Question - Solving a Trigonometric Function Using Identities
mathtrigonometric identitiestrigonometric functionsolving trigonometric expressionsmathematical properties
Solution:
The expression given in the image is a mathematical function involving trigonometric identities:\[ \frac{\tan 315^\circ - \cos 1020^\circ}{\sin 150^\circ - \tan (-135^\circ)} \]Let's solve it step by step using trigonometric identities and properties.Firstly, we can simplify each trigonometric function by using standard angles and periodic properties.The tangent and sine functions have a period of 360°, meaning $$\tan(\theta) = \tan(\theta + k \cdot 360^\circ)$$ and $$\sin(\theta) = \sin(\theta + k \cdot 360^\circ)$$ where $$k$$ is any integer. The cosine function also has the same period.So:- $$\tan 315^\circ$$ is equivalent to $$\tan (360^\circ - 45^\circ)$$, which equals $$\tan (-45^\circ)$$. Since $$\tan$$ is an odd function, $$\tan (-\theta) = -\tan (\theta)$$, so $$\tan 315^\circ = -\tan 45^\circ = -1$$.- $$\cos 1020^\circ$$ is equivalent to $$\cos (3 \cdot 360^\circ - 60^\circ)$$, which equals $$\cos (-60^\circ)$$. Since $$\cos$$ is an even function, $$\cos (-\theta) = \cos (\theta)$$, so $$\cos 1020^\circ = \cos 60^\circ = \frac{1}{2}$$.- $$\sin 150^\circ$$ is equivalent to $$\sin (180^\circ - 30^\circ)$$, which equals $$\sin 30^\circ$$. Thus, $$\sin 150^\circ = \frac{1}{2}$$.- $$\tan (-135^\circ)$$ is equivalent to $$-\tan (135^\circ)$$, which is $$-\tan (180^\circ - 45^\circ)$$ and equals $$-(-1)$$ because $$\tan (180^\circ - \theta) = -\tan (\theta)$$. So, $$\tan (-135^\circ) = 1$$.Now we can plug these values into the original expression:\[ \frac{-1 - \frac{1}{2}}{\frac{1}{2} - 1} = \frac{-\frac{3}{2}}{-\frac{1}{2}} \]Upon simplifying the fraction:\[ \frac{-\frac{3}{2}}{-\frac{1}{2}} = \frac{-3}{2} \div \frac{-1}{2} = \frac{-3}{2} \cdot \frac{-2}{1} = \frac{3 \cdot 2}{2 \cdot 1} = \frac{6}{2} = 3 \]Hence, the simplified value of the given expression is 3.
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