Solving a Trigonometric Function Using Identities
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.
