Animated Solution for Mathematics - Trigonometry: Given both $\theta$ and $\varphi$ are acute angles and $\sin \theta =1/2,\cos \varphi =1/3$, then the value of $\theta +\varphi$ belongs to

Imagine you are standing at the origin of a coordinate plane, looking out into the first quadrant. We are given two angles, $\theta$ and $\varphi$, which are both acute.

This is a vital constraint that locks our angles into the interval $(0,\pi /2)$. When we see the word 'acute' in a trigonometry problem, we restrict our focus to the first quadrant of the unit circle.

We are given that $\sin \theta =1/2$. In the world of trigonometry, this is a classic value.

We know that $\sin (\pi /6)=1/2$. Since we are restricted to the first quadrant, there is no ambiguity here.

Therefore, $\theta$ must be exactly $\pi /6$. We have successfully pinned down one of our variables as a solid, immovable constant.

We are told that $\cos \varphi =1/3$. Unlike $1/2$, the value $1/3$ is not a standard trigonometric ratio that corresponds to a simple angle.

We must use the power of inequalities to bound $\varphi$. We know that $\cos (\pi /3)=1/2$ and $\cos (\pi /2)=0$.

Since $1/3$ sits comfortably between $0$ and $1/2$, we can establish the following inequality:

This translates to the relationship:

This is the moment where many students stumble. We are dealing with the cosine function, which is strictly decreasing in the first quadrant.

As the angle increases, the cosine value drops. Because of this, when we take the inverse cosine to isolate $\varphi$, the inequality signs must flip.

The smaller cosine value ($0$) corresponds to the larger angle ($\pi /2$), and the larger cosine value ($1/2$) corresponds to the smaller angle ($\pi /3$). Thus, our inequality transforms into:

Now, we bring it all together. We have $\theta =\pi /6$ and the range for $\varphi$ as $(\pi /3,\pi /2)$.

To find the range of the sum $\theta +\varphi$, we add $\pi /6$ to every part of our inequality for $\varphi$:

The lower bound simplifies to $\pi /2$, and the upper bound simplifies to $2\pi /3$.

Therefore, the sum $\theta +\varphi$ must lie in the interval $(\pi /2,2\pi /3)$.