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Question: If \(\cos \theta + \sin \theta = \sqrt 2 \sin \theta \) , then \(\sin \theta - \cos \theta \) ?...

If cos⁡θ+sin⁡θ=2sin⁡θ\cos \theta + \sin \theta = \sqrt 2 \sin \theta , then sin⁡θ−cos⁡θ\sin \theta - \cos \theta ?

Explanation

Solution

First, we shall analyze the equation so that we can able to solve the given problem. Here, the given equation is cos⁡θ+sin⁡θ=2sin⁡θ\cos \theta + \sin \theta = \sqrt 2 \sin \theta and we are asked to calculate the value of sin⁡θ−cos⁡θ\sin \theta - \cos \theta
Now before getting into the solution, we need to consider (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2{\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2} . That is, we need to solve (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2{\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2} so that we can able to find sin⁡θ−cos⁡θ\sin \theta - \cos \theta

Formula to be used:
a) The required algebraic identities that are to be applied to the given problem are as follows.
(a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}
(a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}
b) The required trigonometric identities that are to be applied to the given problem are as follows.
sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
1−sin⁡2θ=cos⁡2θ1 - {\sin ^2}\theta = {\cos ^2}\theta

Complete step-by-step answer:
The given equation is cos⁡θ+sin⁡θ=2sin⁡θ\cos \theta + \sin \theta = \sqrt 2 \sin \theta and we are asked to calculate sin⁡θ−cos⁡θ\sin \theta - \cos \theta
Before getting into the solution, we shall consider (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2{\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2}
We shall solve (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2{\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2}to obtain the required answer.
{\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2}$$$ = {\sin ^2}\theta + 2\sin \theta \cos \theta + {\cos ^2}\theta + {\sin ^2}\theta - 2\sin \theta \cos \theta + {\cos ^2}\theta $ (In the above equation, we have applied the algebraic identities ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ ) Then, we have {\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2} = {\sin ^2}\theta + {\cos ^2}\theta + {\sin ^2}\theta + {\cos ^2}\theta $ $ = 2{\sin ^2}\theta + 2{\cos ^2}\theta $ $ = 2\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)$ $ = 2$ (Here we applied ${\sin ^2}\theta + {\cos ^2}\theta = 1$) $${\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2} = 2Sincewearegiventhat Since we are given that\cos \theta + \sin \theta = \sqrt 2 \sin \theta , we put it in the above equation. Thus, we get $${\left( {\sqrt 2 \sin \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2} = 2$$ \Rightarrow {\sqrt 2 ^2}{\sin ^2}\theta + {\left( {\sin \theta - \cos \theta } \right)^2} = 2 \Rightarrow 2{\sin ^2}\theta + {\left( {\sin \theta - \cos \theta } \right)^2} = 2 \Rightarrow {\left( {\sin \theta - \cos \theta } \right)^2} = 2 - 2{\sin ^2}\theta \Rightarrow {\left( {\sin \theta - \cos \theta } \right)^2} = 2\left( {1 - {{\sin }^2}\theta } \right) \Rightarrow {\left( {\sin \theta - \cos \theta } \right)^2} = 2{\cos ^2}\theta (Herewehaveappliedtheidentity(Here we have applied the identity1 - {\sin ^2}\theta = {\cos ^2}\theta )Now,weshalltakethesquarerootsonbothsidesoftheaboveequation.Thus,weget) Now, we shall take the square roots on both sides of the above equation. Thus, we get \sin \theta - \cos \theta = \pm \sqrt {2{{\cos }^2}\theta } \Rightarrow \sin \theta - \cos \theta = \pm \sqrt 2 \cos \theta Therefore,wehavefoundtherequiredanswer Therefore, we have found the required answer\sin \theta - \cos \theta = \pm \sqrt 2 \cos \theta $

Note: If we start with the given equationcos⁡θ+sin⁡θ=2sin⁡θ\cos \theta + \sin \theta = \sqrt 2 \sin \theta , we cannot get the value for sin⁡θ−cos⁡θ\sin \theta - \cos \theta . So, we need to consider (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2{\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2}
And, when we are solving (sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2{\left( {\sin \theta + \cos \theta } \right)^2} + {\left( {\sin \theta - \cos \theta } \right)^2}, we will automatically get the required answer. Also, we should know to use the appropriate trigonometric identities to solve the trigonometric expression or equation. Here, we have used algebraic identities too to obtain the desired answer.