If $\tan2\theta=\frac{b}{a-c}$, then $\cos 2\theta=\frac{a-c}{\sqrt{b^2+(a-c)^2}}$ and $\sin2\theta=\frac{b}{\sqrt{b^2+(a-c)^2}}$

Question

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?

Answer 1

$$\tan(2\theta) = \frac{b}{a-c}$$ 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 $\frac{\text{opp.}}{\text{hyp.}}$

Answer 2

So,$$\dfrac{\sin2\theta}b=\dfrac{\cos2\theta}{a-c}$$

which is equal to $$\pm\sqrt{\dfrac{\cdots ?}{b^2+(a-c)^2}}$$

Answer 3

From

$$\sin^2\alpha+\cos^2\alpha=1,$$

draw

$$\tan^2\alpha+1=\frac1{\cos^2\alpha}$$

and

$$\cos\alpha=\pm\frac1{\sqrt{\tan^2\alpha+1}}.$$

Then

$$\sin\alpha=\tan\alpha\cos\alpha=\pm\frac{\tan\alpha}{\sqrt{\tan^2\alpha+1}}.$$