Question

If the identity
\[ \cos^4\theta=a\cos4\theta+b\cos2\theta+c \] holds for some \(a,b,c\in \mathbb{Q}\), then \((a,b,c)=\)

• A. \(\left(\dfrac18,\dfrac38,\dfrac12\right)\)
• B. \(\left(\dfrac18,\dfrac12,\dfrac38\right)\)
• C. \(\left(\dfrac12,\dfrac18,\dfrac38\right)\)
• D. \(\left(\dfrac12,\dfrac38,\dfrac18\right)\)

Hint:

To express higher powers of trigonometric functions in terms of multiple angles, repeatedly use the identity \(\cos^2x=\frac{1+\cos2x}{2}\).

Answer 1

The Correct Option is B

Step 1: Use the double angle identity.
We know that
\[ \cos^2\theta=\frac{1+\cos2\theta}{2} \]
Therefore,
\[ \cos^4\theta=\left(\frac{1+\cos2\theta}{2}\right)^2 \]
\[ =\frac{1+2\cos2\theta+\cos^22\theta}{4} \]

Step 2: Expand \(\cos^22\theta\).
Again using the identity
\[ \cos^2x=\frac{1+\cos2x}{2}, \] we get
\[ \cos^22\theta=\frac{1+\cos4\theta}{2} \]
Substituting this into the expression,
\[ \cos^4\theta = \frac{1+2\cos2\theta+\frac{1+\cos4\theta}{2}}{4} \]

Step 3: Simplify the expression.
Taking LCM inside the numerator,
\[ \cos^4\theta = \frac{\frac{2+4\cos2\theta+1+\cos4\theta}{2}}{4} \]
\[ = \frac{3+4\cos2\theta+\cos4\theta}{8} \]
Hence,
\[ \cos^4\theta = \frac18\cos4\theta+\frac12\cos2\theta+\frac38 \]
Comparing with
\[ \cos^4\theta=a\cos4\theta+b\cos2\theta+c, \] we get
\[ a=\frac18,\qquad b=\frac12,\qquad c=\frac38 \]

Step 4: Final conclusion.
Thus,
\[ \boxed{\left(\frac18,\frac12,\frac38\right)} \]