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Question: If \[p = \dfrac{1}{2}{\sin ^2}\theta + \dfrac{1}{3}{\cos ^2}\theta \], then 1) \[\dfrac{1}{3} \le...

If p=12sin⁡2θ+13cos⁡2θp = \dfrac{1}{2}{\sin ^2}\theta + \dfrac{1}{3}{\cos ^2}\theta , then

  1. 13⩽p⩽12\dfrac{1}{3} \leqslant p \leqslant \dfrac{1}{2}
  2. p⩽12p \leqslant \dfrac{1}{2}
  3. 2⩽p⩽32 \leqslant p \leqslant 3
  4. −36⩽p⩽36 - \dfrac{{\sqrt 3 }}{6} \leqslant p \leqslant \dfrac{{\sqrt 3 }}{6}
Explanation

Solution

Here in this question we have to determine the range of p, i.e., what are the values the variable p can take. So on considering the trigonometric equation, applying the trigonometric identities which relate to the equation and then on simplifying the equation we obtain the required answer.

Complete step by step answer:
We have to determine the range of the p
Consider the given equation p=12sin⁡2θ+13cos⁡2θp = \dfrac{1}{2}{\sin ^2}\theta + \dfrac{1}{3}{\cos ^2}\theta -------- (1)
As we know the trigonometric identities sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, so we have sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta --(2)
On substituting the equation (2) in the equation (1) we have
⇒p=12(1−cos⁡2θ)+13cos⁡2θ\Rightarrow p = \dfrac{1}{2}\left( {1 - {{\cos }^2}\theta } \right) + \dfrac{1}{3}{\cos ^2}\theta
On simplifying we have
⇒p=12−12cos⁡2θ+13cos⁡2θ\Rightarrow p = \dfrac{1}{2} - \dfrac{1}{2}{\cos ^2}\theta + \dfrac{1}{3}{\cos ^2}\theta
Taking the common term we have
⇒p=12+cos⁡2θ(13−12)\Rightarrow p = \dfrac{1}{2} + {\cos ^2}\theta \left( {\dfrac{1}{3} - \dfrac{1}{2}} \right)
On simplifying we have
⇒p=12+cos⁡2θ(−16)\Rightarrow p = \dfrac{1}{2} + {\cos ^2}\theta \left( { - \dfrac{1}{6}} \right)
So we have
⇒p=12−16cos⁡2θ\Rightarrow p = \dfrac{1}{2} - \dfrac{1}{6}{\cos ^2}\theta
In the above equation there is a minus sign then the value of p will be less than 12\dfrac{1}{2}
So this can be written as
⇒p⩽12\Rightarrow p \leqslant \dfrac{1}{2} _________(3)
Now consider the given equation p=12sin⁡2θ+13cos⁡2θp = \dfrac{1}{2}{\sin ^2}\theta + \dfrac{1}{3}{\cos ^2}\theta -------- (1)
As we know the trigonometric identities sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, so we have cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta --(5)
On substituting the equation (5) in the equation (1) we have
⇒p=12sin⁡2θ+13(1−sin⁡2θ)\Rightarrow p = \dfrac{1}{2}{\sin ^2}\theta + \dfrac{1}{3}\left( {1 - {{\sin }^2}\theta } \right)
On simplifying we have
⇒p=12sin⁡2θ+13−13sin⁡2θ\Rightarrow p = \dfrac{1}{2}{\sin ^2}\theta + \dfrac{1}{3} - \dfrac{1}{3}{\sin ^2}\theta
Taking the common term we have
⇒p=13+sin⁡2θ(12−13)\Rightarrow p = \dfrac{1}{3} + {\sin ^2}\theta \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right)
On simplifying we have
⇒p=13+sin⁡2θ(16)\Rightarrow p = \dfrac{1}{3} + {\sin ^2}\theta \left( {\dfrac{1}{6}} \right)
So we have
⇒p=13+16sin⁡2θ\Rightarrow p = \dfrac{1}{3} + \dfrac{1}{6}{\sin ^2}\theta
In the above equation there is a plus sign then the value of p will be greater than 13\dfrac{1}{3}
So this can be written as
⇒p⩾13\Rightarrow p \geqslant \dfrac{1}{3} _________(4)
From equation (3) and equation (4) we have the range of p and it will be 13⩽p⩽12\dfrac{1}{3} \leqslant p \leqslant \dfrac{1}{2}.
So, the correct answer is “Option 1”.

Note: In a trigonometry we have three identities they are sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta and 1+cot⁡2θ=csc⁡2θ1 + {\cot ^2}\theta = {\csc ^2}\theta . While adding and subtracting the fractions where the denominator values are different then we have to take LCM for them and then on simplification we get the solution for addition and subtraction of fractions.