2. Calculate without using a calculator. a) $$ \cos 30^{\circ} imes \sin 60^{\circ} $$ b) $$ \sec ^{2} 45^{\circ}- an ^{2} 45^{\circ} $$ c) $$ \sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ} $$ d) $$ 2 \cos ^{2} 45^{\circ}-1 $$ e) $$ \operatorname{cosec} 90^{\circ} imes \sin 90^{\circ} $$ f) $$ \cot ^{2} 45^{\circ}+ an ^{2} 60^{\circ}-\sin ^{2} 0^{\circ}-\cos ^{2} 0^{\circ} $$

Question

2. Calculate without using a calculator. a) $$ \cos 30^{\circ} 	imes \sin 60^{\circ} $$ b) $$ \sec ^{2} 45^{\circ}-	an ^{2} 45^{\circ} $$ c) $$ \sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ} $$ d) $$ 2 \cos ^{2} 45^{\circ}-1 $$ e) $$ \operatorname{cosec} 90^{\circ} 	imes \sin 90^{\circ} $$ f) $$ \cot ^{2} 45^{\circ}+	an ^{2} 60^{\circ}-\sin ^{2} 0^{\circ}-\cos ^{2} 0^{\circ} $$

UpStudy AI · Accepted Answer

Alright, I need to calculate these trigonometric expressions without using a calculator. Let's tackle them one by one.

**a) $$ \cos 30^{\circ} \times \sin 60^{\circ} $$**

First, I recall that $$ \cos 30^{\circ} = \frac{\sqrt{3}}{2} $$ and $$ \sin 60^{\circ} = \frac{\sqrt{3}}{2} $$. Multiplying these together:

$$
\cos 30^{\circ} \times \sin 60^{\circ} = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} = \frac{3}{4}
$$

So, the result is $$ \frac{3}{4} $$.

**b) $$ \sec ^{2} 45^{\circ}-\tan ^{2} 45^{\circ} $$**

I know that $$ \sec \theta = \frac{1}{\cos \theta} $$ and $$ \tan \theta = \frac{\sin \theta}{\cos \theta} $$. For $$ 45^{\circ} $$, $$ \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} $$.

Calculating $$ \sec 45^{\circ} $$:

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

Then, $$ \sec^2 45^{\circ} = (\sqrt{2})^2 = 2 $$.

Next, $$ \tan 45^{\circ} = \frac{\sin 45^{\circ}}{\cos 45^{\circ}} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1 $$.

So, $$ \tan^2 45^{\circ} = 1^2 = 1 $$.

Subtracting these:

$$
\sec^2 45^{\circ} - \tan^2 45^{\circ} = 2 - 1 = 1
$$

The result is $$ 1 $$.

**c) $$ \sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ} $$**

Using the values I know:

$$
\sin 60^{\circ} = \frac{\sqrt{3}}{2}, \quad \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \quad \cos 60^{\circ} = \frac{1}{2}, \quad \sin 30^{\circ} = \frac{1}{2}
$$

Plugging these in:

$$
\sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ} = \left( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \times \frac{1}{2} \right) = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
$$

So, the result is $$ \frac{1}{2} $$.

**d) $$ 2 \cos ^{2} 45^{\circ}-1 $$**

First, $$ \cos 45^{\circ} = \frac{\sqrt{2}}{2} $$, so:

$$
\cos^2 45^{\circ} = \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{2}{4} = \frac{1}{2}
$$

Multiplying by 2:

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

Subtracting 1:

$$
2 \cos^2 45^{\circ} - 1 = 1 - 1 = 0
$$

The result is $$ 0 $$.

**e) $$ \operatorname{cosec} 90^{\circ} \times \sin 90^{\circ} $$**

I recall that $$ \operatorname{cosec} \theta = \frac{1}{\sin \theta} $$. For $$ 90^{\circ} $$, $$ \sin 90^{\circ} = 1 $$, so:

$$
\operatorname{cosec} 90^{\circ} = \frac{1}{1} = 1
$$

Multiplying by $$ \sin 90^{\circ} $$:

$$
\operatorname{cosec} 90^{\circ} \times \sin 90^{\circ} = 1 \times 1 = 1
$$

The result is $$ 1 $$.

**f) $$ \cot ^{2} 45^{\circ}+\tan ^{2} 60^{\circ}-\sin ^{2} 0^{\circ}-\cos ^{2} 0^{\circ} $$**

First, let's find each trigonometric value:

- $$ \cot 45^{\circ} = \frac{1}{\tan 45^{\circ}} = 1 $$, so $$ \cot^2 45^{\circ} = 1^2 = 1 $$
- $$ \tan 60^{\circ} = \sqrt{3} $$, so $$ \tan^2 60^{\circ} = (\sqrt{3})^2 = 3 $$
- $$ \sin 0^{\circ} = 0 $$, so $$ \sin^2 0^{\circ} = 0^2 = 0 $$
- $$ \cos 0^{\circ} = 1 $$, so $$ \cos^2 0^{\circ} = 1^2 = 1 $$

Now, plugging these into the expression:

$$
\cot^2 45^{\circ} + \tan^2 60^{\circ} - \sin^2 0^{\circ} - \cos^2 0^{\circ} = 1 + 3 - 0 - 1 = 3
$$

The result is $$ 3 $$.

a) $$ \frac{3}{4} $$ b) $$ 1 $$ c) $$ \frac{1}{2} $$ d) $$ 0 $$ e) $$ 1 $$ f) $$ 3 $$

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