Help with $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$

Question

Assume that $0<\theta<\frac{\pi}{4}$. If $\sin2\theta=\frac{1}{4}$, then $\frac{\sin\theta+\cos\theta}{-\sin\theta+\cos\theta}=?$

My approach with this problem is to rationalize the denominator in the fraction to get this: $$ \frac{\sin2\theta+1}{\cos2\theta}=\frac{\frac{1}{4}+1}{\cos2\theta} $$

After that i got stuck with $\cos2\theta$ any advice how to attack the problem?

Answer 1

Hint: Note that $0 < 2\theta < \dfrac{\pi}{2} \implies \cos(2\theta) = \sqrt{1-\sin^2(2\theta)}$.

Answer 2

Alternative approach:

$$\frac{1}{4} = \sin(2\theta) = 2\sin(\theta)\cos(\theta). \tag1 $$

Also, since $0 < \theta < (\pi/4)$, you have that

$$\cos(2\theta) = \sqrt{1 - \left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4}. \tag2 $$

Finally, you have that

$$\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta) = \frac{\sqrt{15}}{4}. \tag3 $$

---

You have that:

$$\frac{\cos(\theta) + \sin(\theta)}{\cos(\theta) - \sin(\theta)} \times \frac{\cos(\theta) - \sin(\theta)}{\cos(\theta) - \sin(\theta)}$$

$$= ~\frac{\cos^2(\theta) - \sin^2(\theta)}{\cos^2(\theta) - 2\sin(\theta)\cos(\theta) + \sin^2(\theta)}. \tag4 $$

Using (1), (2), and (3), the expression in (4) equals

$$\frac{\frac{\sqrt{15}}{4}}{1 - \frac{1}{4}} = \frac{\sqrt{15}}{3}.$$