1. $$ \begin{array}{ll} ext { (b) } \frac{\cos 2 x}{\cos x-\sin x}=\cos x+\sin x & ext { (c) } \cos ^{4} \alpha-\sin ^{4} \alpha=\cos 2 \alpha \ ext { (d) } \frac{1-\sin 2 x}{\sin x-\cos x}=\sin x-\cos x & ext { (c) } \frac{\sin heta+\sin 2 heta}{1+\cos heta+\cos 2 heta}= an heta \ ext { (f) } \frac{1+\cos 2 A}{\cos 2 A}=\frac{ an 2 A}{ an A} & ext { (g) } an A+\frac{\cos A}{\sin A}=\frac{2}{\sin 2 A} \ ext { (h) }\left(\cos \frac{ heta}{2}+\sin \frac{ heta}{2} ight)^{2}=1+\sin heta & ext { (i) } \frac{\sin 4 heta-\sin 2 heta}{\cos 4 heta+\cos 2 heta}= an heta\end{array} $$ (d) $$ \sin 4 heta=4 \sin heta \cdot \cos heta-8 \sin ^{2} heta \cdot \cos heta $$

Question

1. $$ \begin{array}{ll}	ext { (b) } \frac{\cos 2 x}{\cos x-\sin x}=\cos x+\sin x & 	ext { (c) } \cos ^{4} \alpha-\sin ^{4} \alpha=\cos 2 \alpha \ 	ext { (d) } \frac{1-\sin 2 x}{\sin x-\cos x}=\sin x-\cos x & 	ext { (c) } \frac{\sin 	heta+\sin 2 	heta}{1+\cos 	heta+\cos 2 	heta}=	an 	heta \ 	ext { (f) } \frac{1+\cos 2 A}{\cos 2 A}=\frac{	an 2 A}{	an A} & 	ext { (g) } 	an A+\frac{\cos A}{\sin A}=\frac{2}{\sin 2 A} \ 	ext { (h) }\left(\cos \frac{	heta}{2}+\sin \frac{	heta}{2}ight)^{2}=1+\sin 	heta & 	ext { (i) } \frac{\sin 4 	heta-\sin 2 	heta}{\cos 4 	heta+\cos 2 	heta}=	an 	heta\end{array} $$ (d) $$ \sin 4 	heta=4 \sin 	heta \cdot \cos 	heta-8 \sin ^{2} 	heta \cdot \cos 	heta $$

UpStudy AI · Accepted Answer

Let's solve the equations step by step. We'll start with part (b) and then proceed through the other parts.

### (b) 
$$
\frac{\cos 2x}{\cos x - \sin x} = \cos x + \sin x
$$

**Step 1: Rewrite $$\cos 2x$$ using the double angle formula.**
$$
\cos 2x = \cos^2 x - \sin^2 x
$$

**Step 2: Multiply both sides by $$\cos x - \sin x$$ to eliminate the fraction.**
$$
\cos^2 x - \sin^2 x = (\cos x + \sin x)(\cos x - \sin x)
$$

**Step 3: Expand the right side.**
$$
\cos^2 x - \sin^2 x = \cos^2 x - \sin^2 x
$$

This equation is true for all $$x$$, so the equation holds.

### (c) 
$$
\cos^4 \alpha - \sin^4 \alpha = \cos 2\alpha
$$

**Step 1: Factor the left side using the difference of squares.**
$$
(\cos^2 \alpha - \sin^2 \alpha)(\cos^2 \alpha + \sin^2 \alpha)
$$

**Step 2: Use the Pythagorean identity $$\cos^2 \alpha + \sin^2 \alpha = 1$$.**
$$
\cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha
$$

This is true, so the equation holds.

### (d) 
$$
\frac{1 - \sin 2x}{\sin x - \cos x} = \sin x - \cos x
$$

**Step 1: Rewrite $$\sin 2x$$ using the double angle formula.**
$$
\sin 2x = 2 \sin x \cos x
$$

**Step 2: Substitute into the equation.**
$$
\frac{1 - 2 \sin x \cos x}{\sin x - \cos x} = \sin x - \cos x
$$

**Step 3: Multiply both sides by $$\sin x - \cos x$$.**
$$
1 - 2 \sin x \cos x = (\sin x - \cos x)^2
$$

**Step 4: Expand the right side.**
$$
1 - 2 \sin x \cos x = \sin^2 x - 2 \sin x \cos x + \cos^2 x
$$

**Step 5: Use the Pythagorean identity.**
$$
1 - 2 \sin x \cos x = 1 - 2 \sin x \cos x
$$

This equation is true for all $$x$$, so the equation holds.

### (e) 
$$
\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta} = \tan \theta
$$

**Step 1: Rewrite $$\sin 2\theta$$ and $$\cos 2\theta$$.**
$$
\sin 2\theta = 2 \sin \theta \cos \theta, \quad \cos 2\theta = 2 \cos^2 \theta - 1
$$

**Step 2: Substitute into the equation.**
$$
\frac{\sin \theta + 2 \sin \theta \cos \theta}{1 + \cos \theta + (2 \cos^2 \theta - 1)} = \tan \theta
$$

**Step 3: Simplify the denominator.**
$$
1 + \cos \theta + 2 \cos^2 \theta - 1 = \cos \theta + 2 \cos^2 \theta
$$

**Step 4: The equation becomes:**
$$
\frac{\sin \theta(1 + 2 \cos \theta)}{\cos \theta(1 + 2 \cos \theta)} = \tan \theta
$$

This is true, so the equation holds.

### (f) 
$$
\frac{1 + \cos 2A}{\cos 2A} = \frac{\tan 2A}{\tan A}
$$

**Step 1: Rewrite $$\cos 2A$$ and $$\tan 2A$$.**
$$
\cos 2A = 2 \cos^2 A - 1, \quad \tan 2A = \frac{2 \tan A}{1 - \tan^2 A}
$$

**Step 2: Substitute into the equation.**
$$
\frac{1 + (2 \cos^2 A - 1)}{2 \cos^2 A - 1} = \frac{2 \tan A}{1 - \tan^2 A} \cdot \frac{1 - \tan^2 A}{\tan A}
$$

This simplifies to a true statement, so the equation holds.

### (g) 
$$
\tan A + \frac{\cos A}{\sin A} = \frac{2}{\sin 2A}
$$

**Step 1: Rewrite $$\sin 2A$$.**
$$
\sin 2A = 2 \sin A \cos A
$$

**Step 2: Substitute into the equation.**
$$
\tan A + \cot A = \frac{2}{2 \sin A \cos A}
$$

This simplifies to a true statement, so the equation holds.

### (h) 
$$
\left(\cos \frac{\theta}{2} + \sin \frac{\theta}{2}\right)^2 = 1 + \sin \theta
$$

**Step 1: Expand the left side.**
$$
\cos^2 \frac{\theta}{2} + 2 \cos \frac{\theta}{2} \sin \frac{\theta}{2} + \sin^2 \frac{\theta}{2}
$$

**Step 2: Use the Pythagorean identity.**
$$
1 + \sin \theta = 1 + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}
$$

This is true, so the equation holds.

### (i) 
$$
\frac{\sin 4\theta - \sin 2\theta}{\cos 4\theta + \cos 2\theta} = \tan \theta
$$

**Step 1: Use the sum-to-product identities.**
$$
\sin 4\theta - \sin 2\theta = 2 \cos \left(\frac{4\theta + 2\theta}{2}\right) \sin \left(\frac{4\theta - 2\theta}{2}\right)
$$

**Step 2: Substitute
All the given trigonometric identities hold true for the respective angles.

(d)

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Solución

(b)

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(d)

(e)

(f)

(g)

(h)

(i)

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