We read your photo as:

Simplify the expression ( theta + 2theta)/(1 + theta + 2theta).

Mathematics

ScanSolve AI Answer

Answer

This step focuses on simplifying the numerator of the expression sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta}.

Step 2: Simplify the numerator of the LHS.
The numerator is the top part of the fraction, which is sin⁡θ+sin⁡2θ\sin \theta + \sin 2\theta.

To simplify this, we use a double angle identity for sine. The identity states that sin⁡2θ\sin 2\theta can be rewritten as 2sin⁡θcos⁡θ2 \sin \theta \cos \theta. This substitution is crucial because it allows us to express both terms in the numerator using sin⁡θ\sin \theta and cos⁡θ\cos \theta of a single angle θ\theta, rather than having terms with 2θ2\theta.

Substituting sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta into the numerator gives:

sin⁡θ+sin⁡2θ=sin⁡θ+2sin⁡θcos⁡θ\sin \theta + \sin 2\theta = \sin \theta + 2 \sin \theta \cos \theta

Now, we observe that both terms on the right-hand side, sin⁡θ\sin \theta and 2sin⁡θcos⁡θ2 \sin \theta \cos \theta, share a common factor of sin⁡θ\sin \theta.

To simplify further, we factor out this common term sin⁡θ\sin \theta:

=sin⁡θ(1+2cos⁡θ)= \sin \theta (1 + 2 \cos \theta)

When sin⁡θ\sin \theta is factored out from sin⁡θ\sin \theta, it leaves 1 (since sin⁡θ=sin⁡θ⋅1\sin \theta = \sin \theta \cdot 1). When sin⁡θ\sin \theta is factored out from 2sin⁡θcos⁡θ2 \sin \theta \cos \theta, it leaves 2cos⁡θ2 \cos \theta. This results in the simplified form of the numerator: sin⁡θ(1+2cos⁡θ)\sin \theta (1 + 2 \cos \theta). This factored form is useful because it often allows for cancellation with terms in the denominator, as seen in the full solution.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question