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Question

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

Find the value of sinθ\sin \theta ++ cosθ\cos \theta .

Explanation

Solution

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 sin2θ+cos2θ=1{\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, sin2θ+cos2θ=1{\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=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab . Then after solving it further we are able to get the value of sinθ+cosθ\sin \theta + \cos \theta .

Complete step-by-step answer :
We know that sin2θ+cos2θ{\sin ^2}\theta + {\cos ^2}\theta == 11 -------- (1)(1)
To find the value of sinθ+cosθ\sin \theta + \cos \theta . We will add 2sinθcosθ2\sin \theta \cos \theta on both sides of the equation (1)(1) . Therefore equation (1)(1) becomes;
sin2θ+cos2θ+2sinθcosθ = 1+2sinθcosθ{\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta {\text{ }} = {\text{ }}1 + 2\sin \theta \cos \theta ----------- (2)(2)
Also we know that a2+b2+2ab{a^2} + {b^2} + 2ab == (a+b)2{\left( {a + b} \right)^2} . Therefore equation (2)(2) becomes
(sinθ+cosθ)2{\left( {\sin \theta + \cos \theta } \right)^2} == 1+2sinθcosθ1 + 2\sin \theta \cos \theta --------- (3)\left( 3 \right)
Because 2sinθcosθ2\sin \theta \cos \theta == sin2θ\sin 2\theta , equation (3)\left( 3 \right) becomes
(sinθ+cosθ)2{\left( {\sin \theta + \cos \theta } \right)^2} == 1+sin2θ1 + \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
sinθ+cosθ = (1+sin2θ)12\sin \theta + \cos \theta {\text{ }} = {\text{ }}{\left( {1 + \sin 2\theta } \right)^{\dfrac{1}{2}}}
Or we can write it as
sinθ+cosθ = 1+sin2θ\sin \theta + \cos \theta {\text{ }} = {\text{ }}\sqrt {1 + \sin 2\theta }
Hence the value of sinθ+cosθ\sin \theta + \cos \theta is 1+sin2θ\sqrt {1 + \sin 2\theta } .
So, the correct answer is “ 1+sin2θ\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.