Find the value of `(i) cos 840 "(ii) sin 870

Answer

To solve the problem, we need to find the values of \( \cos 840^\circ \) and \( \sin 870^\circ \).

### Step 1: Finding \( \cos 840^\circ \)

1. **Reduce the angle**:
We can reduce \( 840^\circ \) by subtracting multiples of \( 360^\circ \) (since cosine is periodic with a period of \( 360^\circ \)).
\[ 840^\circ - 720^\circ = 120^\circ \]
Therefore, \( \cos 840^\circ = \cos 120^\circ \).

2. **Evaluate \( \cos 120^\circ \)**:
The angle \( 120^\circ \) is in the second quadrant, where cosine is negative. We can express it as:
\[ \cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ \]
We know that \( \cos 60^\circ = \frac{1}{2} \), thus:
\[ \cos 120^\circ = -\frac{1}{2} \]

### Final Value for \( \cos 840^\circ \):
\[ \cos 840^\circ = -\frac{1}{2} \]

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### Step 2: Finding \( \sin 870^\circ \)

1. **Reduce the angle**:
We can reduce \( 870^\circ \) by subtracting multiples of \( 360^\circ \).
\[ 870^\circ - 720^\circ = 150^\circ \]
Therefore, \( \sin 870^\circ = \sin 150^\circ \).

2. **Evaluate \( \sin 150^\circ \)**:
The angle \( 150^\circ \) is also in the second quadrant, where sine is positive. We can express it as:
\[ \sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ \]
We know that \( \sin 30^\circ = \frac{1}{2} \), thus:
\[ \sin 150^\circ = \frac{1}{2} \]

### Final Value for \( \sin 870^\circ \):
\[ \sin 870^\circ = \frac{1}{2} \]

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### Summary of Results:
- \( \cos 840^\circ = -\frac{1}{2} \)
- \( \sin 870^\circ = \frac{1}{2} \)

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