Chapter 4: Problem 102
In a triangle \(\mathrm{ABC}, 2 \mathrm{abc} \cos \frac{\mathrm{A}}{2} \cos \frac{\mathrm{B}}{2} \cos \frac{\mathrm{C}}{2}\) is equal to (a) \(2 \mathrm{~s}\) (b) \(2 \mathrm{sr}\) (c) \(2 \mathrm{~s}^{2} \mathrm{r}\) (d) \(2 \mathrm{~s} \mathrm{r}^{2}\)
Short Answer
Expert verified
Answer: (b) $2sr$
Step by step solution
01
Define the given expression in the problem
We are given the following expression to work with:
$$2abc \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$
02
Make use of half angle formulas
Recall the half angle formulas for cosine:
$$\cos \frac{A}{2} = \sqrt{\frac{1+\cos A}{2}}$$
and similarly for B and C. Substitute these formulas into our given expression:
$$2abc \sqrt{\frac{1 + \cos A}{2}} \sqrt{\frac{1 + \cos B}{2}} \sqrt{\frac{1 + \cos C}{2}}$$
03
Apply the Law of Cosines
The Law of Cosines states that for a triangle with sides a, b, and c, and angles A, B, and C, the following are true:
$$a^2 = b^2 + c^2 - 2bc\cos A$$
$$b^2 = a^2 + c^2 - 2ac\cos B$$
$$c^2 = a^2 + b^2 - 2ab\cos C$$
Solving for cosines and substituting in the expression:
$$2abc \sqrt{\frac{1 + (b^2 + c^2 - a^2)/(2bc)}{2}} \sqrt{\frac{1 + (a^2 + c^2 - b^2)/(2ac)}{2}} \sqrt{\frac{1 + (a^2 + b^2 - c^2)/(2ab)}{2}}$$
Simplify the expression:
$$2abc\sqrt{\frac{b^2+c^2+2bc-a^2}{4bc^2}}\sqrt{\frac{a^2+c^2+2ac-b^2}{4a^2c^2}}\sqrt{\frac{a^2+b^2+2ab-c^2}{4a^2b^2}}$$
04
Introduce terms of semiperimeter and inradius
Semiperimeter \((s)\) is defined as:
$$s = \frac{a+b+c}{2}$$
The area of the triangle \((A)\) can be described using inradius \((r)\) and semiperimeter:
$$A = rs$$
We also know that the area of the triangle can be expressed using Heron's formula:
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
Now, our aim is to rewrite the expression from step 3 in terms of \(s\) and \(r\).
05
Simplifying and finding the final expression
Let's rewrite our expression from step 3 using semiperimeter:
$$2abc\sqrt{\frac{s(s-a)}{4b^2c^2}}\sqrt{\frac{s(s-b)}{4a^2c^2}}\sqrt{\frac{s(s-c)}{4a^2b^2}}$$
Combine the square roots:
$$2abc\sqrt{\frac{s(s-a)(s-b)(s-c)}{16a^2b^2c^2}}$$
Substitute the area formula:
$$2abc\frac{rs}{8abc}$$
Simplify the expression:
$$2sr$$
Comparing the simplified expression to the given choices, we find that the correct answer is (b): $$2 \mathrm{sr}$$.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometry in Triangles
Understanding trigonometry within the context of triangles is crucial for solving a variety of geometric problems. When dealing with triangles, particularly right-angled and non-right-angled ones, trigonometric ratios such as sine, cosine, and tangent provide a relationship between the angles and the lengths of the sides. In the context of our exercise, we focus on cosine half-angle formulas, which aid in breaking down the properties of the triangle into more manageable pieces.
For example, the half-angle formula \[\begin{equation}\cos \frac{A}{2} = \sqrt{\frac{1+\cos A}{2}}\end{equation}\] helps us relate the standard cosine of an angle to its half-angle counterpart, thereby intertwining the geometry of the triangle with trigonometry. This becomes especially handy when we don’t have direct measures of certain angles or when working with the triangle’s inscribed circle - the circle touching all sides of the triangle, where the triangle's angles bisect the circle's radii.
In addition to half-angle formulas, we can utilize other trigonometric identities and laws, such as the sine and cosine laws, to relate different elements of a triangle, including side lengths, angles, and even area in terms of trigonometric functions.
For example, the half-angle formula \[\begin{equation}\cos \frac{A}{2} = \sqrt{\frac{1+\cos A}{2}}\end{equation}\] helps us relate the standard cosine of an angle to its half-angle counterpart, thereby intertwining the geometry of the triangle with trigonometry. This becomes especially handy when we don’t have direct measures of certain angles or when working with the triangle’s inscribed circle - the circle touching all sides of the triangle, where the triangle's angles bisect the circle's radii.
In addition to half-angle formulas, we can utilize other trigonometric identities and laws, such as the sine and cosine laws, to relate different elements of a triangle, including side lengths, angles, and even area in terms of trigonometric functions.
Law of Cosines
The Law of Cosines is a fundamental theorem that bridges trigonometry and geometry by relating the lengths of a triangle's sides to its angles.
For a triangle with sides labeled a, b, and c, and opposite angles A, B, and C respectively, the Law of Cosines is given by:
This remarkable formula extends the Pythagorean theorem to non-right triangles and is employed to find an unknown side or angle. In our exercise, we apply the Law of Cosines to determine \[\begin{equation}\cos A, \cos B, \end{equation}\] and \[\begin{equation}\cos C\end{equation}\] and use these values within the context of cosine half-angle formulas. It's important for solving complex problems in triangles where you can't apply simpler trigonometric rules. Moreover, the Law of Cosines paves the way to further insights about a triangle, such as determining whether it's obtuse or acute, based on the cosine values of its angles.
For a triangle with sides labeled a, b, and c, and opposite angles A, B, and C respectively, the Law of Cosines is given by:
- \(a^2 = b^2 + c^2 - 2bc\cos A\)
- \(b^2 = a^2 + c^2 - 2ac\cos B\)
- \(c^2 = a^2 + b^2 - 2ab\cos C\)
This remarkable formula extends the Pythagorean theorem to non-right triangles and is employed to find an unknown side or angle. In our exercise, we apply the Law of Cosines to determine \[\begin{equation}\cos A, \cos B, \end{equation}\] and \[\begin{equation}\cos C\end{equation}\] and use these values within the context of cosine half-angle formulas. It's important for solving complex problems in triangles where you can't apply simpler trigonometric rules. Moreover, the Law of Cosines paves the way to further insights about a triangle, such as determining whether it's obtuse or acute, based on the cosine values of its angles.
Heron's Formula
Heron's formula is a gem in the study of triangles as it allows us to calculate the area of a triangle when we know the lengths of all three sides, without the need to identify the measurements of the angles. It is expressed in terms of the semiperimeter (s) of the triangle, which is half of the triangle’s perimeter (the sum of the lengths of its sides).
If a triangle has sides of length a, b, and c, then the semiperimeter is:
\[\begin{equation}s = \frac{a+b+c}{2}\end{equation}\]With the semiperimeter in hand, Heron’s formula states that the area (A) of the triangle is:< br>\[\begin{equation}A = \sqrt{s(s-a)(s-b)(s-c)}\end{equation}\]Heron’s formula is directly linked to our exercise as it plays a critical role in connecting the half-angle cosine product with the area and the radius of the inscribed circle (inradius). It simplifies the complex expression involving cosine half-angles by relating it to the easily computable quantities of semiperimeter and area. Furthermore, Heron's formula underlines the importance of understanding how a triangle's side lengths can comprehensively reveal intrinsic properties like its area, reinforcing the idea that triangles can be studied and understood from multiple perspectives.
If a triangle has sides of length a, b, and c, then the semiperimeter is:
\[\begin{equation}s = \frac{a+b+c}{2}\end{equation}\]With the semiperimeter in hand, Heron’s formula states that the area (A) of the triangle is:< br>\[\begin{equation}A = \sqrt{s(s-a)(s-b)(s-c)}\end{equation}\]Heron’s formula is directly linked to our exercise as it plays a critical role in connecting the half-angle cosine product with the area and the radius of the inscribed circle (inradius). It simplifies the complex expression involving cosine half-angles by relating it to the easily computable quantities of semiperimeter and area. Furthermore, Heron's formula underlines the importance of understanding how a triangle's side lengths can comprehensively reveal intrinsic properties like its area, reinforcing the idea that triangles can be studied and understood from multiple perspectives.