What is the value of tan \( \alpha \) when the expression \( 2 \sin x - \cos x \) is written in the form \( \sqrt{5} \sin(x - \alpha) \)? - HSC - SSCE Mathematics Extension 1 - Question 4 - 2017 - Paper 1

Rewrite the expression as follows:

$$R \sin(x - \alpha) = R \left(\sin x \cos \alpha - \cos x \sin \alpha\right)$$

Here, we equate terms to find:

• Coefficient of ( \sin x ): ( R \cos \alpha = 2 )
• Coefficient of ( \cos x ): ( R \sin \alpha = -1 )

To find ( R ), use:

$$R = \sqrt{(2)^2 + (-1)^2} = \sqrt{5}$$

We find ( \tan \alpha ) using the ratios derived from the sine and cosine relationships:

$$\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-1/R}{2/R} = \frac{-1}{2}$$

Thus, the value of ( \tan \alpha ) is ( -\frac{1}{2} ).