Question

if sin x = siny how is x+y =180 degrees ​

Answer 1

sin x=90 degrees

sin y=90 degrees

so x + y =180 degree

.... 90 degrees + 90 degrees=180 degrees

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Answer 2

Refer the attachment given below also.  

The figure shows a circle centered at O(0, 0) of radius OA = OB = OC = 1 unit and in the circle two right triangles are seen. Here  ∠AOB = ∠COD = x  and  ∠AOC = ∠BOD = 180° - x = y.  

[Note: Don't take ∠BOC = 180° - x = y. Sometimes it may be confused from the figure.]

"Note that  180° - x  is taken instead of  y  since  x + y = 180° ⇒ y = 180° - x"  

Here  B  and  C  are two points on the circle with arbitrary coordinates such that B and C should lie in first quadrant and second quadrant respectively.  

Why it is said that B and C should lie in these two quadrants?!

We have to prove that  x + y = 180°  if  sin x  may be equal to  sin y.  

Since  x + y = 180°,  

    →  both  x  and  y  are less than 180°.  

    →  both  x  and  y  can be equal to 90°.  

    →  x  and  y  can be equal to 0° and 180° respectively and vice versa.  

First condition implies that x and y lie anywhere in the first and second quadrants.  

We can ignore second and third conditions, as they only imply about sin 0°, sin 90° and sin 180°, which does not imply a universal proof for the question.  

But what if  x ≠ y ?!  

$x+y=180\textdegree\ \ \ \Longrightarrow\ \ \ x+y=90\textdegree+k+90\textdegree-k\ \ \ [\textsf{Let}]$

$\boxed{\begin{minipage}{11.44 cm}\large \textbf{\underline{\underline{Case 1:}}}\\ \\ \normalsize\textsf{Let \ x=90\textdegree+k\ \ \ \&\ \ \ y=90\textdegree -k }\\ \\ \Longrightarrow\ x>90\textdegree\ \ \ \&\ \ \ y<90\textdegree\\ \\ \\ \\ \large \textbf{\underline{\underline{Case 2:}}}\\ \\ \normalsize\textsf{Let \ x=90\textdegree-k\ \ \ \&\ \ \ y=90\textdegree +k }\\ \\ \Longrightarrow\ x<90\textdegree\ \ \ \&\ \ \ y>90\textdegree\end{minipage}}$

I just took the second case, thus  x  and  y  lies in first quadrant and second quadrant respectively, and so do points B and C.  

Consider triangles OEB and ODC. They're right angled at E and D respectively.  

$\begin{aligned}&\angle OEB&=&\ \ \angle ODC&=&\ \ 90\textdegree \\ \\ \&\ \ \ \ &\angle EOB&=&\ \ \angle DOC&=&\ \ x\\ \\ \Longrightarrow\ \ &\angle OBE&=&\ \ \angle OCD&=&\ \ 90\textdegree-x\\ \\ \\ &\angle EOB&=&\ \ \angle DOC\\ \\ \&\ \ \ \ &OB&=&\ \ OC\\ \\ \&\ \ \ \ &\angle OBE&=&\ \ \angle OCD\\ \\ \Longrightarrow\ \ &\triangle OEB&\cong &\ \ \triangle ODC\\ \\ \Longrightarrow\ \ &OE&=&\ \ OD&=&\ \ a&&[\textsf{Let}]\\ \\ \&\ \ \ \ &EB&=&\ \ DC&=&\ \ b&&[\textsf{Let}]\end{aligned}$

These thereby imply,  

$B(a,\ b)\\ \\ \\ \&\\ \\ \\ C(-a,\ b)$

And now we remember that,  

In a circle of radius 1 unit, usually centered at the origin, if a leg of an angle at the center of the circle meets the circle at a point, say P, with the other leg along the x axis towards ∞⁺, then the sine and cosine values of that angle will always be the y coordinate and x coordinate of the point P respectively.

In other words,

Assume a circle centered at O(0, 0) of radius 1 unit and let there exists  ∠AOB = θ  where A and B are two points on the circle such that the y coordinate of A is 0 and B has arbitrary coordinates, say B(m, n),  ∀m, n ∈ [-1, 1] ∧ m, n ∈ R. Then,

  →   sin θ = n

  →   cos θ = m

So,

Consider  ∠AOB = ∠EOB = x  and  ∠AOC = ∠EOC = 180° - x.  Since OA and OE lie on x axis, or we can say that since the y coordinates of both A and E are 0, then,

• sin x = b

• sin (180° - x) = b

Both imply,

sin x = sin (180° - x)

Taking  180° - x = y,  we get,

sin x = sin y

Now I proved here that  sin x  will be equal to  sin y  if  x + y = 180°.

Conversely, we can say that, if  x + y = 180°,  then  sin x  'can be' equal to  sin y, 'but not always.'

Hence Proved!