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Question: If we have an expression as \[\sin \theta. \cos \theta = \dfrac{1}{2}\] . then find the value of \[\...

If we have an expression as sin⁡θ.cos⁡θ=12\sin \theta. \cos \theta = \dfrac{1}{2} . then find the value of sin⁡θ+cos⁡θ\sin \theta + \cos \theta

Explanation

Solution

In this question, we need to find the value of sin⁡θ+cos⁡θ\sin \theta + \cos \theta and also given sin⁡θ.cos⁡θ=12\sin \theta. \cos \theta = \dfrac{1}{2} . Sine , cosine and tangent are the basic trigonometric functions . Sine is nothing but a ratio of the opposite side of a right angle to the hypotenuse of the right angle. Similarly, cosine is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle . Here we need to find the value of sin⁡θ+cos⁡θ\sin \theta + \cos \theta . With the help of the Trigonometric functions , we can find the value of sin⁡θ+cos⁡θ\sin \theta + \cos \theta .
Identity used :
sin2θ+cos2θ=1sin^{2}\theta + cos^{2}\theta = 1
Algebraic formula used :
(a+b)2=a2+b2+2ab\left( a + b \right)^{2} = a^{2} + b^{2} + 2ab

Complete step by step solution:
Given,
⇒sin⁡θ+cos⁡θ\Rightarrow \sin \theta + \cos \theta
By squaring ,
We get,
⇒(sin⁡θ+cos⁡θ)2\Rightarrow \left( \sin \theta + \cos \theta \right)^{2}
We know that (a+b)2=a2+b2+2ab\left( a + b \right)^{2} = a^{2} + b^{2} + 2ab
By expanding,
We get,
(sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ\left( \sin \theta + \cos \theta \right)^{2} = \sin^{2}\theta + \cos^{2} \theta + 2\sin \theta \cos \theta
From the trigonometry identity , sin⁡2θ+cos⁡2θ=1\sin^{2} \theta + \cos^{2} \theta = 1
By substituting sin⁡2θ+cos⁡2θ=1\sin^{2}\theta + \cos^{2}\theta = 1
We get,
(sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ\left( \sin \theta + \cos \theta \right)^{2} = 1 + 2\sin \theta \cos \theta
In question, given that sin⁡θ.cos⁡θ=12\sin \theta. \cos \theta = \dfrac{1}{2}
By substituting sin⁡θ.cos⁡θ=12\sin \theta. \cos \theta = \dfrac{1}{2}
We get,
⇒(sin⁡θ+cos⁡θ)2=1+2×12\Rightarrow \left( \sin \theta + \cos \theta \right)^{2} = 1 + 2 \times \dfrac{1}{2}
By simplifying,
We get,
⇒(sin⁡θ+cos⁡θ)2=1+1\Rightarrow \left( \sin \theta + \cos \theta \right)^{2} = 1 + 1
By adding,
We get,
⇒(sin⁡θ+cos⁡θ)2=2\Rightarrow \left( \sin \theta + \cos \theta \right)^{2} = 2
On taking square root on both sides,
We get,
⇒(sin⁡θ+cos⁡θ)2=±2\Rightarrow \sqrt{\left( \sin \theta + \cos \theta \right)^{2}} = \pm \sqrt{2}
Thus we get, sin⁡θ+cos⁡θ=±2\sin \theta + \cos \theta = \pm \sqrt{2}
Final answer :
The value of sin⁡θ+cos⁡θ\sin \theta + \cos \theta is equal to ±2\pm \sqrt{2}

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the algebraic formula with the use of trigonometric functions . Trigonometric functions are also known as circular functions or geometrical functions.