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Question: How do you solve \(\cos (\theta ) - \sin (\theta ) = 1?\)...

How do you solve cos⁡(θ)−sin⁡(θ)=1?\cos (\theta ) - \sin (\theta ) = 1?

Explanation

Solution

Try to convert the equation in a form such that you can apply compound angle formula of trigonometric identities to the given trigonometric equation.
Since, this equation contains sin⁡\sin and cos⁡\cos functions only, so you should divide the equation (i.e. both sides) with (coefficient  of  sin⁡θ)2+(coefficient  of  cos⁡θ)2\sqrt {{{(coefficient\;of\;\sin \theta )}^2} + {{(coefficient\;of\;\cos \theta )}^2}} in order to get the required equation in which you can easily apply compound angle formula of trigonometric identities.

Complete answer:
The given trigonometric equation which has to be solved is
cos⁡(θ)−sin⁡(θ)=1\cos (\theta ) - \sin (\theta ) = 1
Now, we should divide both the sides with (coefficient  of  sin⁡θ)2+(coefficient  of  cos⁡θ)2\sqrt {{{(coefficient\;of\;\sin \theta )}^2} + {{(coefficient\;of\;\cos \theta )}^2}}
As from the equation we came to know that coefficient of sin⁡θ\sin \theta is 11 and coefficient of cos⁡θ\cos \theta is also 11
So the required divisor will be
=(coefficient  of  sin⁡θ)2+(coefficient  of  cos⁡θ)2 =12+12 =1+1 =2  = \sqrt {{{(coefficient\;of\;\sin \theta )}^2} + {{(coefficient\;of\;\cos \theta )}^2}} \\\ = \sqrt {{1^2} + {1^2}} \\\ = \sqrt {1 + 1} \\\ = \sqrt 2 \\\
Now dividing both sides of the equation cos⁡(θ)−sin⁡(θ)=1\cos (\theta ) - \sin (\theta ) = 1 with 2\sqrt 2 , we will get
⇒cos⁡(θ)−sin⁡(θ)2=12 ⇒cos⁡(θ)2−sin⁡(θ)2=12  \Rightarrow \dfrac{{\cos (\theta ) - \sin (\theta )}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow \dfrac{{\cos (\theta )}}{{\sqrt 2 }} - \dfrac{{\sin (\theta )}}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }} \\\
Can we simplify this furthermore? Or write it all in a trigonometric equation?
Yes, we can write this, as we know that
sin⁡π4=cos⁡π4=12\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}
So we can replace 12\dfrac{1}{{\sqrt 2 }} in terms of sin⁡π4\sin \dfrac{\pi }{4} and cos⁡π4\cos \dfrac{\pi }{4}
Now replacing and rewriting the equation as
⇒cos⁡(θ)×cos⁡π4−sin⁡(θ)×sin⁡π4=12\Rightarrow \cos (\theta ) \times \cos \dfrac{\pi }{4} - \sin (\theta ) \times \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}
Are we familiar with this type of equation?
Actually yes, you should have seen this type of trigonometric equation before in the compound angle formula of trigonometric identities.
And this trigonometric equation is looking parallel to this trigonometric identity
cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)\cos x\cos y - \sin x\sin y = \cos (x + y)
Isn’t it similar to our equation,
So let us continue solving our equation with help of the above trigonometric identity.
⇒cos⁡(θ)×cos⁡π4−sin⁡(θ)×sin⁡π4=12 ⇒cos⁡(θ+π4)=12  \Rightarrow \cos (\theta ) \times \cos \dfrac{\pi }{4} - \sin (\theta ) \times \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} \\\ \Rightarrow \cos \left( {\theta + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }} \\\
We know that the general solution of cos⁡x=12\cos x = \dfrac{1}{{\sqrt 2 }} is
x=2nπ±π4x = 2n\pi \pm \dfrac{\pi }{4} , where n∈In \in I
That means,
⇒θ+π4=2nπ±π4\Rightarrow \theta + \dfrac{\pi }{4} = 2n\pi \pm \dfrac{\pi }{4}, where n∈In \in I
⇒θ=2nπ±π4−π4\Rightarrow \theta = 2n\pi \pm \dfrac{\pi }{4} - \dfrac{\pi }{4}, where n∈In \in I
Therefore general solution of cos⁡(θ)−sin⁡(θ)=1\cos (\theta ) - \sin (\theta ) = 1 is θ=2nπ±π4−π4\theta = 2n\pi \pm \dfrac{\pi }{4} - \dfrac{\pi }{4}

Note: General equation of cos⁡x=12\cos x = \dfrac{1}{{\sqrt 2 }} is x=2nπ±π4x = 2n\pi \pm \dfrac{\pi }{4} , where n∈In \in I but if you can’t find it in the options, then convert the terms into sine angles compound formula, i.e. sin⁡xcos⁡y−sin⁡ycos⁡x=sin⁡(x−y)\sin x\cos y - \sin y\cos x = \sin (x - y).