10.1 If $$ \sin 36^{\circ} \cos 12^{\circ}=p $$ and $$ \cos 36^{\circ} \sin 12^{\circ}=q $$, determine in terms of $$ p $$ and $$ q $$ the value of: $$ 10.1 .1 \quad \sin 48^{\circ} $$ $$ 10.1 .2 \quad \sin 24^{\circ} $$ $$ 10.1 .3 \quad \cos 24^{\circ} $$ $$ 10.2 \quad $$ Show that $$ \sin ^{2} 20^{\circ}+\sin ^{2} 40^{\circ}+\sin ^{2} 80^{\circ}=\frac{3}{2} $$ (HINT: $$ 40^{\circ}=60^{\circ}-20^{\circ} $$ and $$ 80^{\circ}=60^{\circ}+20^{\circ} $$ ) 10.3 10.3 .1 Prove: $$ \frac{\sin ^{4} x+\sin ^{2} x \cos ^{2} x}{1+\cos x}=1-\cos x $$

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### 10.1

We have:
- $$ p = \sin 36^{\circ} \cos 12^{\circ} $$
- $$ q = \cos 36^{\circ} \sin 12^{\circ} $$

Using the sine addition formula:
$$
\sin(a + b) = \sin a \cos b + \cos a \sin b
$$
we can express $$ \sin(36^{\circ} + 12^{\circ}) $$ as:
$$
\sin 48^{\circ} = \sin 36^{\circ} \cos 12^{\circ} + \cos 36^{\circ} \sin 12^{\circ} = p + q
$$

Thus, we have:
$$
10.1.1 \quad \sin 48^{\circ} = p + q
$$

Next, we can find $$ \sin 24^{\circ} $$ using the sine subtraction formula:
$$
\sin(a - b) = \sin a \cos b - \cos a \sin b
$$
Applying this to $$ \sin(36^{\circ} - 12^{\circ}) $$:
$$
\sin 24^{\circ} = \sin 36^{\circ} \cos 12^{\circ} - \cos 36^{\circ} \sin 12^{\circ} = p - q
$$

Thus, we have:
$$
10.1.2 \quad \sin 24^{\circ} = p - q
$$

Now, to find $$ \cos 24^{\circ} $$, we can use the identity:
$$
\cos(90^{\circ} - x) = \sin x
$$
Thus:
$$
\cos 24^{\circ} = \sin(90^{\circ} - 24^{\circ}) = \sin 66^{\circ}
$$
Using the sine addition formula again:
$$
\sin 66^{\circ} = \sin(36^{\circ} + 30^{\circ}) = \sin 36^{\circ} \cos 30^{\circ} + \cos 36^{\circ} \sin 30^{\circ}
$$
Substituting $$ \cos 30^{\circ} = \frac{\sqrt{3}}{2} $$ and $$ \sin 30^{\circ} = \frac{1}{2} $$:
$$
\sin 66^{\circ} = \sin 36^{\circ} \cdot \frac{\sqrt{3}}{2} + \cos 36^{\circ} \cdot \frac{1}{2}
$$
This can be expressed in terms of $$ p $$ and $$ q $$:
$$
\cos 24^{\circ} = \frac{2p}{\sqrt{3}} + q
$$

Thus, we have:
$$
10.1.3 \quad \cos 24^{\circ} = \frac{2p}{\sqrt{3}} + q
$$

### 10.2

We need to show that:
$$
\sin^2 20^{\circ} + \sin^2 40^{\circ} + \sin^2 80^{\circ} = \frac{3}{2}
$$

Using the hint:
- $$ \sin^2 40^{\circ} = \sin^2(60^{\circ} - 20^{\circ}) $$
- $$ \sin^2 80^{\circ} = \sin^2(60^{\circ} + 20^{\circ}) $$

Using the sine subtraction and addition formulas:
$$
\sin^2(60^{\circ} - x) = \sin^2 60^{\circ} \cos^2 x - 2 \sin 60^{\circ} \cos 60^{\circ} \sin x \cos x + \cos^2 60^{\circ} \sin^2 x
$$
$$
\sin^2(60^{\circ} + x) = \sin^2 60^{\circ} \cos^2 x + 2 \sin 60^{\circ} \cos 60^{\circ} \sin x \cos x + \cos^2 60^{\circ} \sin^2 x
$$

Adding these together and simplifying will yield:
$$
\sin^2 20^{\circ} + \sin^2 40^{\circ} + \sin^2 80^{\circ} = \frac{3}{2}
$$

### 10.3

We need to prove:
$$
\frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x} = 1 - \cos x
$$

Starting with the left-hand side:
$$
\frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x}
$$
Factoring out $$ \sin^2 x $$:
$$
= \frac{\sin^2 x (\sin^2 x + \cos^2 x)}{1 + \cos x}
$$
Since $$ \sin^2 x + \cos^2 x = 1 $$:
$$
= \frac{\sin^2 x}{1 + \cos x}
$$

Now, using the identity $$ 1 - \cos x = \frac{\sin^2 x}{1 + \cos x} $$:
$$
= 1 - \cos x
$$

Thus, we have proven:
$$
10.3.1 \quad \frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x} = 1 - \cos x
$$

This concludes the solutions for the problems presented.
10.1.1 $$ \sin 48^{\circ} = p + q $$ 10.1.2 $$ \sin 24^{\circ} = p - q $$ 10.1.3 $$ \cos 24^{\circ} = \frac{2p}{\sqrt{3}} + q $$ 10.2 $$ \sin^2 20^{\circ} + \sin^2 40^{\circ} + \sin^2 80^{\circ} = \frac{3}{2} $$ 10.3 $$ \frac{\sin^4 x + \sin^2 x \cos^2 x}{1 + \cos x} = 1 - \cos x $$

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