Question

What is the range of |5cos2x+12sin2x|−4?

Answer 1

Given:

$|5cos2x+12sin2x|-4$

To find:

The range of $|5cos2x+12sin2x|-4$ = ?

Solution:

First of all, let us consider $5cos2x+12sin2x$.

Please refer to the attached image of a right angled triangle which has sides, 5, 12 and 13 (a triplet).

13 is the hypotenuse.

$sinA = \dfrac{5}{13}\\cosA = \dfrac{12}{13}$

Multiply and divide $5cos2x+12sin2x$ by 13

$\dfrac{13}{13}(5cos2x+12sin2x)\\\Rightarrow {13}(\dfrac{5}{13}cos2x+\dfrac{12}{13}sin2x)\\\Rightarrow {13}(sinAcos2x+cosBsin2x)\\\Rightarrow 13sin(A+2x)$

Formula used: $sin(C+D) = sinCcosD+cosCsinD$

Now, putting the value back to given function:

$|13sin(A+2x)|-4$

We know that range of $sin\theta$ is [-1,1]

But we are given modulus here:

So, range of |$sin\theta$| is [0,1]

OR

Range of $|sin(A+2x)|$ is [0,1] ($\theta = A+2x$)

OR

Range of $|13sin(A+2x)|$ is [0,13]

OR

Range of $|13sin(A+2x)|-4$ is [0-4,13-4] or [-4,9]

Both -4 and 9 are inclusive in the range.