# If Sin(theta) - Cos(theta) = -sqrt(2) sin(theta), what is Sin(theta) + Cos (theta)?

##### 2 Answers
Jun 7, 2016

$\sin \theta + \sqrt{2} \sin \theta = \cos \theta$

$\sin \theta \left(1 + \sqrt{2}\right) = \cos \theta$

$1 + \sqrt{2} = \cos \frac{\theta}{\sin} \theta$

By the quotient identity $\cos \frac{\theta}{\sin} \theta = \cot \theta$

$1 + \sqrt{2} = \cot \theta$

We must now determine the values of $\sin \theta$ and $\cos \theta$.

First, we must define $\cot \theta$: $\cot \theta = \text{adjacent"/"opposite}$.

Therefore, the side adjacent $\theta$ measures $1 + \sqrt{2}$ and the side opposite measures $1$. We will need to find the hypotenuse for $\sin \theta$ and $\cos \theta$.

Let y be the hypotenuse. By pythagorean theorem we have

${\left(1 + \sqrt{2}\right)}^{2} + {1}^{2} = {y}^{2}$

$1 + 2 \sqrt{2} + 2 + 1 = {y}^{2}$

$4 + 2 \sqrt{2} = {y}^{2}$

$\sqrt{4 + 2 \sqrt{2}} = y$

The definition of $\sin \theta = \text{opposite"/"hypotenuse}$ and the definition of $\cos \theta = \text{adjacent"/"hypotenuse}$

Therefore, sintheta = 1/(sqrt(4 + 2(sqrt(2))) and costheta = (1 + sqrt(2))/(sqrt(4 + 2(sqrt(2))

Since we have been asked to evaluate $\sin \theta + \cos \theta$ we add our two expressions. Luckily, they are on equal denominators!

$\cos \theta + \sin \theta = \pm \frac{2 + \sqrt{2}}{\sqrt{4 + 2 \left(\sqrt{2}\right)}}$

= +-sqrt(1 + sqrt(2)/2

Hopefully this helps!

Jun 8, 2016

#### Answer:

$\pm \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2}}}$

#### Explanation:

General discussion to determine the sign of $\implies \sin \theta + \cos \theta$

From the given condition

$\sin \theta + \sqrt{2} \sin \theta = \cos \theta$

$\implies \left(\sqrt{2} + 1\right) \sin \theta = \cos \theta$

$\therefore \sin \frac{\theta}{\cos} \theta = \frac{1}{\sqrt{2} + 1} \ldots . \left(1\right)$

$\tan \theta = \frac{1}{\sqrt{2} + 1} > 0$

$\tan \theta$ being positve $\theta$ should be in 1st or 3rd quadrant.when $\theta$ is in the 1st quadrant is positive but in 3rd quadrant $\sin \theta \mathmr{and} \cos \theta$ both negative and the the value $\sin \theta + \cos \theta$should be negative.

So $\textcolor{red}{\sin \theta + \cos \theta \text{ will be either +ve or -ve}}$

Method -I (A tricky approach)

From equation (1) have

$\sin \frac{\theta}{\cos} \theta = \frac{1}{\sqrt{2} + 1} \ldots . \left(1\right)$

and

Inverting and ratiolising numerator of RHS

$\cos \frac{\theta}{\sin} \theta = \sqrt{2} + 1 = \frac{1}{\sqrt{2} - 1} \ldots . \left(2\right)$

Adding equation(1) and equation (2)we get

$\sin \frac{\theta}{\cos} \theta + \cos \frac{\theta}{\sin} \theta = \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{2} - 1}$

$\frac{{\sin}^{2} \theta + {\cos}^{2} \theta}{\sin \theta \cos \theta} = \frac{2 \sqrt{2}}{\left(\sqrt{2} - 1\right) \left(\sqrt{2} + 1\right)} = \frac{2 \sqrt{2}}{1}$

Inverting and rearranging we get

$\frac{2 \sin \theta \cos \theta}{{\sin}^{2} \theta + {\cos}^{2} \theta} = \frac{1}{\sqrt{2}}$

Adding 1 on both sides we get

$1 + \frac{2 \sin \theta \cos \theta}{{\sin}^{2} \theta + {\cos}^{2} \theta} = 1 + \frac{1}{\sqrt{2}}$

${\left(\sin \theta + \cos \theta\right)}^{2} / \left({\sin}^{2} \theta + {\cos}^{2} \theta\right) = \frac{\sqrt{2} + 1}{\sqrt{2}}$

$\implies {\left(\sin \theta + \cos \theta\right)}^{2} = \frac{\sqrt{2} + 1}{\sqrt{2}}$

$\implies \sin \theta + \cos \theta = \pm \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2}}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method - II

Given
$\sin \theta - \cos \theta = - \sqrt{2} \sin \theta$

Dividing both sides by$\sqrt{2} \text{ }$we have

$\implies \frac{1}{\sqrt{2}} \sin \theta - \frac{1}{\sqrt{2}} \cos \theta = - \sin \theta$

(Considering unit of angle in degree.)

$\implies \cos \left(45\right) \sin \theta - \sin \left(45\right) \cos \theta = - \sin \theta \ldots \left(1\right)$

$\implies \sin \left(\theta - 45\right) = \sin \left(- \theta\right)$

$\implies \theta - 45 = - \theta$

$\implies 2 \theta = 45$

$\implies \theta = 22.5$

Another solution is posible satisfying positive value of $\tan \theta$ when $\theta$ is in third quadrant

Then from eq (1)

$\sin \left(\theta - 45\right) = \sin \left(360 - \theta\right)$
$\implies \theta = \frac{405}{2} = 180 + 22.5$

Now

$\sin \theta + \cos \theta$

$= \sqrt{2} \left(\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta\right)$
$= \sqrt{2} \sin \left(\theta + 45\right) \ldots . \left(2\right)$

color (blue)("when " " theta=22.5

Inserting $\theta = 22.5$

$= \sqrt{2} \sin \left(22.5 + 45\right)$

$= \sqrt{2} \sin \left(90 - 22.5\right)$

$= \sqrt{2} \cos \left(22.5\right)$

$= \sqrt{2} \sqrt{\frac{1}{2} \left(1 + \cos 45\right)}$

$= \sqrt{1 + \cos 45}$

$= \sqrt{1 + \frac{1}{\sqrt{2}}}$

$= \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2}}}$

color (green)("Again when " " theta=180+22.5

we put $\theta = 180 + 22.5$

in eq(2)

we get

=>sintheta+costheta =sqrt2sin(theta+45)

$= \sqrt{2} \sin \left(180 + 22.5 + 45\right)$

$= \sqrt{2} \sin \left(180 + 22.5 + 45 + 22.5 - 22.5\right)$

$= \sqrt{2} \sin \left(270 - 22.5\right)$

$= - \sqrt{2} \cos \left(22.5\right)$

$= - \sqrt{2} \left(\sqrt{\frac{1}{2} \left(1 + \cos 45\right)}\right)$

$= - \sqrt{2} \left(\sqrt{\frac{1}{2} \left(1 + \frac{1}{\sqrt{2}}\right)}\right)$

$= - \sqrt{1 + \frac{1}{\sqrt{2}}}$

$= - \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2}}}$

So combining these two we get

$\sin \theta + \cos \theta = \pm \sqrt{\frac{\sqrt{2} + 1}{\sqrt{2}}}$