Question
Question: How do you solve the equation \({{\left( \sin x-\cos x \right)}^{2}}+\sin 2x\)...
How do you solve the equation (sinx−cosx)2+sin2x
Solution
Now to solve the given equation we will first open the bracket by the formula (a−b)2=a2−2ab+b2 Now we know that the value of sin2x+cos2x=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 . 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 (sinx−cosx)2+sin2x .
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 (a−b)2=a2−2ab+b2 .
Here we have a=sinx and b=cosx ,
Hence using this we get the expression as sin2x−2sinxcosx+cos2x+sin2x
Now rearranging the terms of the expression we get,
⇒(sin2x+cos2x)−2sinxcosx+sin2x
Now we know the identity sin2x+cos2x=1 hence using this we get,
⇒1−2sinxcosx+sin2x .
Now let us expand sin2x using the double angle formula. We know that sin2x=2sinxcosx .
Hence using this formula we get the expression as
⇒1−sin2xcos2x+sin2xcos2x
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 Hence on substituting the value of B as A we get the double angle formula for sin. Now we also know that sinθ=hypotenuseopposite side and cosθ=hypotenuseadjacent side . Now considering sin2θ+cos2θ 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.