Question

Find the value of \[\sin \theta\] \[ + \] \[\cos \theta \].

Answer 1

Hint: Let’s look at the given conditions. In this question, clearly there is no constant value and no angle or any other condition is given. So, from this it is clear that we have to use a basic idea and known formulas to find the answer. After seeing this question \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] comes into our mind so we have to try using this formula by further adjusting L.H.S and R.H.S we can get the result. In other words, using one trigonometric identity, \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] . Add a particular term on both sides of the equation to make the left side term of identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] . Then after solving it further we are able to get the value of \[\sin \theta + \cos \theta \] .

Complete step-by-step answer:
We know that \[{\sin ^2}\theta + {\cos ^2}\theta \] \[ = \] \[1\] -------- \[(1)\]
To find the value of \[\sin \theta + \cos \theta \] . We will add \[2\sin \theta \cos \theta \] on both sides of the equation \[(1)\] . Therefore equation \[(1)\] becomes;
\[{\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta {\text{ }} = {\text{ }}1 + 2\sin \theta \cos \theta \] ----------- \[(2)\]
 Also we know that \[{a^2} + {b^2} + 2ab\] \[ = \] \[{\left( {a + b} \right)^2}\] . Therefore equation \[(2)\] becomes
\[{\left( {\sin \theta + \cos \theta } \right)^2}\] \[ = \] \[1 + 2\sin \theta \cos \theta \] --------- \[\left( 3 \right)\]
 Because \[2\sin \theta \cos \theta \] \[ = \] \[\sin 2\theta \] , equation \[\left( 3 \right)\] becomes
\[{\left( {\sin \theta + \cos \theta } \right)^2}\] \[ = \] \[1 + \sin 2\theta \]
As we all know by removing the square from the left hand side we get the square root on the term written on the right hand side. Therefore we get
\[\sin \theta + \cos \theta {\text{ }} = {\text{ }}{\left( {1 + \sin 2\theta } \right)^{\dfrac{1}{2}}}\]
Or we can write it as
\[\sin \theta + \cos \theta {\text{ }} = {\text{ }}\sqrt {1 + \sin 2\theta } \]
Hence the value of \[\sin \theta + \cos \theta \] is \[\sqrt {1 + \sin 2\theta } \] .
So, the correct answer is “ \[\sqrt {1 + \sin 2\theta } \] ”.

Note: Remember all the formulas in your mind carefully. Then whenever you find questions like this just try to solve the left hand side and right hand side by using formulas to get the desired result. And the trigonometric identity we used here is very easy to remember and used in many numericals.