Little Morrison
05/28/2023 · Elementary School
5) \( \sin ^{2} 20^{\circ}+\cos ^{2} 160^{\circ}= \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the expression \( \sin^2 20^\circ + \cos^2 160^\circ \), we can use the identity that states \( \cos(180^\circ - x) = -\cos x \).
First, we can simplify \( \cos 160^\circ \):
\[
\cos 160^\circ = \cos(180^\circ - 20^\circ) = -\cos 20^\circ
\]
Now, we can substitute this into our expression:
\[
\cos^2 160^\circ = (-\cos 20^\circ)^2 = \cos^2 20^\circ
\]
Now we can rewrite the original expression:
\[
\sin^2 20^\circ + \cos^2 160^\circ = \sin^2 20^\circ + \cos^2 20^\circ
\]
Using the Pythagorean identity, we know that:
\[
\sin^2 x + \cos^2 x = 1
\]
Thus, we have:
\[
\sin^2 20^\circ + \cos^2 20^\circ = 1
\]
Therefore, the final result is:
\[
\sin^2 20^\circ + \cos^2 160^\circ = 1
\]
So, the answer is:
\[
\boxed{1}
\]
Quick Answer
The answer is 1.
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