Hello Tweety Originally Posted by
Tweety Yes thanks,
But how did you know you had to use $\displaystyle \frac{1}{\sqrt{2}} $ ?
I knew because I've done it so often! But the general method is:
Let $\displaystyle \sin\theta + \cos\theta = R\sin(\theta +A) = R\sin\theta\cos A + R\cos\theta\sin A$.
Then, comparing coefficients of $\displaystyle \sin\theta$ and $\displaystyle \cos\theta$: $\displaystyle 1 = R\cos A$ and $\displaystyle 1 = R\sin A$
If we now divide the second of these equations by the first, we get:$\displaystyle \frac{R\sin A}{R\cos A}=\tan A = 1$
$\displaystyle \Rightarrow A = 45^o$
And if we square and add the two equations:$\displaystyle 1^2 + 1^2 = R^2(\cos^2A + \sin^2A)$
$\displaystyle \Rightarrow R^2=2$, since $\displaystyle \cos^2A + \sin^2A=1$
$\displaystyle \Rightarrow R = \sqrt2$
So there we have it:$\displaystyle \sin\theta + \cos\theta = \sqrt2\sin(\theta+45^o)$
Grandad