If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ wher $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find value of $\tan(a+2b)$ and $\tan(2a+b)$

Question

If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ where $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find the value of $\tan(a+2b)$ and $\tan(2a+b)$.

I have tried to apply the formula for $\tan(a+(a+b))$ but couldn't reach anywhere.

THANKS

Answer 1

Hint:

$$\sin(a + b) = 1$$

$$\sin(a + b) = \sin \frac π2$$

$$a + b = \frac π2 = 90^\circ$$

Also,

$$\sin(a - b) = \frac 12$$

$$\sin(a - b) = \sin \frac π6$$

$$a - b = \frac π6 = 30^\circ$$

Solve equations to find $a$, $b$.

$$a = \frac π3 = 60^\circ$$

$$b = \frac π6 = 30^\circ$$

Put values of $a$ and $b$ to solve.

Answer 2

You can immediately determine that $$a+b=\pi/2$$ $$a-b=\pi/6$$ So $a=\pi/3$ and $b=\pi/6$ and so you can calculate the required values directly.