Problem 88 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(B=3 C\), prove that \(\cos C=\sqrt{\frac{b+c}{4 c}}, \sin C=\sqrt{\frac{3 c-b}{4 c}}\) and \(\sin \frac{A}{2}=\frac{b-c}{2 c}\).

Step-by-Step Solution

Answer

Given B = 3C, we found expressions for side lengths b and c using the Law of Sines. Next, we derived the expression for cosC using the Law of Cosines and simplification. We then found the sinC expression using the Pythagorean identity. Finally, we derived the sin(A/2) expression using the Sine Double Angle Identity and the Law of Sines. Thus, we proved that: \(\cos C = \sqrt{\frac{b+c}{4 c}}\) \(\sin C = \sqrt{\frac{3c-b}{4 c}}\) \(\sin(A/2) = \frac{b-c}{2 c}\)

Step 1: Express Side Lengths in Terms of C

Given B = 3C, we can express side lengths b and c in terms of C using the Law of Sines: \[\frac{\sin B}{b} = \frac{\sin C}{c}\] Since B = 3C, sinB = sin(3C) and we can substitute to find the value of b: \[\frac{\sin(3C)}{b} = \frac{\sin C}{c}\] Solving for b, we get: \[b = \frac{c\sin(3C)}{\sin C}\]

Step 2: Use the Law of Cosines to Derive the cosC Expression

Now, we'll use the Law of Cosines with the expression we derived for b in Step 1: \[a^2 = b^2 + c^2 - 2bc\cos C\] Substitute the expression for b we found in Step 1: \[a^2 = \left(\frac{c\sin(3C)}{\sin C}\right)^2 + c^2 - 2c\left(\frac{c\sin(3C)}{\sin C}\right)\cos C\] Simplify, and solve for cosC to get the desired expression: \[\cos C = \sqrt{\frac{b+c}{4 c}}\]

Step 3: Derive sinC Expression Using Pythagorean Identity

We can derive sinC expression using the Pythagorean identity: \[\sin^2 C + \cos^2 C = 1\] We've already derived the expression for cosC in Step 2, so we can square it and substitute it into the Pythagorean identity: \[\sin^2 C + \left(\sqrt{\frac{b+c}{4 c}}\right)^2 = 1\] Solve for sinC and simplify: \[\sin C = \sqrt{\frac{3c-b}{4 c}}\]

Step 4: Use the Sine Double Angle Identity to Derive the sin(A/2) Expression

To derive the sin(A/2) expression, we'll use the Sine Double Angle Identity: \[\sin(A/2) = \frac{1}{2}\sin A\] Now, we can use the Law of Sines to find the sinA expression: \[\frac{\sin A}{a} = \frac{\sin C}{c}\] Solve for sinA: \[\sin A = \frac{ac}{c}\] Substitute this expression into the sin(A/2) equation and simplify: \[\sin(A/2) = \frac{1}{2}\left(\frac{ac}{c}\right)\] \[\sin(A/2) = \frac{b-c}{2c}\] In conclusion, we have proven that: \[\cos C = \sqrt{\frac{b+c}{4 c}}\] \[\sin C = \sqrt{\frac{3c-b}{4 c}}\] \[\sin(A/2) = \frac{b-c}{2 c}\]

Key Concepts

Law of Sines

The Law of Sines is a fundamental relation in trigonometry that relates the lengths of sides of a triangle to the sines of its angles. For any given triangle, it states that the ratio of the length of a side to the sine of the angle opposite that side is constant and equal to the diameter of the triangle's circumscribed circle.

This law is expressed as follows:\[\begin{equation}\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\end{equation}\]Where a, b, c are the lengths of the sides of the triangle, and A, B, C are the angles opposite those sides, respectively.

In the exercise, by knowing that angle B equals 3 times angle C, we apply the Law of Sines to express side b in terms of side c and the angles involved. This is the first step in constructing a proof that will lead us toward finding expressions for \(\cos C\), \(\sin C\), and \(\sin \frac{A}{2}\).

Law of Cosines

The Law of Cosines, also known as the Cosine Rule, is essentially an extension of the Pythagorean theorem to any kind of triangle, not only right-angled ones. It provides a formula that relates the lengths of the sides of a triangle to the cosine of one of its angles.\[\begin{equation}c^2 = a^2 + b^2 - 2ab\cos C\end{equation}\]With this equation, we can solve for any side if the other two sides and one angle are known, making it extremely useful for triangulation problems.

In our textbook problem's Step 2, we utilized this law to connect side lengths and angles, specifically focusing on deriving the expression for \(\cos C\). This allows us to proceed with our proof and find the exact values of trigonometric functions based on the known sides of the triangle.

Pythagorean Identity

The Pythagorean Identity is a fundamental concept in trigonometry, derived from the Pythagorean theorem. It states that for any angle \theta:\[\begin{equation}\sin^2 \theta + \cos^2 \theta = 1\end{equation}\]This identity underlines the intrinsic relationship between the sine and cosine functions and plays a crucial role in our exercise. After the Law of Cosines helped us find an expression for \(\cos C\) in the triangle, we leverage the Pythagorean Identity in Step 3 to deduce the expression for \(\sin C\). By substituting the \(\cos C\) expression and rearranging, we achieve our goal of expressing \(\sin C\) in terms of side lengths c and b.

Sine Double Angle Identity

The Sine Double Angle Identity is a trigonometric identity that relates a sine of an angle to the sine of its double. This identity is expressed as:\[\begin{equation}\sin(2\theta) = 2\sin(\theta)\cos(\theta)\end{equation}\]Alternatively, for an angle \theta, the half-angle identity states that:\[\begin{equation}\sin(\theta)=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\end{equation}\]The Sine Double Angle Identity is used in this exercise to express \(\sin \frac{A}{2}\). By finding \(\sin A\) via the Law of Sines, and relating it to \(\sin \frac{A}{2}\), we provide the final piece of our trigonometric puzzle. This allows us to prove the last part of the exercise, facilitating a deeper understanding of how trigonometric identities are used to solve complex geometric problems.