6. Solve: a) $$ \sin heta \cdot \cos 30^{\circ}+\cos heta \cdot \sin 30^{\circ}=0,5 $$ b) $$ \sin heta \cdot \sin 20^{\circ}+\cos heta \cdot \cos 20^{\circ}=\sin 50^{\circ} $$ Find $$ heta \in\left[-180^{\circ} ; 180^{\circ} ight] \operatorname{such} $$ that: a) $$ \sin 5 heta \cdot \cos 20^{\circ}-\cos 5 heta \cdot \sin 20^{\circ}=1 $$ b) $$ 2 \cos 3 heta \cdot \cos 30^{\circ}-2 \sin 3 heta \cdot \sin 30^{\circ}=1 $$

Question

6. Solve: a) $$ \sin 	heta \cdot \cos 30^{\circ}+\cos 	heta \cdot \sin 30^{\circ}=0,5 $$ b) $$ \sin 	heta \cdot \sin 20^{\circ}+\cos 	heta \cdot \cos 20^{\circ}=\sin 50^{\circ} $$ Find $$ 	heta \in\left[-180^{\circ} ; 180^{\circ}ight] \operatorname{such} $$ that: a) $$ \sin 5 	heta \cdot \cos 20^{\circ}-\cos 5 	heta \cdot \sin 20^{\circ}=1 $$ b) $$ 2 \cos 3 	heta \cdot \cos 30^{\circ}-2 \sin 3 	heta \cdot \sin 30^{\circ}=1 $$

UpStudy AI · Accepted Answer

We first note that both parts consist of rewriting the given expressions by using trigonometric addition formulas.

---

### 1(a)

We are to solve
$$
\sin \theta \cdot \cos 30^{\circ} + \cos \theta \cdot \sin 30^{\circ} = 0.5.
$$
Using the sine addition formula
$$
\sin(\theta + 30^{\circ}) = \sin \theta \cos 30^{\circ} + \cos \theta \sin 30^{\circ},
$$
the equation becomes
$$
\sin (\theta + 30^{\circ}) = 0.5.
$$

Since $$\sin \alpha = 0.5$$ when
$$
\alpha = 30^{\circ} + 360^{\circ} k\quad \text{or}\quad \alpha = 150^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z},
$$
we set
$$
\theta + 30^{\circ} = 30^{\circ} + 360^{\circ} k\quad \text{or}\quad \theta + 30^{\circ} = 150^{\circ} + 360^{\circ} k.
$$
Solving these:

1. For $$\theta + 30^{\circ} = 30^{\circ} + 360^{\circ} k$$:
   $$
   \theta = 360^{\circ} k.
   $$

2. For $$\theta + 30^{\circ} = 150^{\circ} + 360^{\circ} k$$:
   $$
   \theta = 120^{\circ} + 360^{\circ} k.
   $$

Thus, the solutions are
$$
\theta = 360^{\circ} k \quad \text{or} \quad \theta = 120^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}.
$$

---

### 1(b)

We now solve
$$
\sin \theta \cdot \sin 20^{\circ} + \cos \theta \cdot \cos 20^{\circ}=\sin 50^{\circ}.
$$
Note that the cosine addition formula is
$$
\cos(\theta - 20^{\circ}) = \cos \theta \cos 20^{\circ} + \sin \theta \sin 20^{\circ}.
$$
Thus, the left-hand side becomes
$$
\cos(\theta - 20^{\circ}).
$$
So the equation is
$$
\cos(\theta - 20^{\circ}) = \sin 50^{\circ}.
$$

Recognize that $$\sin50^{\circ}=\cos40^{\circ}$$. Therefore, the equation is
$$
\cos(\theta - 20^{\circ}) = \cos 40^{\circ}.
$$

For $$\cos \alpha = \cos \beta$$, the general solution is
$$
\alpha = \beta + 360^{\circ} k \quad \text{or} \quad \alpha = -\beta + 360^{\circ} k,\quad k \in\mathbb{Z}.
$$
Thus, setting $$\alpha = \theta-20^{\circ}$$ and $$\beta = 40^{\circ}$$:

1. $$\theta - 20^{\circ} = 40^{\circ} + 360^{\circ} k$$ gives
   $$
   \theta = 60^{\circ} + 360^{\circ} k.
   $$

2. $$\theta - 20^{\circ} = -40^{\circ} + 360^{\circ} k$$ gives
   $$
   \theta = -20^{\circ} + 360^{\circ} k.
   $$

Thus, the solutions are
$$
\theta = 60^{\circ} + 360^{\circ} k\quad \text{or}\quad \theta = -20^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}.
$$

---

Now we turn to the next two problems where we also require that $$\theta \in [-180^{\circ}, 180^{\circ}]$$.

---

### 2(a)

We wish to find $$\theta \in [-180^{\circ}, 180^{\circ}]$$ such that
$$
\sin 5\theta \cdot \cos 20^{\circ} - \cos 5\theta \cdot \sin 20^{\circ} = 1.
$$
Recognize that
$$
\sin 5\theta \cos 20^{\circ} - \cos 5\theta \sin 20^{\circ} = \sin(5\theta - 20^{\circ})
$$
by the sine subtraction formula. Thus, the equation becomes
$$
\sin(5\theta - 20^{\circ}) = 1.
$$

We know that $$\sin \alpha = 1$$ when
$$
\alpha = 90^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z}.
$$
So set
$$
5\theta - 20^{\circ}= 90^{\circ} + 360^{\circ} k.
$$
Solving for $$\theta$$:
$$
5\theta = 110^{\circ} + 360^{\circ} k,
$$
$$
\theta = 22^{\circ} + 72^{\circ} k.
$$

We must choose integers $$k$$ so that $$\theta$$ lies in $$[-180^{\circ}, 180^{\circ}]$$:

• For $$k = -2$$: $$\theta = 22 - 144 = -122^{\circ}$$.  
• For $$k = -1$$: $$\theta = 22 - 72 = -50^{\circ}$$.  
• For $$k = 0$$: $$\theta = 22^{\circ}$$.  
• For $$k = 1$$: $$\theta = 22 + 72 = 94^{\circ}$$.  
• For $$k = 2$$: $$\theta = 22 + 144 = 166^{\circ}$$.

Checking $$k = -3$$ gives $$-194^{\circ}$$ (which is less than $$-180^{\circ}$$) and $$k = 3$$ gives $$238^{\circ}$$ (which is larger than $$180^{\circ}$$).

Thus, the solutions are
$$
\theta = -122^{\circ},\quad -50^{\circ},\quad 22^{\circ},\quad 94^{\circ},\quad 166^{\circ}.
$$

---

### 2(b)

We now need to find $$\theta \in [-180^{\circ}, 180^{\circ}]$$ such that
$$
2 \cos 3\theta \cdot \cos 30^{\circ} - 2 \sin 3\theta \cdot \sin 30^{\circ} = 1.
$$

Factor out the 2:
$$
2 \left(\cos 3\theta \cdot \cos 30^{\circ} - \sin 3\theta \cdot \sin 30^{\circ}\right) = 1.
$$
Recall that
$$
\cos 3\theta \cos 30^{\circ} - \sin 3\theta \sin 30^{\circ} = \cos(3\theta + 30^{\circ}).
$$
Thus, the equation becomes
$$
2 \cos(3\theta + 30^{\circ}) = 1,
$$
or
$$
\cos(3\theta + 30^{\circ}) = \frac{1}{2}.
$$

Since $$\cos \alpha = \frac{1}{2}$$ when
$$
\alpha = 60^{\circ} + 360^{\circ} k \quad \text{or} \quad \alpha = 300^{\circ} + 360^{\circ} k,\quad k\in\mathbb{Z},
$$
we set

1. $$3\theta + 30^{\circ} = 60^{\circ} + 360^{\circ} k$$:
   $$
   3\theta = 30^{\circ} + 360^{\circ} k,\quad \theta = 10^{\circ} + 120^{\circ} k.
   $$

2. $$3\theta + 30^{\circ} = 300^{\circ} + 360^{\circ} k$$:
   $$
   3\theta = 270^{\circ} + 360^{\circ} k,\quad \theta = 90^{\circ} + 120^{\circ} k.
   $$

Now choose the values such that $$\theta \in [-180^{\circ}, 180^{\circ}]$$.

For $$\theta = 
Solutions for the equations are: 1(a) $$ \theta = 360^{\circ} k $$ or $$ \theta = 120^{\circ} + 360^{\circ} k $$, where $$ k $$ is an integer. 1(b) $$ \theta = 60^{\circ} + 360^{\circ} k $$ or $$ \theta = -20^{\circ} + 360^{\circ} k $$, where $$ k $$ is an integer. 2(a) $$ \theta = -122^{\circ}, -50^{\circ}, 22^{\circ}, 94^{\circ}, 166^{\circ} $$. 2(b) $$ \theta = 10^{\circ} + 120^{\circ} k $$ or $$ \theta = 90^{\circ} + 120^{\circ} k $$, where $$ k $$ is an integer.

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