In this section we have seen that the cosines of the angles...

In this section we have seen that the cosines of the angles in a triangle can be expressed in terms of the lengths of the sides. For instance, for $\cos A$ in $\triangle A B C$, we obtained $\cos A=\left(b^{2}+c^{2}-a^{2}\right) / 2 b c .$ This exercise shows how to derive corresponding expressions for the sines of the angles. For ease of notation in this exercise, let us agree to use the letter $T$ to denote the following quantity: $$T=2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)$$Then the sines of the angles in $\triangle A B C$ are given by$$\sin A=\frac{\sqrt{T}}{2 b c} \quad \sin B=\frac{\sqrt{T}}{2 a c} \quad \sin C=\frac{\sqrt{T}}{2 a b}$$ In the steps that follow, we'll derive the first of these three formulas, the derivations for the other two being entirely similar. (a) In $\triangle A B C$, why is the positive root always appropriate in the formula $\sin A=\sqrt{1-\cos ^{2} A} ?$ (b) In the formula in part (a), replace cos $A$ by $\left(b^{2}+c^{2}-a^{2}\right) / 2 b c$ and show that the result can be written $$ \sin A=\frac{\sqrt{4 b^{2} c^{2}-\left(b^{2}+c^{2}-a^{2}\right)^{2}}}{2 b c} $$ (c) On the right-hand side of the equation in part (b), carry out the indicated multiplication. After combining like terms, you should obtain $\sin A=\sqrt{T / 2 b c, \text { as required. }}$

In this section we have seen that the cosines of the angles in a triangle can be expressed in terms of the lengths of the sides. For instance, for $\cos A$ in $\triangle A B C$, we obtained $\cos A=\left(b^{2}+c^{2}-a^{2}\right) / 2 b c .$ This exercise shows how to derive corresponding expressions for the sines of the angles. For ease of notation in this exercise, let us agree to use the letter $T$ to denote the following quantity:
$$T=2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)$$Then the sines of the angles in $\triangle A B C$ are given by$$\sin A=\frac{\sqrt{T}}{2 b c} \quad \sin B=\frac{\sqrt{T}}{2 a c} \quad \sin C=\frac{\sqrt{T}}{2 a b}$$
In the steps that follow, we'll derive the first of these three formulas, the derivations for the other two being entirely similar.
(a) In $\triangle A B C$, why is the positive root always appropriate in the formula $\sin A=\sqrt{1-\cos ^{2} A} ?$
(b) In the formula in part (a), replace cos $A$ by $\left(b^{2}+c^{2}-a^{2}\right) / 2 b c$ and show that the result can be written
$$
\sin A=\frac{\sqrt{4 b^{2} c^{2}-\left(b^{2}+c^{2}-a^{2}\right)^{2}}}{2 b c}
$$
(c) On the right-hand side of the equation in part (b), carry out the indicated multiplication. After combining like terms, you should obtain $\sin A=\sqrt{T / 2 b c, \text { as required. }}$

Key Concepts

Positivity of the Sine Function in Triangles

In any triangle, the angles lie between 0 and ? radians, ensuring that the sine of any angle is positive. This fact justifies the use of the positive square root when applying the Pythagorean identity to determine the sine from the cosine value, which is particularly important in triangle-related problems where the physical interpretation of sine must represent a positive quantity.

Algebraic Manipulation in Trigonometric Derivations

Derivations in trigonometry often require expanding, simplifying, and factoring algebraic expressions. These manipulations are essential for expressing trigonometric functions in alternative forms—such as rewriting the sine of an angle solely in terms of the side lengths of a triangle. Mastery of these techniques allows one to seamlessly transition from geometric relationships to more complex algebraic forms.

Law of Cosines

The Law of Cosines establishes a relationship between the angles and side lengths in a triangle. It allows one to express the cosine of an angle in terms of the three sides, and it is a critical tool for deriving other trigonometric expressions about triangles. This relationship is particularly useful when the triangle's side lengths are known but its angles are not, forming a foundation for many proofs and derivations in trigonometry.

Pythagorean Trigonometric Identity

The Pythagorean identity, which states that sin²? + cos²? = 1, is fundamental in trigonometry. This identity is used to determine one trigonometric function when the other is known. In derivations involving triangles, it is common to solve for the sine of an angle by taking the square root of 1 minus the square of the cosine of that angle.

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