Question

If $(\sin \theta + \cos \theta )$and $(3,2)$lie on the line $x + y = 1$then find the value of$\theta$.
(a) $\left( {0,0} \right)$
(b) $\left( {0,\dfrac{\pi }{4}} \right)$
(c) $\left( {\dfrac{\pi }{4},0} \right)$
(d) None of the above

Answer 1

The given problem revolves around the concepts of trigonometric equations. Since, we are assuming some variables for the given line (with given certain points). After certain mathematical calculations i.e. substituting the values in the given line equation then putting the values of sine and cosine terms using the factorization formula of trigonometric ratios, the desired solution can be obtained.

Complete step by step answer:
Since, the given equation,
$x + y = 1 \\\ \Rightarrow x + y - 1 = 0 \\\$
Satisfies the given points $(\sin \theta ,\cos \theta )$as well as $(3,2)$respectively,
Therefore, assuming this respective equation as line ‘$L$’, we get
$\Rightarrow L = x + y - 1$
Line ‘${L_1}$’ satisfies the points$(3,2)$on the line ‘$L$’$x + y - 1 = 0$,
Hence, substituting the points in the equation, we get

$$\Rightarrow {L_1} = 3 + 2 - 1 \\\ \Rightarrow {L_1} = 4 \\\$$

Which, seems the boundary conditions may be greater than or equal to${L_1}$,
$\therefore {L_1} \geqslant 4 > 0$
Similarly,
Line ‘${L_2}$’ satisfies the points$(\sin \theta ,\cos \theta )$on the line ‘$L$’$x + y - 1 = 0$,
Hence, substituting the points in the equation, we get
$\Rightarrow {L_2} = \sin \theta + \cos \theta - 1$
Which, seems the boundary conditions may be greater than or equal to${L_2}$,
$\therefore {L_2} \geqslant \sin \theta + \cos \theta - 1 > 0$
Since, the given both the points satisfies on same line, we get
$\therefore {L_1}{L_2} > 0$
Substituting the values ${L_1}$and ${L_2}$ in above equation, we get

$$\Rightarrow 4(\sin \theta + \cos \theta - 1) > 0 \\\ \Rightarrow \sin \theta + \cos \theta - 1 > 0 \\\$$

Solving the equation mathematically, we get
$\Rightarrow \sin \theta + \cos \theta > 1$
Now, since we can predict that to get the value of L.H.S. as greater than$> 1$, the value should be greater than one,
Hence, multiplying and dividing the above equation by$\sqrt 2$, we get

$$\Rightarrow \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin \theta + \dfrac{1}{{\sqrt 2 }}\cos \theta } \right) > 1 \\\ \Rightarrow \dfrac{1}{{\sqrt 2 }}\sin \theta + \dfrac{1}{{\sqrt 2 }}\cos \theta > \dfrac{1}{{\sqrt 2 }} \\\$$

But, we know that$\sin {45^ \circ } = \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$and$\cos {45^ \circ } = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$,
$\therefore$The equation can also written as,
$\Rightarrow \sin \dfrac{\pi }{4}\sin \theta + \cos \dfrac{\pi }{4}\cos \theta > \dfrac{1}{{\sqrt 2 }}$
By using factorisation formula for trigonometric ratios i.e. $\sin (A + B) = \sin A\sin B + \cos A\cos B$, we get,

$$\Rightarrow \sin \left( {\dfrac{\pi }{4} + \theta } \right) > \sin \dfrac{\pi }{4} \\\ \Rightarrow \dfrac{\pi }{4} + \theta > \dfrac{\pi }{4} \\\$$

By solving the equation mathematically, we get
$\Rightarrow \theta > 0$… (i)
That is,
$= \sin \theta = \sin 0 = 0{\text{ also,}} \\\ = \cos \theta = \cos 0 = {90^ \circ } = {\dfrac{\pi }{2}{rad}} \\\$
As a result, equations (i) revolves within these values, we get

$$\Rightarrow \dfrac{\pi }{2} > \left( {\dfrac{\pi }{4} + \theta } \right) > 0 \\\ {\text{ = }}0 < \left( {\dfrac{\pi }{4} + \theta } \right) < \dfrac{\pi }{2} \\\$$

Hence, taking the term $\left( {\dfrac{\pi }{4}} \right)$to R.H.S. i.e. subtracting $\dfrac{\pi }{4}$from$\dfrac{\pi }{2}$, the equation becomes

$$\Rightarrow 0 < \theta < \left( {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right) \\\ = 0 < \theta < \dfrac{\pi }{4} \\\$$

Hence, $\theta$ lies between $0{\text{ and }}\dfrac{\pi }{4}$respectively.

So, the correct answer is “Option b”.

Note:
One must know the trigonometric values for ‘sine’ and ‘cosine’ terms respectively. Then, considering their trigonometric ratios (formulae) like, $\sin (A + B) = \sin A\sin B + \cos A\cos B$ so as to distinguish the solution accurately. Also, we should know all the required values of standard angles say, ${0^o},{30^o},{45^o},{60^o},{90^o},{180^o},{270^o},{360^o}$respectively for each trigonometric term such as$\sin ,\cos ,\tan ,\cot ,\sec ,\cos ec$, etc. We should take care of the calculations to convert the angles from ‘degrees’ to ‘radian’ form say, ${30^ \circ } = {30^ \circ } \times \dfrac{\pi }{{180}} = {\dfrac{\pi }{6}{radian}}$ so as to be sure of our final answer.