Question
Given Sinlb \( =p \) Determine the following in term of \( p \) withait using a colculato Q 1. Sin \( 196^{\circ} \) 2. \( \operatorname{Cos} 16^{\circ} \)
Ask by Fitzgerald Flynn. in South Africa
Feb 23,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
sin 196° = –p
cos 16° = √(1 – p²)
Solution
We are given that sin 16° = p. Notice that
196° = 180° + 16°.
Since sine is negative in the third quadrant and sin(180° + θ) = –sin θ, we have:
sin 196° = –sin 16° = –p.
Now, using the Pythagorean identity for the angle 16°:
sin²16° + cos²16° = 1 ⇒ p² + cos²16° = 1.
Since 16° is in the first quadrant, where cosine is positive, we solve for cos 16° as follows:
cos 16° = √(1 – p²).
Thus, expressed in terms of p:
1. sin 196° = –p.
2. cos 16° = √(1 – p²).
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Simplify this solution The Deep Dive
To find \(\sin 196^\circ\), we can use the identity \(\sin(180^\circ + \theta) = -\sin \theta\). Here, \(196^\circ = 180^\circ + 16^\circ\), so \(\sin 196^\circ = -\sin 16^\circ\). Since \(\sin 16^\circ\) is given as \(p\), we have \(\sin 196^\circ = -p\). For \(\cos 16^\circ\), we can use the Pythagorean identity relating sine and cosine: \(\cos^2 \theta + \sin^2 \theta = 1\). Since \(\sin 16^\circ = p\), it follows that \(\cos^2 16^\circ = 1 - p^2\), which gives us \(\cos 16^\circ = \sqrt{1 - p^2}\). So, the answers are: 1. \(\sin 196^\circ = -p\) 2. \(\cos 16^\circ = \sqrt{1 - p^2}\)
