Question

If sin 17 degree equal to x by y then the value of 617 degree minus sin 73 degree is

Answer 1

Explanation:

By the cofunction identity:
sin(73∘)=cos(90∘−73∘)=cos(17∘)sin(73∘)=cos(90∘-73∘)=cos(17∘)

So we have sec(17∘)−cos(17∘)sec(17∘)-cos(17∘). We know that sec(x)=1cos(x)sec(x)=1cos(x) so we can rewrite this expression:

1cos(17∘)−cos(17∘)=1−cos2(17∘)cos(17∘)1cos(17∘)-cos(17∘)=1-cos2(17∘)cos(17∘)

By the Pythagorean ID, sin2(x)+cos2(x)=1sin2(x)+cos2(x)=1and so 1−cos2(x)=sin2(x)1-cos2(x)=sin2(x):

sin2(17∘)cos(17∘)sin2(17∘)cos(17∘)

Rearranging sin2(x)+cos2(x)=1sin2(x)+cos2(x)=1 we can also see that cos(x)=√1−sin2(x)cos(x)=1-sin2(x).

(Since 17∘17∘ is in QI all trig functions are positive.)

sin2(17∘)√1−sin2(17∘)sin2(17∘)1-sin2(17∘).

Now we substitute xyxy for sin(17∘)sin(17∘):

(xy)2√1−(xy)2(xy)21-(xy)2.

Now we do some algebra:

x2y2√1−x2y2=x2y2√y2−x2y2x2y21-x2y2=x2y2y2-x2y2

=x2y2(1y)√y2−x2=y⋅(x2y2)√y2−x2=x2y2(1y)y2-x2=y⋅(x2y2)y2-x2

=x2y√y2−x2=x2y⋅√y2−x2=x2yy2-x2=x2y⋅y2-x2, as required