Question

\(cos^{-1}(cos\frac{4π}{3})+sin^{-1}(sin\frac{4π}{3}) = \ldots\)

• A. \(\frac{4π}{3}\)
• B. \(\frac{8π}{3}\)
• C. \(\frac{π}{3}\)
• D. \(\frac{3π}{2}\)

Hint:

Express altitudes with area and use the cosine and sine rules.

Answer 1

The Correct Option is C

Step 1: Express the altitudes:
\(P_1 = \frac{2\Delta}{a}\), \(P_2 = \frac{2\Delta}{b}\), \(P_3 = \frac{2\Delta}{c}\). So \(\frac{\cos A}{P_1} = \frac{a\cos A}{2\Delta}\).

Step 2: Sum the terms:
\[ \sum\frac{a\cos A}{2\Delta} = \frac{a\cos A + b\cos B + c\cos C}{2\Delta} \]
Using \(a = 2R\sin A\), we have \(a\cos A = R\sin2A\), so the numerator is \(R(\sin2A + \sin2B + \sin2C) = 4R\sin A\sin B\sin C\).

Step 3: Use the area formula:
\(\Delta = 2R^2\sin A\sin B\sin C\), so
\[ \frac{4R\sin A\sin B\sin C}{2\times 2R^2\sin A\sin B\sin C} = \frac{1}{R} \]
The dimension also checks out, because a ratio of a cosine to a length must have dimension 1/length, as does \(\frac{1}{R}\).

Final Answer:
The sum equals \(\frac{1}{R}\), option (B). \[ \boxed{\frac{1}{R}} \]