Question

PT $\frac{Sin 7\theta - Sin 5\theta}{Cos 7\theta + Cos 5\theta}$ = Tan$\theta$

Answer 1

Proven

Answer 2

To prove the identity $\frac{\sin 7\theta - \sin 5\theta}{\cos 7\theta + \cos 5\theta} = \tan \theta$, we will use the sum-to-product trigonometric formulas.

The relevant formulas are:

1. $\sin A - \sin B = 2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$
2. $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$

Let's apply these to the Left Hand Side (LHS) of the given identity. For the numerator, $\sin 7\theta - \sin 5\theta$: Here, $A = 7\theta$ and $B = 5\theta$. $\frac{A+B}{2} = \frac{7\theta + 5\theta}{2} = \frac{12\theta}{2} = 6\theta$ $\frac{A-B}{2} = \frac{7\theta - 5\theta}{2} = \frac{2\theta}{2} = \theta$ So, the numerator becomes:

$$\sin 7\theta - \sin 5\theta = 2 \cos 6\theta \sin \theta$$

For the denominator, $\cos 7\theta + \cos 5\theta$: Here, $A = 7\theta$ and $B = 5\theta$. $\frac{A+B}{2} = \frac{7\theta + 5\theta}{2} = \frac{12\theta}{2} = 6\theta$ $\frac{A-B}{2} = \frac{7\theta - 5\theta}{2} = \frac{2\theta}{2} = \theta$ So, the denominator becomes:

$$\cos 7\theta + \cos 5\theta = 2 \cos 6\theta \cos \theta$$

Now, substitute these expressions back into the LHS:

$$\text{LHS} = \frac{2 \cos 6\theta \sin \theta}{2 \cos 6\theta \cos \theta}$$

Assuming $\cos 6\theta \neq 0$, we can cancel out the common terms ($2 \cos 6\theta$):

$$\text{LHS} = \frac{\sin \theta}{\cos \theta}$$

We know that $\frac{\sin \theta}{\cos \theta} = \tan \theta$.

$$\text{LHS} = \tan \theta$$

This is equal to the Right Hand Side (RHS) of the given identity. Hence, the identity is proven.

Explanation of the solution: The identity is proven by applying sum-to-product formulas for sine and cosine to the numerator and denominator respectively. The common term $2 \cos 6\theta$ cancels out, simplifying the expression to $\frac{\sin \theta}{\cos \theta}$, which is $\tan \theta$.