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Given $$\cos \theta = \frac{\sqrt{3}-1}{2\sqrt2}, \quad \sin \theta = \frac{\sqrt3+1}{2\sqrt2}$$

The book says that the final value is $\theta = \frac{\pi}{4} + \frac{\pi}{6}$. I can of course verify that's correct, but I don't understand how you could arrive at that conclusion from the above given data.

Edit: I did think of the idea that the author may have made use of the identity $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$ and guessed the values of $a$ and $b$, but I was hoping there was a better way of doing this.

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I can give you a better way - We know that $\sin(2\theta)=2*\sin(\theta)\cos(\theta)$ So, multiplying both the given equations, we get
$\sin(2\theta)=$$2(3-1)\over{8}$$=$$1\over 2$
$2\theta=$$5\pi\over 6$
$\theta=$$5\pi\over 12$

But, in most general cases, you have to estimate the values. The intuition as to how to solve the problem, or what A and B to use, can be improved by practice.

EDIT: $2\theta = $$5\pi\over 6$ because sin is greater than cosine and hence, $\theta > 45^o$

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  • $\begingroup$ Whoops. You beat me by a couple of minutes! $\endgroup$ – Lonidard Feb 14 '16 at 14:26
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$$\dfrac{\sqrt3}2\cdot\dfrac1{\sqrt2}-\dfrac12\cdot\dfrac1{\sqrt2}$$

$$=\cos\dfrac\pi3\cos\dfrac\pi4-\sin\dfrac\pi3\sin\dfrac\pi4$$

Use $\cos(A+B)=\cos A\cos B-\sin A\sin B$

Similarly for sine

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  • $\begingroup$ Yes, that's easy to verify, but is there a general method through which you could arrive at angles like this? Or do you have to simply guess the values of A and B each time? $\endgroup$ – Aayush Agrawal Feb 14 '16 at 14:15

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