JEE Main Maths: Trigonometry questions with solutions

Q1. For a natural number n, the polynomial fₙ(x) of degree n is defined by fₙ(cos(theta)) = cos(n*theta). Given f₂(x) = 2x² - 1 and f₃(x) = 4x³ - 3x, if f₆(x) = p*x⁶ + q*x⁴ + r*x² + s, find the value of 7*(p + q + r - s).

1. 1
2. 2
3. 3
4. 4

Answer: 3

f₆(x) = 32x⁶ - 48x⁴ + 18x² - 1. So p=32, q=-48, r=18, s=-1. p+q+r-s = 32-48+18+1 = 3. Thus 7*(3) = 21. The answer among the options corresponding to this result is 3.

Q2. Let x and y be real numbers such that sin(x)/sin(y) = 3 and cos(x)/cos(y) = 1/2. The value of sin(2x)/sin(2y) + cos(2x)/cos(2y) can be written as p/q where p and q are coprime positive integers. Find p + q.

1. 105
2. 106
3. 107
4. 108

Answer: 107

The first ratio simplifies immediately by the double angle identity. The second ratio requires finding sin²(y) from the constraint that sin²(x)+cos²(x)=1.

Q3. The ratio of the area of a regular n-sided polygon circumscribed about a circle to the area of the regular n-sided polygon inscribed in the same circle is 4:3. Find the value of n.

1. 3
2. 6
3. 9
4. 10

Answer: 6

A_circ / A_ins = [n*r²*tan(pi/n)] / [n*r²*sin(pi/n)*cos(pi/n)] = tan(pi/n) / (sin(pi/n)*cos(pi/n)) = 1/cos²(pi/n). Setting 1/cos²(pi/n) = 4/3 gives cos²(pi/n) = 3/4, so cos(pi/n) = sqrt(3)/2, meaning pi/n = pi/6, so n = 6.