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Question: How do you verify the identity\({{\left( \sin \theta +\cos \theta \right)}^{2}}-1=\sin 2\theta \)?...

How do you verify the identity(sin⁡θ+cos⁡θ)2−1=sin⁡2θ{{\left( \sin \theta +\cos \theta \right)}^{2}}-1=\sin 2\theta ?

Explanation

Solution

Before solving the above question let's take a brief discussion on trigonometric functions. In mathematics the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. The trigonometric functions most widely used are the sine, the cosine, and the tangent. They are used in navigation, solid mechanics, celestial, and many others.

Complete step by step solution:
The given trigonometric equation is (sin⁡θ+cos⁡θ)2−1=sin⁡2θ{{\left( \sin \theta +\cos \theta \right)}^{2}}-1=\sin 2\theta . In this equation we have sin and cosine functions. In trigonometry there are many formulas related to the function. The given equation is the left and right part. We have to solve one part of the equation and if it equals to the other part then our answer is correct.
Let`s we will solve left part of the equation which is given as:
⇒(sin⁡θ+cos⁡θ)2−1\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1
First we will open the bracket by using the formula(a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab then we get,
⇒(sin⁡θ+cos⁡θ)2−1=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ−1\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1
We know sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 now use this in above equation ,we get
⇒(sin⁡θ+cos⁡θ)2−1=1+2sin⁡θcos⁡θ−1 ⇒(sin⁡θ+cos⁡θ)2−1=2sin⁡θcos⁡θ \begin{aligned} & \Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1=1+2\sin \theta \cos \theta -1 \\\ & \Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1=2\sin \theta \cos \theta \\\ \end{aligned}
Now we know that we can write 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta then we get,
⇒(sin⁡θ+cos⁡θ)2−1=2sin⁡θcos⁡θ=sin⁡2θ\Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}-1=2\sin \theta \cos \theta =\sin 2\theta
Hence after simplify we get that left part of the equation is become equals to the right part that is (sin⁡θ+cos⁡θ)2−1=sin⁡2θ{{\left( \sin \theta +\cos \theta \right)}^{2}}-1=\sin 2\theta .

Note: We can verify our answer whether it is correct or not by solving or simplifying the right part of the equation.
The right part of the given equation issin⁡2θ\sin 2\theta , now simplifying it we get,
We know that we can write 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta =\sin 2\theta then,
⇒sin⁡2θ=2sin⁡θcos⁡θ\Rightarrow \sin 2\theta =2\sin \theta \cos \theta
Now add and subtract 11 in the above equation, we get
⇒sin⁡2θ=2sin⁡θcos⁡θ−1+1\Rightarrow \sin 2\theta =2\sin \theta \cos \theta -1+1
We know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 now in the place of 11 write sin⁡2θ+cos⁡2θ{{\sin }^{2}}\theta +{{\cos }^{2}}\theta then we get,
⇒sin⁡2θ=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ−1\Rightarrow \sin 2\theta ={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1
Now we write (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab, then we get
⇒sin⁡2θ=(sin⁡θ+cos⁡θ)2−1\Rightarrow \sin 2\theta ={{\left( \sin \theta +\cos \theta \right)}^{2}}-1
Hence we get the left part of the equation(sin⁡θ+cos⁡θ)2−1{{\left( \sin \theta +\cos \theta \right)}^{2}}-1.