5.1 Prove the identity: $$ \frac{1-\sin 2 \mathrm{~A}}{\sin \mathrm{~A}-\cos \mathrm{A}}=\sin \mathrm{A}-\cos \mathrm{A} $$ 5.2 Prove that: $$ \cos \left(30^{\circ}+ heta ight)-\cos \left(30^{\circ}- heta ight)=-\sin heta $$ 5.3 Consider the Geometric series: $$ \cos \alpha+\sin 2 \alpha+4 \sin ^{2} \alpha \cdot \cos \alpha+\ldots $$, where $$ \alpha $$ is an acute angle. Determine the value of $$ r $$, the constant ratio of the series.

Question

5.1 Prove the identity: $$ \frac{1-\sin 2 \mathrm{~A}}{\sin \mathrm{~A}-\cos \mathrm{A}}=\sin \mathrm{A}-\cos \mathrm{A} $$ 5.2 Prove that: $$ \cos \left(30^{\circ}+	hetaight)-\cos \left(30^{\circ}-	hetaight)=-\sin 	heta $$ 5.3 Consider the Geometric series: $$ \cos \alpha+\sin 2 \alpha+4 \sin ^{2} \alpha \cdot \cos \alpha+\ldots $$, where $$ \alpha $$ is an acute angle. Determine the value of $$ r $$, the constant ratio of the series.

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**5.1**  
We start with the left-hand side of the identity:  
$$
\frac{1-\sin 2A}{\sin A-\cos A}
$$  
Recall that  
$$
\sin 2A = 2 \sin A \cos A.
$$  
Thus,  
$$
1-\sin 2A = 1-2\sin A\cos A.
$$  
Notice that using the Pythagorean identity,  
$$
\sin^2 A + \cos^2 A = 1,
$$  
we can write:  
$$
1-2\sin A\cos A = \sin^2 A + \cos^2 A - 2\sin A\cos A = (\sin A - \cos A)^2.
$$  
Then, the expression becomes:  
$$
\frac{(\sin A-\cos A)^2}{\sin A-\cos A}.
$$  
For $$\sin A \neq \cos A$$, we cancel a factor of $$\sin A-\cos A$$ to obtain:  
$$
\sin A-\cos A.
$$  
Thus, the identity is proven.

---

**5.2**  
We need to show:  
$$
\cos\left(30^\circ+\theta\right)-\cos\left(30^\circ-\theta\right)=-\sin\theta.
$$  
We use the cosine subtraction formula:  
$$
\cos (\alpha+\beta) - \cos (\alpha-\beta) = -2\sin \alpha \sin \beta.
$$  
Let $$\alpha=30^\circ$$ and $$\beta=\theta$$. Then:  
$$
\cos\left(30^\circ+\theta\right)-\cos\left(30^\circ-\theta\right)=-2\sin 30^\circ \sin\theta.
$$  
Since $$\sin 30^\circ=\frac{1}{2}$$, we have:  
$$
-2\left(\frac{1}{2}\right)\sin\theta=-\sin\theta.
$$  
Thus, the identity is proven.

---

**5.3**  
We are given the geometric series:  
$$
\cos \alpha+\sin 2\alpha+4\sin^2 \alpha\cdot\cos \alpha+\ldots
$$  
Let the first term be $$a=\cos \alpha$$ and the common ratio be $$r$$. The second term then must equal $$a \cdot r$$. Hence,  
$$
\sin 2\alpha =\cos \alpha\cdot r \quad\Longrightarrow\quad r=\frac{\sin2\alpha}{\cos\alpha}.
$$
  
To verify that this $$r$$ works for the third term as well, notice that the third term is given as $$4\sin^2 \alpha\cdot\cos \alpha$$. In a geometric series, the third term is also $$a\cdot r^2$$. Thus, we require:  
$$
\cos\alpha\cdot r^2 = 4\sin^2\alpha\cdot\cos\alpha.
$$
  
Dividing both sides by $$\cos \alpha$$ (assuming $$\cos\alpha\neq 0$$), we have:  
$$
r^2 = 4\sin^2\alpha.
$$
  
Now, substitute the expression for $$r$$ found earlier:  
$$
\left(\frac{\sin2\alpha}{\cos\alpha}\right)^2 = 4\sin^2\alpha.
$$
  
Recall that  
$$
\sin2\alpha=2\sin\alpha\cos\alpha,
$$  
so  
$$
\left(\frac{2\sin\alpha\cos\alpha}{\cos\alpha}\right)^2 = \left(2\sin\alpha\right)^2 = 4\sin^2\alpha.
$$
  
This confirms the value of $$r$$. Therefore,  
$$
r=\frac{\sin2\alpha}{\cos\alpha}=2\sin\alpha.
$$
**5.1** The identity is proven by simplifying the left-hand side to equal the right-hand side. **5.2** The identity is proven using the cosine subtraction formula. **5.3** The common ratio $$ r $$ of the geometric series is $$ 2\sin \alpha $$.

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