$$ 2 A=2 \sin A \cdot \cos A \quad \cos 2 A=\left\{\begin{array}{l} 2 \cos ^{2} A-1 \ 1-2 \sin ^{2} A \end{array}\right. $$ RCISE 4 Evaluate without using a calculator: (a) $$ \frac{\sin 40^{\circ} \cdot \cos 40^{\circ}}{\sin 100^{\circ}} $$ (b) $$ \frac{\cos ^{2} 22,5+\sin ^{2} 22,5}{\cos ^{2} 22,5-\sin ^{2} 22,5} $$ (c) $$ 2 \sin 105^{\circ} \cdot \cos 255^{\circ} $$ (d) $$ 2 \cos 15^{\circ} \cdot \cos 75^{\circ} $$ (e) $$ 1-4 \sin ^{2} 30^{\circ} $$ (f) $$ \cos 40^{\circ}+2 \sin ^{2} 200 $$ (g) $$ \tan 135^{\circ}+2 \cos ^{2} 15^{\circ} $$ (h) $$ \tan 15^{\circ}+\frac{\cos 15^{\circ}}{\sin 15^{\circ}} $$ (i) $$ \cos ^{2} 15^{\circ}+\sin 22,5^{\circ} \cdot \cos 22,5^{\circ}-\sin ^{2} 15^{\circ} $$ Evaluate without using a calculator: $$ \frac{\sin 72^{\circ}}{\sin 24^{\circ}}-\frac{\cos 72^{\circ}}{\cos 24^{\circ}} $$ Calculate: $$ \sin 110^{\circ}+\cos 70^{\circ} \cdot \tan 190^{\circ} $$ Rewrite $$ \cos 3 \mathrm{~A} $$ in terms of $$ \cos \mathrm{A} $$ (a) Show that $$ \sin \left(45^{\circ}-\alpha\right)=\frac{\sqrt{2}(\cos \alpha-\sin \alpha)}{2} $$ (b) Hence prove that $$ \sin 2 \alpha+2 \sin ^{2}\left(45^{\circ}-\alpha\right)=1 $$

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UpStudy AI · Accepted Answer

Let's evaluate each part step by step.

### Part (a)
Evaluate $$ \frac{\sin 40^{\circ} \cdot \cos 40^{\circ}}{\sin 100^{\circ}} $$.

Using the identity $$ \sin 100^{\circ} = \sin (180^{\circ} - 80^{\circ}) = \sin 80^{\circ} $$, we can rewrite the expression:

$$
\frac{\sin 40^{\circ} \cdot \cos 40^{\circ}}{\sin 100^{\circ}} = \frac{\sin 40^{\circ} \cdot \cos 40^{\circ}}{\sin 80^{\circ}}
$$

Using the double angle identity $$ \sin 2A = 2 \sin A \cos A $$:

$$
\sin 80^{\circ} = \sin (2 \cdot 40^{\circ}) = 2 \sin 40^{\circ} \cos 40^{\circ}
$$

Thus, we have:

$$
\frac{\sin 40^{\circ} \cdot \cos 40^{\circ}}{\sin 80^{\circ}} = \frac{\sin 40^{\circ} \cdot \cos 40^{\circ}}{2 \sin 40^{\circ} \cos 40^{\circ}} = \frac{1}{2}
$$

### Part (b)
Evaluate $$ \frac{\cos^{2} 22.5^{\circ} + \sin^{2} 22.5^{\circ}}{\cos^{2} 22.5^{\circ} - \sin^{2} 22.5^{\circ}} $$.

Using the Pythagorean identity $$ \cos^{2} A + \sin^{2} A = 1 $$:

$$
\frac{1}{\cos^{2} 22.5^{\circ} - \sin^{2} 22.5^{\circ}}
$$

Using the identity $$ \cos 2A = \cos^{2} A - \sin^{2} A $$:

$$
\cos 45^{\circ} = \frac{\sqrt{2}}{2}
$$

Thus, we have:

$$
\frac{1}{\cos 45^{\circ}} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}
$$

### Part (c)
Evaluate $$ 2 \sin 105^{\circ} \cdot \cos 255^{\circ} $$.

Using the identity $$ \cos 255^{\circ} = \cos (270^{\circ} - 15^{\circ}) = \sin 15^{\circ} $$:

$$
2 \sin 105^{\circ} \cdot \sin 15^{\circ}
$$

Using $$ \sin 105^{\circ} = \sin (90^{\circ} + 15^{\circ}) = \cos 15^{\circ} $$:

$$
2 \cos 15^{\circ} \cdot \sin 15^{\circ} = \sin 30^{\circ} = \frac{1}{2}
$$

### Part (d)
Evaluate $$ 2 \cos 15^{\circ} \cdot \cos 75^{\circ} $$.

Using the identity $$ \cos 75^{\circ} = \sin 15^{\circ} $$:

$$
2 \cos 15^{\circ} \cdot \sin 15^{\circ} = \sin 30^{\circ} = \frac{1}{2}
$$

### Part (e)
Evaluate $$ 1 - 4 \sin^{2} 30^{\circ} $$.

Since $$ \sin 30^{\circ} = \frac{1}{2} $$:

$$
1 - 4 \left(\frac{1}{2}\right)^{2} = 1 - 4 \cdot \frac{1}{4} = 1 - 1 = 0
$$

### Part (f)
Evaluate $$ \cos 40^{\circ} + 2 \sin^{2} 200^{\circ} $$.

Using $$ \sin 200^{\circ} = -\sin 20^{\circ} $$:

$$
2 \sin^{2} 200^{\circ} = 2 \sin^{2} 20^{\circ}
$$

Thus, we have:

$$
\cos 40^{\circ} + 2 \sin^{2} 20^{\circ}
$$

Using $$ \sin^{2} 20^{\circ} = \frac{1 - \cos 40^{\circ}}{2} $$:

$$
\cos 40^{\circ} + 2 \cdot \frac{1 - \cos 40^{\circ}}{2} = \cos 40^{\circ} + 1 - \cos 40^{\circ} = 1
$$

### Part (g)
Evaluate $$ \tan 135^{\circ} + 2 \cos^{2} 15^{\circ} $$.

Using $$ \tan 135^{\circ} = -1 $$:

$$
-1 + 2 \cos^{2} 15^{\circ}
$$

Using $$ \cos^{2} 15^{\circ} = \frac{1 + \cos 30^{\circ}}{2} = \frac{1 + \frac{\sqrt{3}}{2}}{2} = \frac{2 + \sqrt{3}}{4} $$:

Thus, we have:

$$
-1 + 2 \cdot \frac{2 + \sqrt{3}}{4} = -1 + \frac{2 + \sqrt{3}}{2} = \frac{-2 + 2 + \sqrt{3}}{2} = \frac{\sqrt{3}}{2}
$$

### Part (h)
Evaluate $$ \tan 15^{\circ} + \frac{\cos 15^{\circ}}{\sin 15^{\circ}} $$.

Using $$ \frac{\cos 15^{\circ}}{\sin 15^{\circ}} = \cot 15^{\circ} $$:

$$
\tan 15^{\circ} + \cot 15^{\circ} = \frac{\sin^{2} 15^{\circ} + \cos^{2} 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}} = \frac{1}{\sin 15^{\circ} \cos 15^{\circ}} = \frac{2}{\sin 30^{\circ}} = 4
$$

### Part (i)
Evaluate $$ \cos^{2} 15^{\circ} + \sin 22
(a) $$ \frac{1}{2} $$ (b) $$ \sqrt{2} $$ (c) $$ \frac{1}{2} $$ (d) $$ \frac{1}{2} $$ (e) $$ 0 $$ (f) $$ 1 $$ (g) $$ \frac{\sqrt{3}}{2} $$ (h) $$ 4 $$ (i) $$ \cos^{2} 15^{\circ} + \sin 22.5^{\circ} \cdot \cos 22.5^{\circ} - \sin^{2} 15^{\circ} $$ Evaluate $$ \frac{\sin 72^{\circ}}{\sin 24^{\circ}} - \frac{\cos 72^{\circ}}{\cos 24^{\circ}} $$: $$ \frac{\sin 72^{\circ}}{\sin 24^{\circ}} - \frac{\cos 72^{\circ}}{\cos 24^{\circ}} = 2 $$ Calculate $$ \sin 110^{\circ} + \cos 70^{\circ} \cdot \tan 190^{\circ} $$: $$ \sin 110^{\circ} + \cos 70^{\circ} \cdot \tan 190^{\circ} = 1 $$ Rewrite $$ \cos 3A $$ in terms of $$ \cos A $$: $$ \cos 3A = 4 \cos^{3} A - 3 \cos A $$ (a) $$ \sin(45^{\circ} - \alpha) = \frac{\sqrt{2}(\cos \alpha - \sin \alpha)}{2} $$ (b) $$ \sin 2\alpha + 2 \sin^{2}(45^{\circ} - \alpha) = 1 $$

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