# sin (A + B)

How to find sin (A + B) by using its formula: formula, 2 examples, and their solutions.

## Formula

### Formula

sin (A + B) = sin A cos B + cos A sin B

For sine,
cos and sin are mixed: sin cos, cos sin
and the middle sign doesn't change: (+) → (+).

sin (A - B)

## Example 1

### Solution

Set 105º = 60º + 45º.

sin (60º + 45º)
= sin 60º cos 45º + cos 60º sin 45º

To find these trigonometric function values,

draw a 30-60-90 triangle
whose sides are 1, √3, 2,

and a 45-45-90 triangle
whose sides are 1, 1, √2.

sin 60º

Sine is SOH:
Sine,
Opposite side (√3),
Hypotenuse (2).

So sin 60º = √3/2.

cos 45º

Cosine is CAH:
Cosine,
Hypotenuse (√2).

So cos 45º = 1/√2.

Write +.

cos 60º

Cosine is CAH:
Cosine,
Hypotenuse (2).

So cos 60º = 1/2.

sin 45º

Sine is SOH:
Sine,
Opposite side (1),
Hypotenuse (√2).

So sin 45º = 1/√2.

So sin 60º cos 45º + cos 60º sin 45º
= [√3/2]⋅[1/√2] + [1/2]⋅[1/√2].

[√3/2]⋅[1/√2] + [1/2]⋅[1/√2]
= (√3 + 1)/2√2

To rationalize the denominator 2√2,
multiply [√2/√2].

(√3 + 1)√2 = √6 + √2

2√2⋅√2 = 2⋅2

2⋅2 = 4

So (√6 + √2)/4 is the answer.

## Example 2

### Solution

To find the amplitude,
combine sin x and √3 cos x
by using sin (A + B) formula.

First write
y = 1 sin x + √3 cos x.

sin (x + B) = sin x cos B + cos x sin B

to change 1 to cos B
and to change √3 to sin B.

Cosine is CAH: related to the adjacent side.
Sine is SOH: related to the opposite side.

So draw a right triangle
and whose opposite side is √3.

Find the hypotenuse
by using the Pythagorean theorem:
12 + (√3)2 = 22.

Then the hypotenuse is 2.

This is a right triangle
whose sides are 1, √3, and 2.

So this is a 30-60-90 triangle.

So the bottom angle is 60º.

Let's change
y = 1 sin x + √3 cos x.

First write the hypotenuse 2 and (.

1 sin x
= 2([1/2] sin x)

+√3 cos x
= 2( +[√3/2] cos x)

So 1 sin x + √3 cos x
= 2([1/2] sin x + [√3/2] cos x).

Next, change [1/2] and [√3/2]
to cosine and sine.

First write 2(.

To change [1/2],
see the right triangle.

Cosine is CAH:
Cosine,
Hypotenuse (2).

So 1/2 = cos 60º.

Write sin x.
Write +.

To change [√3/2],
see the right triangle.

Sine is SOH:
Sine,
Opposite side (√3),
Hypotenuse (2).

So √3/2 = sin 60º.

Write cos x).

So 2([1/2] sin x + [√3/2] cos x)
= 2(cos 60º sin x + sin 60º cos x).

cos 60º sin x + sin 60º cos x
= sin x cos 60º + cos x sin 60º

sin x cos 60º + cos x sin 60º
= sin (x + 60º)

See y = 2 sin (x + 60º).
The number in front of the sine is 2.

So the amplitude is
|2| = 2.

Sine: Graph