1. ## Trigonometry

What does $\sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi \right)$ equal when convering it using the form:

$a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)$?

2. Originally Posted by Air
What does $\sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi \right)$ equal when convering it using the form:

$a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)$?
$\sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi \right) = \sqrt{2}\sin \left(x+\frac14k\pi + \frac{\pi}4\right) = \sqrt{2}\sin \left(x+\frac{(k+1)\pi}4\right)$

3. Originally Posted by Air
What does $\sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi \right)$ equal when convering it using the form:

$a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)$?
Expand the right hand side of $a\sin\theta+b\cos\theta = R\sin(\theta+\alpha)$ using the compound angle formula and compare each side:

$a = R \cos \alpha$ .... (1)

$b = R \sin \alpha$ .... (2)

$(1)^2 + (2)^2 \Rightarrow R = \sqrt{a^2 + b^2}$.

$\frac{(2)}{(1)} \Rightarrow \tan \alpha = \frac{b}{a}$.

In your case, a = b = 1.

Note: $\theta = x + \frac{1}{4} k \pi$.