<s>[INST] If $\tan2\theta=\frac{b}{a-c}$, then $\cos 2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}}$ and $\sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}}$ I am studying general equation of the second degree. While studying that chapter I came across 
$$\tan2\theta=\frac{b}{a-c}   \tag{1}$$
Now from (1), the author computed 
$$\cos2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}} \quad\text{and}\quad \sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}} \tag{2}$$ 
I don't understand how did the author compute these terms? 
If any member knows the correct answer may reply to this question? 
 [/INST] $$\tan(2\theta) = \frac{b}{a-c}$$
So we have a right triangle where the legs of the right triangle are of lengths $b$ and $a-c$, which means the hypotenuse should be $\sqrt{b^2 + (a-c)^2}$. We can, thus, compute the sine and cosine of the concerned angle using $\frac{\text{adj.}}{\text{hyp.}}$ and $\frac{\text{opp.}}{\text{hyp.}}$
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