Question:
Given
\[
\cos6x=(a)\cos^6x+(b)\cos^4x+(c)\cos^2x+(d)
\]
for any real number \(x\), where \(a,b,c,d\) are constants. Find the value of \(a+b+c\).
Given
\[
\cos6x=(a)\cos^6x+(b)\cos^4x+(c)\cos^2x+(d)
\]
for any real number \(x\), where \(a,b,c,d\) are constants. Find the value of \(a+b+c\).
Show Hint
Memorize the multiple-angle identity for \(\cos6x\). It frequently appears in algebraic and trigonometric simplification problems.
Updated On: Jun 11, 2026
- \(7\)
- \(2\)
- \(49\)
- \(13\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Concept:
The standard identity is
\[
\cos6x
=
32\cos^6x
-48\cos^4x
+18\cos^2x
-1
\]
Step 1: Compare coefficients.
Therefore, \[ a=32,\qquad b=-48,\qquad c=18 \]
Step 2: Find \(a+b+c\).
\[ a+b+c = 32-48+18 \] \[ =2 \] Thus \[ \boxed{2} \]
Step 1: Compare coefficients.
Therefore, \[ a=32,\qquad b=-48,\qquad c=18 \]
Step 2: Find \(a+b+c\).
\[ a+b+c = 32-48+18 \] \[ =2 \] Thus \[ \boxed{2} \]
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