Smplify: a) $$ \frac{\sin \left(360^{\circ}- heta ight) \cdot \cos \left(180^{\circ}+ heta ight)}{\sin 270^{\circ} \cdot \cos \left(360^{\circ}- heta ight) \cdot \sin \left(90^{\circ}+ heta ight)} $$ b) $$ \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} $$

Question

Smplify: a) $$ \frac{\sin \left(360^{\circ}-	hetaight) \cdot \cos \left(180^{\circ}+	hetaight)}{\sin 270^{\circ} \cdot \cos \left(360^{\circ}-	hetaight) \cdot \sin \left(90^{\circ}+	hetaight)} $$ b) $$ \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} $$

UpStudy AI · Accepted Answer

**a)**

We start with
$$
\frac{\sin \left(360^\circ-\theta\right) \cdot \cos \left(180^\circ+\theta\right)}{\sin 270^\circ \cdot \cos \left(360^\circ-\theta\right) \cdot \sin \left(90^\circ+\theta\right)}.
$$

1. Note that
   $$
   \sin \left(360^\circ - \theta\right) = \sin(-\theta) = -\sin \theta.
   $$
2. Also,
   $$
   \cos \left(180^\circ + \theta\right) = -\cos \theta.
   $$
3. In the denominator, we have
   $$
   \sin 270^\circ = -1.
   $$
4. Next,
   $$
   \cos \left(360^\circ - \theta\right) = \cos \theta.
   $$
5. And,
   $$
   \sin \left(90^\circ + \theta\right) = \cos \theta.
   $$

Substitute these into the expression:

- The numerator becomes:
  $$
  (-\sin\theta)(-\cos\theta) = \sin\theta\,\cos\theta.
  $$
- The denominator becomes:
  $$
  (-1) \cdot \cos\theta \cdot \cos\theta = -\cos^2\theta.
  $$

Thus, the expression simplifies to
$$
\frac{\sin\theta\,\cos\theta}{-\cos^2\theta} = -\frac{\sin\theta}{\cos\theta} = -\tan\theta.
$$

---

**b)**

We need to simplify
$$
\sin^2 108^\circ + \sin^2 18^\circ.
$$

Recognize that
$$
\sin 108^\circ = \sin(180^\circ - 108^\circ) = \sin 72^\circ.
$$
Hence,
$$
\sin^2 108^\circ + \sin^2 18^\circ = \sin^2 72^\circ + \sin^2 18^\circ.
$$

We use the identity
$$
\sin^2 \theta = \frac{1 - \cos 2\theta}{2}.
$$

Apply the formula to each term:

- For $$\theta = 108^\circ$$:
  $$
  \sin^2 108^\circ = \frac{1 - \cos 216^\circ}{2}.
  $$
- For $$\theta = 18^\circ$$:
  $$
  \sin^2 18^\circ = \frac{1 - \cos 36^\circ}{2}.
  $$

Thus, their sum is
$$
\frac{1 - \cos 216^\circ}{2} + \frac{1 - \cos 36^\circ}{2} = \frac{2 - \left(\cos 216^\circ+\cos 36^\circ\right)}{2}.
$$

Now, note that
$$
\cos 216^\circ = \cos(180^\circ+36^\circ) = -\cos 36^\circ.
$$
So,
$$
\cos 216^\circ+\cos 36^\circ = -\cos 36^\circ + \cos 36^\circ = 0.
$$

Therefore, the sum becomes
$$
\frac{2}{2} = 1.
$$

---

**Final Answers:**

a) $$ -\tan \theta $$

b) $$ 1 $$
a) $$ -\tan \theta $$ b) $$ 1 $$

Smplify: a) \( \frac{\sin \left(360^{\circ}-\theta\right) \cdot \cos \left(180^{\circ}+\theta\right)}{\sin 270^{\circ} \cdot \cos \left(360^{\circ}-\theta\right) \cdot \sin \left(90^{\circ}+\theta\right)} \) b) \( \sin ^{2} 108^{\circ}+\sin ^{2} 18^{\circ} \)

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Trigonometry Questions