Problem 29 - Algebra and Trigonometry

Question

\(23-40\) . Prove the identity. $$ \cos \left(x+\frac{\pi}{6}\right)+\sin \left(x-\frac{\pi}{3}\right)=0 $$

Step-by-Step Solution

Answer

The identity is verified since all terms cancel out to equal zero.

Step 1: Use Angle Addition Formulas

Break down the given trigonometric identity using the angle addition formulas: \[ \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) \\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)\]Apply these to the identity: \[\cos \left(x+\frac{\pi}{6} \right) = \cos x \cdot \cos \frac{\pi}{6} - \sin x \cdot \sin \frac{\pi}{6}\]\[\sin \left(x-\frac{\pi}{3} \right) = \sin x \cdot \cos \frac{\pi}{3} - \cos x \cdot \sin \frac{\pi}{3}\]

Step 2: Simplify Using Known Values

Substitute known trigonometric values: \[ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad \sin \frac{\pi}{6} = \frac{1}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}, \quad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \]Substitute into the expressions: \[\cos(x+\frac{\pi}{6}) = \cos x \cdot \frac{\sqrt{3}}{2} - \sin x \cdot \frac{1}{2}\]\[\sin(x-\frac{\pi}{3}) = \sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2}\]

Step 3: Combine and Simplify

Combine the expressions from Step 2 and set them equal to zero:\[\left(\cos x \cdot \frac{\sqrt{3}}{2} - \sin x \cdot \frac{1}{2}\right) + \left(\sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2}\right) = 0\]Simplify:\[\cos x \cdot \frac{\sqrt{3}}{2} - \sin x \cdot \frac{1}{2} + \sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2} = 0\]All terms cancel out, confirming the identity:\[0 = 0\]

Key Concepts

Angle Addition Formulas

Trigonometric Angle Addition Formulas are essential tools in simplifying and manipulating expressions involving angles. They help us break down complex trigonometric expressions into simpler components that are easier to solve or verify. For angles expressed as a sum or difference like \(a + b\) or \(a - b\), these formulas become incredibly handy.

• The cosine of an angle sum is: \(\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)\).
• The sine of an angle difference is: \(\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)\).

In the exercise, we apply these formulas to the expressions \(\cos(x + \frac{\pi}{6})\) and \(\sin(x - \frac{\pi}{3})\). By breaking these down, each part becomes a combination of basic sine and cosine functions, making the process of proving identities manageable and straightforward.

Simplifying Trigonometric Expressions

Simplifying trigonometric expressions involves replacing complex parts of an equation with simpler, equivalent values. This is often done after we've used the angle addition formulas.
Once you've expanded your trigonometric functions with the formulas, simplification involves algebraically manipulating those expanded forms. In our problem, the process was:

• Write \(\cos(x + \frac{\pi}{6})\) as \(\cos x \cdot \cos \frac{\pi}{6} - \sin x \cdot \sin \frac{\pi}{6}\)
• Write \(\sin(x - \frac{\pi}{3})\) as \(\sin x \cdot \cos \frac{\pi}{3} - \cos x \cdot \sin \frac{\pi}{3}\)

Then, substitute these back into the original identity and combine similar terms. The goal of each step is to bring the expression closer to a form that can be simplified all the way to zero or any other provided equality.

Trigonometric Values

Trigonometric values for standard angles like \(\frac{\pi}{6}\), \(\frac{\pi}{3}\), and others are fundamental in trigonometry. Knowing these allows quick substitution in expressions, a practice very useful in simplification.

• \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\)
• \(\sin \frac{\pi}{6} = \frac{1}{2}\)
• \(\cos \frac{\pi}{3} = \frac{1}{2}\)
• \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)

When substituting these values in your equations, it not only makes solving easier but also confirms the correctness of the step-by-step process. During our exercise, after substituting these known values, further simplification was performed to verify the trigonometric identity, ultimately demonstrating that the initial equation balances to zero, thereby proving the identity.