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

Ask by Franklin Joseph. in South Africa
Mar 11,2025

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Answer

The simplified expressions are: 1. \( \cos x + \frac{1}{2} \sin x - \sin x \cdot \frac{\sin^{2} x}{\cos^{2} x} \) 2. \( \sin^{2} x = \frac{1}{\cos x} \)

Solution

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!

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Mind Expander

Did you know that sine and cosine are not just tools for solving mathematical problems, but they've been used since ancient times for navigation and astronomy? The ancient Greeks, for example, used trigonometric ratios to create accurate maps and understand the heavens, making sense of the stars with the very functions you’re working with today. If you’re diving into trigonometric identities, a common mistake is forgetting the periodic properties of the sine and cosine functions. For instance, \(\sin(360^\circ - x) = -\sin(x)\) can trip you up! Remember that many angles have equivalent values due to their periodic nature, so being mindful of those can save you from unnecessary confusion in your calculations. Happy sinning and cosining!

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