Answer :
Let's solve the problem step-by-step.
We are given:
[tex]\[ \cos A = \frac{1}{2} \][/tex]
[tex]\[ \cos B = \frac{\sqrt{3}}{2} \][/tex]
First, we'll calculate [tex]\( \sin A \)[/tex] and [tex]\( \sin B \)[/tex] using the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
[tex]\[ \sin^2 B + \cos^2 B = 1 \][/tex]
### Calculating [tex]\( \sin A \)[/tex]:
[tex]\[ \begin{align*} \cos^2 A &= \left( \frac{1}{2} \right)^2 \\ &= \frac{1}{4} \end{align*} \][/tex]
[tex]\[ \begin{align*} \sin^2 A &= 1 - \cos^2 A \\ &= 1 - \frac{1}{4} \\ &= \frac{3}{4} \end{align*} \][/tex]
[tex]\[ \sin A = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \][/tex]
### Calculating [tex]\( \sin B \)[/tex]:
[tex]\[ \begin{align*} \cos^2 B &= \left( \frac{\sqrt{3}}{2} \right)^2 \\ &= \frac{3}{4} \end{align*} \][/tex]
[tex]\[ \begin{align*} \sin^2 B &= 1 - \cos^2 B \\ &= 1 - \frac{3}{4} \\ &= \frac{1}{4} \end{align*} \][/tex]
[tex]\[ \sin B = \sqrt{\frac{1}{4}} = \frac{1}{2} \][/tex]
### Calculating [tex]\( \cos(A + B) \)[/tex]:
Using the angle addition formula for cosine:
[tex]\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \][/tex]
[tex]\[ \cos A = \frac{1}{2}, \quad \cos B = \frac{\sqrt{3}}{2}, \quad \sin A = \frac{\sqrt{3}}{2}, \quad \sin B = \frac{1}{2} \][/tex]
[tex]\[ \begin{align*} \cos(A + B) &= \left( \frac{1}{2} \times \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{3}}{2} \times \frac{1}{2} \right) \\ &= \left( \frac{\sqrt{3}}{4} \right) - \left( \frac{\sqrt{3}}{4} \right) \\ &= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\ &= 0 \end{align*} \][/tex]
So the value of [tex]\( \cos(A + B) \)[/tex] is [tex]\( 0 \)[/tex].
The correct answer is (a) 0.
We are given:
[tex]\[ \cos A = \frac{1}{2} \][/tex]
[tex]\[ \cos B = \frac{\sqrt{3}}{2} \][/tex]
First, we'll calculate [tex]\( \sin A \)[/tex] and [tex]\( \sin B \)[/tex] using the Pythagorean identity:
[tex]\[ \sin^2 A + \cos^2 A = 1 \][/tex]
[tex]\[ \sin^2 B + \cos^2 B = 1 \][/tex]
### Calculating [tex]\( \sin A \)[/tex]:
[tex]\[ \begin{align*} \cos^2 A &= \left( \frac{1}{2} \right)^2 \\ &= \frac{1}{4} \end{align*} \][/tex]
[tex]\[ \begin{align*} \sin^2 A &= 1 - \cos^2 A \\ &= 1 - \frac{1}{4} \\ &= \frac{3}{4} \end{align*} \][/tex]
[tex]\[ \sin A = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \][/tex]
### Calculating [tex]\( \sin B \)[/tex]:
[tex]\[ \begin{align*} \cos^2 B &= \left( \frac{\sqrt{3}}{2} \right)^2 \\ &= \frac{3}{4} \end{align*} \][/tex]
[tex]\[ \begin{align*} \sin^2 B &= 1 - \cos^2 B \\ &= 1 - \frac{3}{4} \\ &= \frac{1}{4} \end{align*} \][/tex]
[tex]\[ \sin B = \sqrt{\frac{1}{4}} = \frac{1}{2} \][/tex]
### Calculating [tex]\( \cos(A + B) \)[/tex]:
Using the angle addition formula for cosine:
[tex]\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \][/tex]
[tex]\[ \cos A = \frac{1}{2}, \quad \cos B = \frac{\sqrt{3}}{2}, \quad \sin A = \frac{\sqrt{3}}{2}, \quad \sin B = \frac{1}{2} \][/tex]
[tex]\[ \begin{align*} \cos(A + B) &= \left( \frac{1}{2} \times \frac{\sqrt{3}}{2} \right) - \left( \frac{\sqrt{3}}{2} \times \frac{1}{2} \right) \\ &= \left( \frac{\sqrt{3}}{4} \right) - \left( \frac{\sqrt{3}}{4} \right) \\ &= \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} \\ &= 0 \end{align*} \][/tex]
So the value of [tex]\( \cos(A + B) \)[/tex] is [tex]\( 0 \)[/tex].
The correct answer is (a) 0.
