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Question: How do you find the rectangular coordinate of \(\left( {\sqrt 2 ,{{315}^ \circ }} \right)\)?...

How do you find the rectangular coordinate of (2,315∘)\left( {\sqrt 2 ,{{315}^ \circ }} \right)?

Explanation

Solution

Here, we are given the polar coordinates. We are asked to convert them into rectangular coordinates. For this, we need to use the relation between polar and rectangular coordinates.
Formula used:
x=rcos⁡θx = r\cos \theta
y=rsin⁡θy = r\sin \theta
Where, (x,y)\left( {x,y} \right) are the rectangular coordinates and (r,θ)\left( {r,\theta } \right) are the polar coordinates.

Complete step by step answer:
We are given the polar coordinates (2,315∘)\left( {\sqrt 2 ,{{315}^ \circ }} \right).
Therefore, we can say that (r,θ)=(2,315∘)\left( {r,\theta } \right) = \left( {\sqrt 2 ,{{315}^ \circ }} \right) and we have to find the rectangular coordinates (x,y)\left( {x,y} \right).
Now, we will use the formulas for converting polar coordinates into rectangular coordinates.
x=rcos⁡θ ⇒x=2cos⁡315∘  x = r\cos \theta \\\ \Rightarrow x = \sqrt 2 \cos {315^ \circ } \\\
We know that cos⁡315∘=cos⁡(360−45)=cos⁡45=12\cos {315^ \circ } = \cos \left( {360 - 45} \right) = \cos 45 = \dfrac{1}{{\sqrt 2 }}.
⇒x=2×12=1\Rightarrow x = \sqrt 2 \times \dfrac{1}{{\sqrt 2 }} = 1
Similarly,

y=rsin⁡θ ⇒y=2sin⁡315∘  y = r\sin \theta \\\ \Rightarrow y = \sqrt 2 \sin {315^ \circ } \\\

We know that sin⁡315∘=sin⁡(360−45)=−sin⁡45=−12\sin {315^ \circ } = \sin \left( {360 - 45} \right) = - \sin 45 = - \dfrac{1}{{\sqrt 2 }}.
⇒y=2×(−12)=−1\Rightarrow y = \sqrt 2 \times \left( { - \dfrac{1}{{\sqrt 2 }}} \right) = - 1
Therefore, by converting the polar coordinates (2,315∘)\left( {\sqrt 2 ,{{315}^ \circ }} \right)in rectangular coordinates, we get (1,−1)\left( {1, - 1} \right)
(1,−1)\left( {1, - 1} \right) as our final answer.

Note: While solving this type of question, we need to be very careful with the sign of sine and cosine function. Here, we have taken cos⁡315∘=cos⁡(360−45)=cos⁡45=12\cos {315^ \circ } = \cos \left( {360 - 45} \right) = \cos 45 = \dfrac{1}{{\sqrt 2 }} , because 315 comes in the fourth quadrant where the cosine function is positive. Whereas the sine function is negative in the fourth quadrant and therefore we have taken sin⁡315∘=sin⁡(360−45)=−sin⁡45=−12\sin {315^ \circ } = \sin \left( {360 - 45} \right) = - \sin 45 = - \dfrac{1}{{\sqrt 2 }} .
In short we have to keep in mind these four points related to the signs of the main trigonometric functions based on the location or the quadrants.
In the first quadrant, all the trigonometric functions are positive.
In the second quadrant, only sine function is positive.
In the third quadrant, only tangent function is positive.
In the fourth quadrant, only cosine function is positive.