180^{\circ}ight) \cos (90+x)=\frac{1}{\cos x} $$

UpStudy AI · Accepted Answer

Let's break down the problem step by step. We have the following expressions to analyze:

1. $$ \sin (90-x) \cdot \cos (360^{\circ} x) + \tan x \cos (-x) \sin (150) - \sin (360-x) \cdot \tan^{2} x $$
2. $$ \sin (x-180^{\circ}) \cos (90+x) = \frac{1}{\cos x} $$

### Step 1: Simplifying the First Expression

We will simplify the first expression step by step.

1. **Using Trigonometric Identities:**
   - $$ \sin(90 - x) = \cos x $$
   - $$ \cos(360^{\circ} x) = \cos(0) = 1 $$ (since cosine is periodic with a period of $$360^{\circ}$$)
   - $$ \tan x = \frac{\sin x}{\cos x} $$
   - $$ \cos(-x) = \cos x $$
   - $$ \sin(150^{\circ}) = \frac{1}{2} $$
   - $$ \sin(360 - x) = -\sin x $$ (since sine is periodic with a period of $$360^{\circ}$$)

Now substituting these identities into the expression:

$$
\cos x \cdot 1 + \frac{\sin x}{\cos x} \cdot \cos x \cdot \frac{1}{2} - (-\sin x) \cdot \tan^{2} x
$$

This simplifies to:

$$
\cos x + \frac{1}{2} \sin x - \sin x \cdot \tan^{2} x
$$

2. **Substituting $$ \tan^{2} x = \frac{\sin^{2} x}{\cos^{2} x} $$:**

$$
\cos x + \frac{1}{2} \sin x - \sin x \cdot \frac{\sin^{2} x}{\cos^{2} x}
$$

This can be further simplified, but let's keep it in this form for now.

### Step 2: Simplifying the Second Expression

Now let's simplify the second expression:

$$
\sin (x-180^{\circ}) \cos (90+x) = \frac{1}{\cos x}
$$

1. **Using Trigonometric Identities:**
   - $$ \sin(x - 180^{\circ}) = -\sin x $$
   - $$ \cos(90 + x) = -\sin x $$

Substituting these identities gives:

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

This simplifies to:

$$
\sin^{2} x = \frac{1}{\cos x}
$$

### Step 3: Final Formulation

Now we have two simplified expressions:

1. From the first expression: 
   $$
   \cos x + \frac{1}{2} \sin x - \sin x \cdot \frac{\sin^{2} x}{\cos^{2} x}
   $$

2. From the second expression:
   $$
   \sin^{2} x = \frac{1}{\cos x}
   $$

### Conclusion

The first expression can be further simplified or evaluated based on specific values of $$x$$, while the second expression gives a relationship between $$\sin^{2} x$$ and $$\cos x$$. If you have specific values for $$x$$ or further instructions on what to do with these expressions, please let me know!
The simplified expressions are: 1. $$ \cos x + \frac{1}{2} \sin x - \sin x \cdot \frac{\sin^{2} x}{\cos^{2} x} $$ 2. $$ \sin^{2} x = \frac{1}{\cos x} $$

\( \sin (90-x) \cdot \cos \left(360^{\circ} x\right)+\tan x \cos (-x) \sin (150- \) Shat that \( \sin (360-x) \cdot \tan ^{2} x \) \( \sin \left(x-180^{\circ}\right) \cos (90+x)=\frac{1}{\cos x} \)

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