Question

Verify that each equation is an identity.​​(sin 3t + sin 2t)/(sin 3t - sin 2t ) = tan (5t/2)/(tan (t/2))

Accepted Answer

Start by recalling the sum-to-product identities for sine: \sin A + \sin B = 2 \sin \$$\frac{A+B}{2}\$$ \cos \$$\frac{A-B}{2}\$$ and \sin A - \sin B = 2 \cos \$$\frac{A+B}{2}\$$ \sin \$$\frac{A-B}{2}\. $$Apply these to the numerator and denominator of the left-hand side (LHS) expression: \$$\frac{\sin 3t + \sin 2t}{\sin 3t - \sin 2t}\. $$Using the identities, rewrite the numerator as 2 \sin \$$\frac{3t + 2t}{2}\$$ \cos \$$\frac{3t - 2t}{2}\ = 2$$ \sin \$$\frac{5t}{2}\$$ \cos \$$\frac{t}{2}\$$, and the denominator as 2 \cos \$$\frac{3t + 2t}{2}\$$ \sin \$$\frac{3t - 2t}{2}\ = 2$$ \cos \$$\frac{5t}{2}\$$ \sin \$$\frac{t}{2}\. $$Simplify the fraction by canceling the common factor 2, so the LHS becomes \$$\frac{\sin $\frac{5t}{2}\$$ \cos \$$\frac{t}{2}\$$}{\cos \$$\frac{5t}{2}\$$ \sin \$$\frac{t}{2}\$$}$. Rewrite the fraction as a product of two fractions: \$$\frac{\sin $\frac{5t}{2}\$$}{\cos \$$\frac{5t}{2}\$$} \times \frac{\cos \$$\frac{t}{2}\$$}{\sin \$$\frac{t}{2}\$$}$. Recognize that \$$\frac{\sin x}{\cos x} = \tan x\$$ and \$$\frac{\cos x}{\sin x} = \cot x\. $$Express the LHS as \$$\tan $\frac{5t}{2}\$$ \times \cot \$$\frac{t}{2}\$$$. Since \cot x = \frac{1}{\tan x}\), rewrite this as \$$\frac{\tan $\frac{5t}{2}\$$}{\tan \$$\frac{t}{2}\$$}$, which matches the right-hand side (RHS) of the original equation, thus verifying the identity.

Verify that each equation is an identity.
(sin 3t + sin 2t)/(sin 3t - sin 2t ) = tan (5t/2)/(tan (t/2))

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Key Concepts

Trigonometric Identities

Sum-to-Product Formulas

Tangent Function and Angle Relationships