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# Calculus/pre cal 12th Circular functions II

Question
prove the identities

sin (x+y)- sin (x-y) =2 cos x sin y

Find the exact value of cos 75°

First you must know that :

sin (x+y) = sin(x) cos(y) + cos(x) sin(y)  and
sin (x-y) = sin(x) cos(y) - cos(x) sin(y).

Knowing the above two identities, we see that

sin (x+y)- sin (x-y)
= (  sin(x) cos(y) + cos(x) sin(y)) -  (sin(x) cos(y) - cos(x) sin(y))
= sin(x) cos(y) + cos(x) sin(y) - sin(x) cos(y) + cos(x) sin(y)
=          cos(x) sin(y)          + cos(x) sin(y)
= 2 cos(x) sin(y)

Now substitute x = 75°  and y = 15°,  we get

sin (75°+15°) - sin (75°-15°) = 2 cos ( 75°) * sin (15°)
that implies

sin (90°)      - sin (60°)      =  2 cos ( 75°) * sin (15°)

We know sin (90°) and  sin (60°) because both 60° and 90° are special angles.

But how about sin (15°)?  There are two different ways:

I) We can use the power reducing formula that says:
2 sin^2 (15°) = 1- cos( 30°)

Since 30° is a special angle and we know  cos( 30°) = sqrt(3)/2, therefore

sin(15°) = sqrt( (1-  sqrt(3)/2 )/2  )

So now we have

1         -  sqrt(3)/2      = 2 cos ( 75°) * sqrt( (1-sqrt(3)/2 )/2)

I leave it to you to simplify and get solve for cos(75°).

II)  75° and  15° are complementary because 75° and  15° together make a right angle.
So  sin (15°) = cos ( 75°) .

So we have from the equation:

1         -  sqrt(3)/2      = 2 cos ( 75°) *  cos ( 75°)

So that  (cos(75°))^2 = 1  -  sqrt(3)/2 )/2.

Solving for cos(75°) by taking the positive square root yield your answer.

Calculus

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