JEE - Trigonometry With k

As the final answer is just a number and independent of theta, that means that the whole expression is true for any and all theta.. So just take theta = pi/2 where sin(theta) =1 and cos(theta) = 0 so u get 1/4-1/6 = 1/12 = M therefore 1/M=12

let $\sin(t) = s$ ; $\cos(t) = c$ . $\begin{aligned}\dfrac{1}{4}(s^4 + c^4) - \dfrac{1}{6}(s^6+c^6) &= \dfrac{1}{4}((s^2+c^2)^2 - 2s^2c^2) - \dfrac{1}{6}((s^2+c^2)^3-3s^2c^2(s^2+c^2)) \\ &= \dfrac{1}{4}-\dfrac{1}{6} \\ &= 1/12\end{aligned}$

4 f4 = (sin(r))^4 + (cos(r))^4 6 f6 = (sin(r))^6 + (cos(r))^6

4 f4 can be written as: ((sin(r))^2 + (cos(r))^2)^2 - 2(sinr cosr)^2 = 1 - 2(sinr cosr)^2 6f6 can be written as: ((sin(r))^2 + (cos(r))^2)^3 - 3(sinr cosr)^2 ((sin(r))^2 + (cos(r))^2) = 1 - 3(sinr cosr)^2

[ as (a+b)^3 = a^3 + b^3 +3ab(a+b)]

hence f4 = 1/4-1/2 (sinr cosr)^2 and f6 = 1/6 - 1/2 (sinr cosr)^2

subtracting f6 from f4 we get 1/12

solution is 1/12

It's very simple untill you don't apply the algebric formulae... just use (a+b)^2-2ab=a^2+b^2 and (a+b)^3-3ab(a+b)=a^3+b^3

$\boxed{f_{k}(\theta ) = \frac{1}{k}(\sin^{k}{\theta} + \cos^{k}{\theta})}$

$f_{4}(\theta ) = \frac{1}{4}(\sin^{4}{\theta} + \cos^{4}{\theta})$

$f_{4}(\theta ) = \frac{1}{4}(1-\frac{\sin^22\theta}{2})$

$Similarly,$

$f_{6}(\theta ) = \frac{1}{6}(\sin^6\theta + \cos^6\theta)$

$f_{6}(\theta ) = \frac{1}{6}(1-\frac{3\sin^2 2\theta}{4})$

$Subtracting,$

$f_{4}(\theta ) - f_{6}(\theta)= \begin{bmatrix} \frac{1}{4} - \frac{\sin^2 2\theta}{8} - \frac{1}{6} + \frac{\sin^2 2\theta}{8} \end{bmatrix}$

$f_{4}(\theta ) - f_{6}(\theta)= \frac{1}{4} - \frac{1}{6}$

$\boxed{f_{4}(\theta ) - f_{6}(\theta)= \frac{1}{12}}$

let sint = a ; cost = b ;then ,(a^4+b^2)/4 - (a^6+b^6)/6 = (a^4+b^4)/4 - (a^2+b^2)(a^4+b^4-a^2 b^2)/6 =(a^4+b^4)/4 - (a^4+b^4)/6 + (a^2 b^2)/6 =(a^4+b^4+2 a^2 b^2)/12 = (a^2+b^2)^2/12 = 1/12 so M = 1/12 1/M = 12

take any value of theta..(take pi/2 or 0 for convenience) and solve the both eqns.(for k=4,k=6)

to all those who are saying that it can be solved by substituting theta as pi/2,would you do the same thing if in the equation there was a minus rather than the plus.

Jee is especially known for giving shortcut in its question or a trap simply put ø 0 and find answer if u r not satisfied putø π÷2

The solution is 12 and it is independent of Theta. This can be solved by rewriting cos and sin with e^(i \theta) and e^(-i \theta). Now use binomial expansion for functions f 4 and f 6 . the rest is straight forward. 6/32-20/192=1/12=M. therefore inverse is 12.

Substitute \sin ^{ 2 }{ \theta } = x and \cos ^{ 2 }{ \theta } = 1.x. Then solve.

People , Just simplify it , everything cancels out and you get (cos^(theta)+sin^2(theta))/12 = 1/12=M so , M=12.

for values of theta when sin theta is 1 and cos theta is 0, or vice verse we get a rational M, which is 1/12.. for other values of theta, we get M={1-(some irrational number)}/12 = a irrational number.. So the M only takes a constant rational value which is 1/12. .

using sin^2 (theta)=(1-cos 2(theta))/(2) and cos^2 (theta)=(1+cos 2(theta))/(2)

I found that derivative of function is 0, so the function is constant. Then I put theta =0. This is a joke for now (but could be useful sometimes). Be happy, do math :)

as the function is dependant on theta n k. but question is such that the value of theta doesnt matter. so just put theta=0 n substitute