Question

If $\sin \theta +\cos \theta =\sqrt{2}$, find the value of $\sin \theta \times \cos \theta$.
(a) 2
(b) $\sqrt{2}-1$
(c) $\dfrac{1}{2}$
(d) $2\sec \theta$

Answer 1

Hint: Square the given equation on both sides and use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ to expand the equation. Use another trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and rearrange the terms to find the value of $\sin \theta \times \cos \theta$.

Complete step-by-step answer:
We have the trigonometric expression $\sin \theta +\cos \theta =\sqrt{2}$. We have to calculate the value of $\sin \theta \times \cos \theta$.
We will begin by squaring the equation $\sin \theta +\cos \theta =\sqrt{2}$ on both sides.
Thus, we have ${{\left( \sin \theta +\cos \theta \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}}$.
We know the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Substituting $a=\sin \theta ,b=\cos \theta$ in the above expression, we have ${{\left( \sin \theta +\cos \theta \right)}^{2}}={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \times \cos \theta$.
Thus, we have ${{\left( \sin \theta +\cos \theta \right)}^{2}}={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \times \cos \theta ={{\left( \sqrt{2} \right)}^{2}}$.
So, we have ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \times \cos \theta =2$.
We know the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Thus, we have ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \times \cos \theta =2\Rightarrow 1+2\sin \theta \times \cos \theta =2$.
Simplifying the above expression, we have $2\sin \theta \times \cos \theta =2-1=1$.
Thus, we have $\sin \theta \times \cos \theta =\dfrac{1}{2}$.
Hence, if $\sin \theta +\cos \theta =\sqrt{2}$, then the value of $\sin \theta \times \cos \theta$ is $\dfrac{1}{2}$, which is option (c).
Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.

Note: We can also solve this question by trying to find the value of $\theta$ which satisfies the given equation. We observe that $\theta =\dfrac{\pi }{4}$ satisfies the given equation. Thus, we can calculate the value of $\sin \theta \times \cos \theta$ by substituting $\theta =\dfrac{\pi }{4}$.