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Solve in english

ScanSolve AI Tutor · Accepted Answer

Here's the proof for the given trigonometric identity:

We need to prove that:

\frac{2\sin(2\theta) - 3\cos(2\theta) + 3}{\sin\theta} = 4\cos\theta + 6\sin\theta

Start with the Left Hand Side (LHS) of the equation.
$\text{LHS} = \frac{2\sin(2\theta) - 3\cos(2\theta) + 3}{\sin\theta}$
Apply double angle identities for $\sin(2\theta)$ and $\cos(2\theta)$ .
We use the identities:
- $\sin(2\theta) = 2\sin\theta\cos\theta$
- $\cos(2\theta) = 1 - 2\sin^2\theta$ (This form is chosen to help simplify the constant terms in the numerator).
Substitute these identities into the LHS:
$\text{LHS} = \frac{2(2\sin\theta\cos\theta) - 3(1 - 2\sin^2\theta) + 3}{\sin\theta}$
Expand and simplify the numerator.
$\text{LHS} = \frac{4\sin\theta\cos\theta - 3 + 6\sin^2\theta + 3}{\sin\theta}$
The constant terms $-3$ and $+3$ cancel each other out:
$\text{LHS} = \frac{4\sin\theta\cos\theta + 6\sin^2\theta}{\sin\theta}$
Factor out $\sin\theta$ from the numerator.
$\text{LHS} = \frac{\sin\theta(4\cos\theta + 6\sin\theta)}{\sin\theta}$
Cancel out $\sin\theta$ (assuming $\sin\theta \ne 0$ ).
$\text{LHS} = 4\cos\theta + 6\sin\theta$
This is equal to the Right Hand Side (RHS) of the equation.
$\text{LHS} = \text{RHS}$
Therefore, the identity is proven.
$\boxed{\frac{2\sin(2\theta) - 3\cos(2\theta) + 3}{\sin\theta} = 4\cos\theta + 6\sin\theta}$

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