# Thread: solve trig equation help.

1. ## solve trig equation help.

Solve, in the interval 0° ≤ θ<360°, the following equations. Give answers to the nearest 0.1°.

$sin \theta + cos \theta = 1$

2. Hello Tweety
Originally Posted by Tweety
Solve, in the interval 0° ≤ θ<360°, the following equations. Give answers to the nearest 0.1°.

$sin \theta + cos \theta = 1$
Use the fact that $\sin 45^o=\cos 45^o = \frac{1}{\sqrt2}$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$ to write
$\sin\theta+\cos\theta =\sqrt2 \sin(\theta + 45^o)=1$

$\Rightarrow \sin(\theta + 45^o) = \frac{1}{\sqrt2}$
Can you complete it now?

Hello TweetyUse the fact that $\sin 45^o=\cos 45^o = \frac{1}{\sqrt2}$ and $\sin(A+B) = \sin A \cos B + \cos A \sin B$ to write
$\sin\theta+\cos\theta =\sqrt2 \sin(\theta + 45^o)=1$

$\Rightarrow \sin(\theta + 45^o) = \frac{1}{\sqrt2}$
Can you complete it now?

Yes thanks,

But how did you know you had to use $\frac{1}{\sqrt{2}}$ ?

4. Hello Tweety
Originally Posted by Tweety
Yes thanks,

But how did you know you had to use $\frac{1}{\sqrt{2}}$ ?
I knew because I've done it so often! But the general method is:

Let $\sin\theta + \cos\theta = R\sin(\theta +A) = R\sin\theta\cos A + R\cos\theta\sin A$.

Then, comparing coefficients of $\sin\theta$ and $\cos\theta$:
$1 = R\cos A$ and $1 = R\sin A$
If we now divide the second of these equations by the first, we get:
$\frac{R\sin A}{R\cos A}=\tan A = 1$

$\Rightarrow A = 45^o$
And if we square and add the two equations:
$1^2 + 1^2 = R^2(\cos^2A + \sin^2A)$

$\Rightarrow R^2=2$, since $\cos^2A + \sin^2A=1$

$\Rightarrow R = \sqrt2$
So there we have it:
$\sin\theta + \cos\theta = \sqrt2\sin(\theta+45^o)$