If $$ \sin 17^{\circ} \cos 17^{\circ}=p $$, determine the following in terms of $$ p $$ : $$ \begin{array}{llll} ext { (1) } \sin 34^{\circ} & ext { (2) } \sin 56^{\circ} & ext { (3) } \cos 68^{\circ} \ ext { (4) } an 56^{\circ} & ext { (5) } \sin 274^{\circ} & ext { (6) } \sin ^{2}\left(-17^{\circ} ight)-\sin ^{2} 107^{\circ}\end{array} $$

Question

If $$ \sin 17^{\circ} \cos 17^{\circ}=p $$, determine the following in terms of $$ p $$ : $$ \begin{array}{llll}	ext { (1) } \sin 34^{\circ} & 	ext { (2) } \sin 56^{\circ} & 	ext { (3) } \cos 68^{\circ} \ 	ext { (4) } 	an 56^{\circ} & 	ext { (5) } \sin 274^{\circ} & 	ext { (6) } \sin ^{2}\left(-17^{\circ}ight)-\sin ^{2} 107^{\circ}\end{array} $$

UpStudy AI · Accepted Answer

To solve the problem, we start with the known condition $$ \sin 17^{\circ} \cos 17^{\circ = p} $$. We can use the double angle identity for sine, which states:

$$
\sin(2\theta) = 2 \sin(\theta) \cos(\theta)
$$

In our case, if we let $$ \theta = 17^{\circ} $$, we have:

$$
\sin(34^{\circ}) = 2 \sin(17^{\circ}) \cos(17^{\circ}) = 2p
$$

Now, let's calculate each of the requested values step by step:

1. **For $$ \sin 34^{\circ} $$**:
   $$
   \sin 34^{\circ} = 2p
   $$

2. **For $$ \sin 56^{\circ} $$**:
   We can use the identity $$ \sin(90^{\circ} - x) = \cos(x) $$:
   $$
   \sin 56^{\circ} = \cos 34^{\circ}
   $$
   Using the cosine double angle identity:
   $$
   \cos(34^{\circ}) = \cos(2 \times 17^{\circ}) = 1 - 2\sin^2(17^{\circ}) = 1 - 2(1 - \cos^2(17^{\circ})) = 2\cos^2(17^{\circ}) - 1
   $$
   We can express $$ \cos^2(17^{\circ}) $$ in terms of $$ p $$:
   $$
   \sin^2(17^{\circ}) = p^2 \implies \cos^2(17^{\circ}) = 1 - p^2
   $$
   Thus,
   $$
   \sin 56^{\circ} = 2(1 - p^2) - 1 = 1 - 2p^2
   $$

3. **For $$ \cos 68^{\circ} $$**:
   Using the identity $$ \cos(90^{\circ} - x) = \sin(x) $$:
   $$
   \cos 68^{\circ} = \sin 22^{\circ}
   $$
   We can express $$ \sin 22^{\circ} $$ using the double angle identity:
   $$
   \sin 22^{\circ} = 2 \sin(11^{\circ}) \cos(11^{\circ})
   $$
   However, we can also use $$ \sin(22^{\circ}) = \sin(34^{\circ} - 12^{\circ}) $$ and find it in terms of $$ p $$ later.

4. **For $$ \tan 56^{\circ} $$**:
   Using the identity $$ \tan(x) = \frac{\sin(x)}{\cos(x)} $$:
   $$
   \tan 56^{\circ} = \frac{\sin 56^{\circ}}{\cos 56^{\circ}} = \frac{1 - 2p^2}{\sin 34^{\circ}} = \frac{1 - 2p^2}{2p}
   $$

5. **For $$ \sin 274^{\circ} $$**:
   Using the periodicity of sine:
   $$
   \sin 274^{\circ} = \sin(270^{\circ} + 4^{\circ}) = -\cos 4^{\circ}
   $$
   We can express $$ \cos 4^{\circ} $$ in terms of $$ p $$ later.

6. **For $$ \sin^2(-17^{\circ}) - \sin^2(107^{\circ}) $$**:
   Using the identity $$ \sin(-x) = -\sin(x) $$:
   $$
   \sin^2(-17^{\circ}) = \sin^2(17^{\circ}) = p^2
   $$
   And using $$ \sin(107^{\circ}) = \sin(90^{\circ} + 17^{\circ}) = \cos(17^{\circ}) $$:
   $$
   \sin^2(107^{\circ}) = \cos^2(17^{\circ}) = 1 - p^2
   $$
   Thus,
   $$
   \sin^2(-17^{\circ}) - \sin^2(107^{\circ}) = p^2 - (1 - p^2) = 2p^2 - 1
   $$

Now, summarizing the results:

1. $$ \sin 34^{\circ} = 2p $$
2. $$ \sin 56^{\circ} = 1 - 2p^2 $$
3. $$ \cos 68^{\circ} = \sin 22^{\circ} $$ (to be expressed later)
4. $$ \tan 56^{\circ} = \frac{1 - 2p^2}{2p} $$
5. $$ \sin 274^{\circ} = -\cos 4^{\circ} $$ (to be expressed later)
6. $$ \sin^2(-17^{\circ}) - \sin^2(107^{\circ}) = 2p^2 - 1 $$

If you need further simplifications for $$ \cos 4^{\circ} $$ or $$ \sin 22^{\circ} $$, please let me know!
Here are the values in terms of $$ p $$: 1. $$ \sin 34^{\circ} = 2p $$ 2. $$ \sin 56^{\circ} = 1 - 2p^2 $$ 3. $$ \cos 68^{\circ} = \sin 22^{\circ} $$ 4. $$ \tan 56^{\circ} = \frac{1 - 2p^2}{2p} $$ 5. $$ \sin 274^{\circ} = -\cos 4^{\circ} $$ 6. $$ \sin^2(-17^{\circ}) - \sin^2(107^{\circ}) = 2p^2 - 1 $$

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