Trigonometry problem how to find the value

Question

If $0\leq \theta\leq \frac{\pi}{4}$ and $\sin2\theta = \frac{4}{5}$, find the value of $\tan{\theta}$.

Answer 1

As a hint : use this formula
$$\tan (\theta)=\frac{sin(2\theta)}{1+cos(2\theta)}=\frac{\frac45}{1+cos(2\theta)}\\ \frac{2sin(\theta)cos(\theta)}{2cos^2(\theta)}$$ and $$ sin^2(2\theta)+cos^2(2\theta)=1$$ $$sin(2\theta)=\frac45 \to (\frac45)^2+cos^2(2\theta)=1 $$

Answer 2

We know that if $$\tan(\frac x2)=t$$ then

$$\sin(x)=\frac{2t}{1+t^2}$$

thus

$$\sin(2\theta)=\frac{2\tan(\theta)}{1+\tan^2(\theta)}=\frac 45$$

hence

$$4(1+\tan^2(\theta))=10\tan(\theta)$$

now solve the quadratic using that $$0\le \tan(\theta) \le 1$$

If you do not find that $\tan(\theta)=\frac 12$, you made a mistake.

Answer 3

If $\sin2\phi$ is known and $2\phi$ is acute, then you can find $\cos2\phi$ from $\cos^22\phi+\sin^22\phi=1.$ Hence, you can find $\tan 2\phi.$

Then use the fact that $$\tan 2\phi=\frac{2\tan\phi}{1-\tan^2\phi},$$ to find $\tan\phi.$ Note that perforce $\phi$ is also acute.