# Sin Cos Formulas

Sin Cos formulas are based on the sides of the right-angled triangle.Â Sin and Cos are basic trigonometric functions along with tan function, in trigonometry. The sine of an angle is equal to the ratio of the opposite side to the hypotenuse whereas the cosine of an angle is equal to the ratio of the adjacent side to the hypotenuse.

• Sin Î¸ = Opposite side/Hypotenuse
• Cos Î¸ = Adjacent side/ Hypotenuse

## Basic Trigonometric Identities for Sin and Cos

These formulas help in giving a name to each side of the right triangle and these are also used in trigonometric formulas for class 11. Letâ€™s learn the basic sin and cos formulas.

• cos2(A) + sin2(A) = 1

## Sine and Cosine Formulas

To get help in solving trigonometric functions, you need to know the trigonometry formulas.

### Half-angle formulas

Sin
$$\begin{array}{l}\frac{A}{2}\end{array}$$
=
$$\begin{array}{l}\pm \sqrt{\frac{1- Cos A}{2}}\end{array}$$
• If A/2 is in the first or second quadrants, the formula uses the positive sign.
• If A/2 is in the third or fourth quadrants, the formula uses the negative sign
Cos
$$\begin{array}{l}\frac{A}{2}\end{array}$$
=
$$\begin{array}{l}\pm \sqrt{\frac{1+ Cos A}{2}}\end{array}$$
• If A/2 is in the first or fourth quadrants, the formula uses the positive sign.
• If A/2 is in the second or third quadrants, the formula uses the negative sign

### Double and Triple angle formulas

• Sin 2A = 2Sin A Cos A
• Cos 2A = Cos2A – Sin2A = 2 Cos2Â A- 1 = 1- Sin2A
• Sin 3A = 3Sin A – 4 Sin 3A
• Cos 3A = 4 Cos3A – 3CosA
• Sin2A =
$$\begin{array}{l}\frac{1 – Cos(2A)}{2}\end{array}$$
• Cos2A =
$$\begin{array}{l}\frac{1 + Cos(2A)}{2}\end{array}$$

### Sum and Difference of Angles

• sin(A + B) = sin(A).cos(B) + cos(A)sin(B)
• sin(Aâˆ’B)=sin(A)â‹…cos(B)âˆ’cos(A)â‹…sin(B)
• cos(A+B)=cos(A)â‹…cos(B)âˆ’sin(A)â‹…sin(B)
• cos(Aâˆ’B)=cos(A)â‹…cos(B)+sin(A)â‹…sin(B)
• sin(A+B+C)=sinAâ‹…cosBâ‹…cosC+cosAâ‹…sinBâ‹…cosC+cosAâ‹…cosBâ‹…sinCâˆ’sinAâ‹…sinBâ‹…sinC
• cos (A + B +C) = cos A cos B cos C- cos A sin B sin C â€“ sin A cos B sin C â€“ sin A sin B cos C
• Sin A + Sin B = 2Sin
$$\begin{array}{l}\frac{(A+B)}{2}\end{array}$$
Cos
$$\begin{array}{l}\frac{(A-B)}{2}\end{array}$$
• Sin A – Sin B = 2Sin
$$\begin{array}{l}\frac{(A-B)}{2}\end{array}$$
Cos
$$\begin{array}{l}\frac{(A+B)}{2}\end{array}$$
• Cos A + Cos B = 2Cos
$$\begin{array}{l}\frac{(A+B)}{2}\end{array}$$
Cos
$$\begin{array}{l}\frac{(A-B)}{2}\end{array}$$
• Cos A – Cos B = -2Sin
$$\begin{array}{l}\frac{(A+B)}{2}\end{array}$$
Sin
$$\begin{array}{l}\frac{(A-B)}{2}\end{array}$$

### Multiple Angle Formulas

Sin (2Î¸) = 2 sin Î¸ cos Î¸

Cos (2Î¸) = cos2Î¸ – sin2Î¸ =2 cos2Î¸ -1 = 1- 2 sin2Î¸

### Product To Sum Formulas

2 cos Î¸ cos Ï† = cos (Î¸-Ï†) + cos (Î¸+Ï†)

2 sinÎ¸ sin Ï† = cos (Î¸-Ï†) – cos (Î¸+Ï†)

2 sin Î¸ cos Ï† = sinÂ  (Î¸+Ï†) + sin (Î¸-Ï†)

2 cos Î¸ sin Ï† = sin (Î¸+Ï†) – sin (Î¸-Ï†)

### Periodic Identities

 Sin ((Ï€/2) – x) = cos x Cos ((Ï€/2) – x) = sin x Sin ((Ï€/2) + x) = cos x Cos ((Ï€/2) + x) = -sin x Sin ((3Ï€/2) – x) = -cos x cos ((3Ï€/2) – x) = -sin x Sin ((3Ï€/2) + x) = -cos x Cos ((3Ï€/2) + x) =Â  sin x Sin (Ï€ – x) = sin x Cos (Ï€ – x) = – cos x Sin (Ï€ + x) = -sin x Cos (Ï€ + x) = -cos x Sin (2Ï€ – x) = -sin x Cos (2Ï€ – x) = cos x Sin (2Ï€ + x) = sin x Cos (2Ï€ + x) = cos x

### Sine Law

$$\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\end{array}$$

### Cosine Law

a2 = b2+c2 – 2bc cos A

b2Â = c2+ a2Â – 2ca cos B

c2Â = a2+b2Â – 2ab cos C

### Product Identities

Sin x cos y = (Â½)[sin (x+y) + sin (x-y)]

Cos xÂ  sin y = (Â½) [sin (x+y)- sin (x-y)]

Cos x cos y = (Â½)[cos(x-y) + cos (x+y)]

Sin x sin y = (Â½) [cos (x-y) – cos (x+y)]

### Example on Sin Cos Formula

Example:

Find the value of sin 20Â° sin 40Â° sin 60Â° sin 80Â°.

Solution:

Given: sin 20Â° sin 40Â° sin 60Â° sin 80Â°

This can be written as:

â‡’ sin 60Â° sin 20Â° sin 40Â° sin 80Â°

Substitute sin 60Â° = âˆš3/2

â‡’ (âˆš3/2) sin 20Â° (sin 60Â°-20Â°) sin (60Â°+20Â°)

â‡’ (âˆš3/2) sin 20Â° (sin2 60Â° – sin2 20Â°)

â‡’ (âˆš3/2)(Â¼) [3 sin 20Â° – 4 sin3 20Â°]

Simplifying 3 sin 20Â° – 4 sin3 20Â°, we get sin 60Â°

Now substitute the values,

â‡’(âˆš3/2)(Â¼)(âˆš3/2)

â‡’(Â¾)(Â¼)

â‡’ 3/16

Hence, the value of sin 20Â° sin 40Â° sin 60Â° sin 80Â° is 3/16.

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