To solve this problem, we need to express [tex]\(\cos 5\theta\)[/tex] in terms of powers of [tex]\(\cos \theta\)[/tex]. We will use the trigonometric identity for the cosine of a multiple angle.
The general form we have is:
[tex]\[
\cos 5\theta = a_0 + a_1 \cos \theta + a_2 \cos^2 \theta + a_3 \cos^3 \theta + a_4 \cos^4 \theta + a_5 \cos^5 \theta
\][/tex]
We are asked to identify the coefficients [tex]\(a_0, a_1, a_2, a_3, a_4, a_5\)[/tex].
One of the standard identities for multiple angle cosines, specifically cosine quintuple angle, is:
[tex]\[
\cos 5\theta = 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta
\][/tex]
Comparing this with:
[tex]\[
\cos 5\theta = a_0 + a_1 \cos \theta + a_2 \cos^2 \theta + a_3 \cos^3 \theta + a_4 \cos^4 \theta + a_5 \cos^5 \theta,
\][/tex]
we can identify the coefficients directly:
- For [tex]\(\cos \theta\)[/tex], we compare [tex]\(5 \cos \theta\)[/tex], thus [tex]\(a_1 = 5\)[/tex].
- For [tex]\(\cos^3 \theta\)[/tex], we compare [tex]\(-20 \cos^3 \theta\)[/tex], thus [tex]\(a_3 = -20\)[/tex].
- For [tex]\(\cos^5 \theta\)[/tex], we compare [tex]\(16 \cos^5 \theta\)[/tex], thus [tex]\(a_5 = 16\)[/tex].
- Notice there are no pure [tex]\(\cos^2 \theta\)[/tex] or [tex]\(\cos^4 \theta\)[/tex] terms in the standard identity, so [tex]\(a_2 = 0\)[/tex] and [tex]\(a_4 = 0\)[/tex].
- There is no constant term, so [tex]\(a_0 = 0\)[/tex].
By examining these comparisons, we can create the pairs for each quantity:
- [tex]\(a_0:\; s \;(\text{as } a_0 = 0)\)[/tex]
- [tex]\(a_1:\; r \;(\text{as } a_1 = 5)\)[/tex]
- [tex]\(a_2:\; s \;(\text{as } a_2 = 0)\)[/tex]
- [tex]\(a_5:\; q \;(\text{as } a_5 = 16)\)[/tex]
So the corresponding matching from Column-I to Column-II is:
- a) [tex]\(a_0 \leftrightarrow s\)[/tex]
- b) [tex]\(a_1 \leftrightarrow r\)[/tex]
- c) [tex]\(a_2 \leftrightarrow s\)[/tex]
- d) [tex]\(a_5 \leftrightarrow q\)[/tex]
