Question

In a triangle $\Delta ABC$, the value of $\cos \left(\frac{B+C}{2}\right)$ in terms of angle $A$.

• A. $\sqrt{\sin \frac{A}{2}}$
• B. $\sqrt{A/2}$
• C. $\sin \frac{A}{2}$
• D. $\sqrt{2A}$

Hint:

Whenever you see a sum of two angles divided by two in a triangle problem, immediately think of $90^\circ$ minus the half-angle of the third vertex. This substitution simplifies almost all triangle-based trig identities.

Answer 1

The Correct Option is C

Similar to the previous problem, we rely on the relationship between the interior angles of a triangle. 1. Angle Relationship: Starting with the sum of angles: $$A + B + C = 180^\circ \implies B + C = 180^\circ - A$$ Dividing by 2 gives: $$\frac{B+C}{2} = 90^\circ - \frac{A}{2}$$

2. Applying the Cosine Function: Apply the cosine ratio to both sides of the equation: $$\cos\left(\frac{B+C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right)$$

3. Using Trigonometric Identities: Using the complementary angle identity $\cos(90^\circ - \theta) = \sin \theta$: $$\cos\left(90^\circ - \frac{A}{2}\right) = \sin \frac{A}{2}$$ Thus, the value of $\cos \left(\frac{B+C}{2}\right)$ is exactly $\sin \frac{A}{2}$.