Sin(X-y)sin(x+y)=sin^2 x - sin^2 y
work on one side only...so i worked on the right
=(sinx-siny)(sinx+siny) does that equal
sin(x-y)sin(x+y)??? help!
To determine if (sinx-siny)(sinx+siny) is equal to sin(x-y)sin(x+y), we can use a trigonometric identity.
The identity that involves the product of two sine functions is:
sin A sin B = 1/2 [cos(A-B) - cos(A+B)]
Using this identity, we can rewrite the expression sin(x-y)sin(x+y):
sin(x-y)sin(x+y) = 1/2 [cos(x-y-x-y) - cos(x-y+x+y)]
Simplifying this further, we get:
1/2 [cos(-2y) - cos(2x)]
Now, let's simplify the expression (sinx-siny)(sinx+siny):
(sin x - sin y)(sin x + sin y) = sin^2 x - sin^2 y
We can see that the two expressions match:
1/2 [cos(-2y) - cos(2x)] = sin^2 x - sin^2 y
Thus, we have shown that (sinx-siny)(sinx+siny) is equal to sin(x-y)sin(x+y).
To determine if the expression (sinx - siny)(sinx + siny) is equal to sin(x - y)sin(x + y), we can expand both expressions and compare the results.
Expanding (sinx - siny)(sinx + siny):
= sinx*sinx + sinx*(-siny) + (-siny)*sinx + (-siny)*(-siny)
= sin^2 x - sinx*siny - sinx*siny + sin^2 y
= sin^2 x - 2sinx*siny + sin^2 y
Expanding sin(x - y)sin(x + y):
= (sinx*cosy - cosx*siny)(sinx*cosy + cosx*siny)
= sinx*cosy*sinx*cosy + sinx*cosy*cosx*siny - cosx*siny*sinx*cosy - cosx*siny*cosx*siny
= sin^2 x*cos^2 y + sinx*cosy*cosx*siny - cosx*siny*sinx*cosy - cos^2 x*sin^2 y
As we can see, the expanded forms of (sinx - siny)(sinx + siny) and sin(x - y)sin(x + y) are not equal. Therefore, (sinx - siny)(sinx + siny) does not equal sin(x - y)sin(x + y).
Use the formulas for sin (A-B) and sin (A+B)
The left side is
(sinx cos y - siny cosx)(sinx cosy + siny cosx)
= sin^2x cos^2y - sin^2y cos^2 x
=(1 - cos^2x)(cos^2y)-sin^2y cos^2x
= cos^2y -cos^2y cos^2x - sin^2y cos^2x
= cos^2y -cos^2x
= 1-sin^2y -1 + sin^2x
= sin^2x - sin^2y