The law of cosines reduces to the Pythagorean theorem whenever the triangle is acute.
Check the final answer first, then review the worked steps.
Problem
The law of cosines reduces to the Pythagorean theorem whenever the triangle is acute.
Answer
False
Step-by-step solution
- Recall the Law of Cosines: The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. For a triangle with sides $a$, $b$, and $c$, and with angle $C$ opposite side $c$, the law is stated as: $$c^2 = a^2 + b^2 - 2ab \cos(C)$$
- Recall the Pythagorean Theorem: The Pythagorean Theorem applies to right-angled triangles and states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. For a right-angled triangle with sides $a$, $b$, and hypotenuse $c$, the theorem is: $$c^2 = a^2 + b^2$$
- Analyze the condition for reduction: The Law of Cosines reduces to the Pythagorean Theorem when the term $-2ab \cos(C)$ becomes zero. This happens when $\cos(C) = 0$.
- Determine the angle for $\cos(C) = 0$: The cosine of an angle is zero when the angle is $90^\circ$ (or $\frac{\pi}{2}$ radians). Therefore, $\cos(C) = 0$ when $C = 90^\circ$.
- Interpret the angle: An angle of $90^\circ$ means the triangle is a right-angled triangle. The Pythagorean Theorem is specifically for right-angled triangles.
- Evaluate the statement: The statement claims the Law of Cosines reduces to the Pythagorean Theorem whenever the triangle is acute. An acute triangle is a triangle where all three angles are less than $90^\circ$. If all angles are less than $90^\circ$, then $\cos(C)$ will be positive for all angles $C$ (since $0^\circ < C < 90^\circ$). This means the term $-2ab \cos(C)$ will be negative, not zero. Therefore, the Law of Cosines does not reduce to the Pythagorean Theorem for acute triangles. It reduces to the Pythagorean Theorem for right-angled triangles.