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Question: How do you solve the equation \({{\left( \sin x-\cos x \right)}^{2}}+\sin 2x\)...

How do you solve the equation (sinxcosx)2+sin2x{{\left( \sin x-\cos x \right)}^{2}}+\sin 2x

Explanation

Solution

Now to solve the given equation we will first open the bracket by the formula (ab)2=a22ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} Now we know that the value of sin2x+cos2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 . Now we will substitute the value obtained in the given equation. Now again we will use the double angle formula for sin which is sin2x=2sinxcosx\sin 2x=2\sin x\cos x . Now we will substitute the value of sin2x and simplify the equation hence we will get the value of the required equation.

Complete step by step solution:
Now consider the given expression (sinxcosx)2+sin2x{{\left( \sin x-\cos x \right)}^{2}}+\sin 2x .
To solve the given equation we will use the algebraic properties and trigonometric properties of sin and cos.
First we will open the square bracket and simplify. We know that (ab)2=a22ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} .
Here we have a=sinxa=\sin x and b=cosxb=\cos x ,
Hence using this we get the expression as sin2x2sinxcosx+cos2x+sin2x{{\sin }^{2}}x-2\sin x\cos x+{{\cos }^{2}}x+\sin 2x
Now rearranging the terms of the expression we get,
(sin2x+cos2x)2sinxcosx+sin2x\Rightarrow \left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)-2\sin x\cos x+\sin 2x
Now we know the identity sin2x+cos2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 hence using this we get,
12sinxcosx+sin2x\Rightarrow 1-2\sin x\cos x+\sin 2x .
Now let us expand sin2x using the double angle formula. We know that sin2x=2sinxcosx\sin 2x=2\sin x\cos x .
Hence using this formula we get the expression as
1sin2xcos2x+sin2xcos2x\Rightarrow 1-\sin 2x\cos 2x+\sin 2x\cos 2x
Hence on subtraction we get the value of the expression as 1.
Hence the value of the given expression is equal to 1.

Note:
Now note that for sin function we have sin(A+B)=sinAcosB+cosAsinB\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B Hence on substituting the value of B as A we get the double angle formula for sin. Now we also know that sinθ=opposite sidehypotenuse\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}} and cosθ=adjacent sidehypotenuse\cos \theta =\dfrac{\text{adjacent side}}{\text{hypotenuse}} . Now considering sin2θ+cos2θ{{\sin }^{2}}\theta +{{\cos }^{2}}\theta and using the Pythagoras theorem we will get the value as 1. Hence we can easily derive the two formulas used to solve the given problem.