Question

If $a\sin x + b\cos (x + \theta ) + b\cos (x - \theta ) = d$, then the value of $\left| {\cos \theta } \right|$ is equal to.

Answer 1

We need to simplify such equation using trigonometric functions
Sum of the trigonometric functions can be formulated as:
$Cos(A + B) + \operatorname{Cos} (A - B) = 2\operatorname{Cos} A\operatorname{Cos} B$
We know that the value of a $\operatorname{Sin} \theta + B\operatorname{Cos} \theta$ lies between: $- \sqrt {{a^2} + {b^2}} \leqslant a\operatorname{Sin} \theta + b\cos \theta \leqslant \sqrt {{a^2} + {b^2}}$.

Complete step-by- step solution:
We have,
$\Rightarrow a\operatorname{Sin} \theta + b\cos (x + \theta ) + b\cos (x - \theta ) - d.............eqn(1)$
Taking out b common from$eqn(1)$, we get
$\Rightarrow a\operatorname{Sin} \theta + b\left[ {\cos (x + \theta ) + \cos (x - \theta )} \right] - d...............eqn(2)$
We know, $Cos(x - \theta ) + \operatorname{Cos} (x - \theta ) = 2Cosx\,Cos\theta ...........eqn(3)$
Using the value of $eqn(3)$in$eqn(2)$, we get
$a\sin x + b\left[ {2Cosx\,\operatorname{Cos} \theta } \right] = d$
$\Rightarrow a\sin x + b\left[ {2bCosx\,\operatorname{Cos} \theta } \right] = d........eqn(4)$
As $a\sin \theta + b\operatorname{Cos} \theta$ lies between $- \sqrt {{a^2} + {b^2}} \leqslant a\operatorname{Sin} \theta + b\operatorname{Cos} \theta \leqslant \sqrt {{a^2} + {b^2}}$
$\Rightarrow \left| {a\sin \theta + b\cos \theta } \right| \leqslant \sqrt {{a^2} + {b^2}} ........eqn(5)$
Compare$a\sin \theta + b\cos \theta$ with $a\sin x + (2b\operatorname{Cos} \theta )\operatorname{Cos} x$
We get, $a = a$ and $b = 2b\operatorname{Cos} \theta ........eqn(6)$
Using the value of $eqn$ (6) and (4) in (5), we have a\sin \theta + b\cos \theta = d,$$$$a = aand $b = 2b\operatorname{Cos} \theta$
$\Rightarrow \left| d \right| \leqslant \sqrt {{a^2} + {{(2b\cos \theta )}^2}}$
$\Rightarrow \left| d \right| \leqslant \sqrt {{a^2} + 4{b^2}Co{s^2}\theta }$
On squaring both sides
$\Rightarrow {d^2} \leqslant {a^2} + 4{b^2}Co{s^2}\theta$
$\Rightarrow {d^2} - {a^2} \leqslant 4{b^2}Co{s^2}\theta$
$\Rightarrow \dfrac{{{d^2} - {a^2}}}{{4{b^2}}} \leqslant Co{s^2}\theta$
Taking under root on both sides
.$\left| {\operatorname{Cos} \theta } \right| \geqslant \dfrac{{\sqrt {{d^2} - {a^2}} }}{{\sqrt {4{b^2}} }}$

\\\ \left| {\operatorname{Cos} \theta } \right| \geqslant \dfrac{{\sqrt {{d^2} - {a^2}} }}{{2\left| b \right|}} \\\ \end{gathered} $$ Hence $$\left| {\operatorname{Cos} \theta } \right| = \dfrac{1}{{2\left| b \right|}}\sqrt {{d^2} - {a^2}} $$ **Note:** Recall that in its basic form $$\,f(x) = \,|x|,\,$$ The absolute value function is one of our tool kit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems. Modulus operation on function converts negative function values to positive function values with equal magnitude. As such, we draw a graph of the modulus function by taking a mirror image of the corresponding core graph in x-axis. We need under that if x lies between $$ - a < x < a$$ then $$\left| a \right| \leqslant x.$$