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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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