IIT-JEE 1987Question 1
The number of all possible triplets $(a_1,a_2,a_3)$ such that $a_1+a_2\cos(2x)+a_3\sin^{2}(x)=0$ for all $x$ is
Trigonometric Ratio & Identities
Solution
Concept. Write $\cos2x$ in $\sin^2x$; an identity in $\sin^2x$ needs both coefficients to be zero. Formulas used. (1) $\cos2x=1-2\sin^2x$Step 1 - Rewrite. By (1), $$a_1+a_2\cos2x+a_3\sin^2x=(a_1+a_2)+(a_3-2a_2)\sin^2x.$$ Step 2 - Identically zero. $\sin^2x$ takes many values, so $a_1+a_2=0$ and $a_3=2a_2$. Step 3 - Count. All triplets $(-t,\,t,\,2t)$, $t\in\mathbb R$, work: infinitely many. Answer: (D)