write the expression in terms of sine only. sin(8x) - cos(8x)

Explanation:

Step1: Use the identity $a\sin\theta + b\cos\theta=A\sin(\theta+\varphi)$

We have $a = 1$, $b=- 1$ and $\theta = 8x$. First, find $A=\sqrt{a^{2}+b^{2}}$. $A=\sqrt{1^{2}+(-1)^{2}}=\sqrt{1 + 1}=\sqrt{2}$.

Step2: Find $\varphi$

We know that $a = A\cos\varphi$ and $b=A\sin\varphi$. Since $a = 1$, $b=-1$ and $A=\sqrt{2}$, then $\cos\varphi=\frac{a}{A}=\frac{1}{\sqrt{2}}$ and $\sin\varphi=-\frac{1}{\sqrt{2}}$. So, $\varphi=-\frac{\pi}{4}$.

Step3: Rewrite the expression

$\sin(8x)-\cos(8x)=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin(8x)-\frac{1}{\sqrt{2}}\cos(8x)\right)=\sqrt{2}\sin\left(8x-\frac{\pi}{4}\right)$.

Answer:

$\sqrt{2}\sin\left(8x - \frac{\pi}{4}\right)$