NIMCET 2026 — Mathematics PYQ

To find the value of $a+b+c$, we need to expand $\cos 6x$ into a polynomial in terms of $\cos x$.

Step 1: Use the triple angle formula

Recall the trigonometric identity for triple angles:

$$\cos 3\theta =4{\cos }^3\theta -3\cos \theta$$

By setting $\theta =2x$, we can write $\cos 6x$ as:

$$\cos 6x=\cos 3(2x)=4{\cos }^3(2x)-3\cos (2x)$$

Step 2: Use the double angle formula

Recall that $\cos 2x=2{\cos }^{2}x-1$. Substitute this into our expression:

$$\cos 6x=4(2{\cos }^{2}x-1{)}^3-3(2{\cos }^{2}x-1)$$

Step 3: Expand the cubic term

Let $u={\cos }^{2}x$. The expression becomes $4(2u-1{)}^3-3(2u-1)$.

Expanding $(2u-1{)}^3$:

$$(2u-1{)}^3=(2u{)}^3-3(2u{)}^2(1)+3(2u)(1{)}^2-1^3$$

$$=8u^3-12u^2+6u-1$$

Now substitute this back into the equation:

$$\cos 6x=4(8u^3-12u^2+6u-1)-3(2u-1)$$

$$\cos 6x=32u^3-48u^2+24u-4-6u+3$$

$$\cos 6x=32u^3-48u^2+18u-1$$

Step 4: Identify constants and calculate the sum

Replacing $u$ with ${\cos }^{2}x$:

$$\cos 6x=32{\cos }^{6}x-48{\cos }^{4}x+18{\cos }^{2}x-1$$

Comparing this to the original form $\cos 6x=a{\cos }^{6}x+b{\cos }^{4}x+c{\cos }^{2}x+d$, we get:

• $a=32$

• $b=-48$

• $c=18$

• $d=-1$

The question asks for the sum $a+b+c$:

$$a+b+c=32+(-48)+18$$

$$a+b+c=32-48+18$$

$$a+b+c=2$$

Correct Option: 4. $2$