Question

If sin (45 -x) = cos (y + 45), then the value of sin (x -y) is​

Answer 1

Answer:

Step-by-step explanation:

sin (45 -x) = cos (y + 45)

We know that cos(90-A)=sin A

RHS cos(y+45)

=cos ( 90-45+y)

=cos { 90-(45-y) )

=sin(45-y)

Now putting  LHS = RHS

sin(45-x)= sin(45-y)

or 45-x=45-y

x-y=0

sin(x-y)= sin 0=0

Answer 2

Answer: The value of $sin (x -y)$ is​ 0.

Given: $sin (45 -x) = cos (y + 45)$

To Find: The value of $sin (x -y)$.

Explanation:

Here, $sin (45 -x) = cos (y + 45)$

It is known that $cos(90-A)=sin A$

Then, $cos(y+45) =cos ( 90-45+y)$

$=cos ( 90-(45-y) )$

$=sin(45-y)$

When substituting the value of $cos (y + 45)$ in $(45 -x) = cos (y + 45)$, we get

$sin(45-x)= sin(45-y)$

⇒$45-x=45-y$

⇒$x-y=0$

⇒$sin(x-y)= sin 0=0$

Therefore, $sin(x-y)=0$.

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