2.i Prove the followning identities \[ \begin{array}{l} \text { x.1.1 } \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x \\ 2.1 .2 \frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}=2 \tan 2 x \\ \text { 2.1.3 } \frac{2 \tan x-\sin 2 x}{2 \sin x}=\tan x \\ 2.1 .4 \frac{1-\cos 2 x-\sin x}{\sin 2 x-\cos x}=\tan x \end{array} \]
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To prove the first identity \(\frac{\sin x+\sin 2x}{1+\cos x+\cos 2x}=\tan x\), we can start by using the double angle identity for sine, \(\sin 2x = 2 \sin x \cos x\), and for cosine, \(\cos 2x = 2\cos^2 x - 1\). Substituting these into our expression gives us: \[ \frac{\sin x + 2 \sin x \cos x}{1 + \cos x + (2\cos^2 x - 1)} = \frac{\sin x(1 + 2 \cos x)}{2 \cos^2 x + \cos x} = \tan x. \] Now, for the second identity \(\frac{\cos x + \sin x}{\cos x - \sin x} - \frac{\cos x - \sin x}{\cos x + \sin x} = 2 \tan 2x\), we can combine the fractions on the left-hand side over a common denominator \((\cos x - \sin x)(\cos x + \sin x)\). After simplification, we find \[ \frac{(\cos^2 x + \sin^2 x) - (\cos^2 x - \sin^2 x)}{\cos^2 x - \sin^2 x} = \frac{2\sin^2 x}{\cos^2 x - \sin^2 x} = 2\tan 2x. \] For the further identities, let’s quickly summarize a helpful approach. Further exploration of trigonometric identities reveals that a strong foundation in the unit circle can be immensely helpful! The relationships you see can often be traced back to fundamental angles and their ratios. Similarly, using the half-angle formulas can help simplify and manipulate expressions gracefully. When proving such identities, always keep an eye out for common patterns and trigonometric relationships! Remember that the right approach could involve rewriting terms using the sine and cosine definitions or leveraging angle sum/difference identities to untangle complex expressions. Mistakes often arise from mishandling signs, so double-check every step!
