Trigonometry

**Simplicity** · June 14th 2008, 02:56 AM

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

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

**Isomorphism** · June 14th 2008, 03:27 AM

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:

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

$\sin \left(x+\frac14k\pi\right)+\cos\left(x+\frac14k\pi \right) = \sqrt{2}\sin \left(x+\frac14k\pi + \frac{\pi}4\right) = \sqrt{2}\sin \left(x+\frac{(k+1)\pi}4\right)$

**mr fantastic** · June 14th 2008, 03:34 AM

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

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