Sin Cos Formulas

The basic trigonometric functions are the sin and cos formulas which relate to the angles and the ratios of the sides of a right-angled triangle. The sine of an angle is the ratio of the opposite side and the hypotenuse and the cosine of an angle is the ratio of the adjacent side and the hypotenuse. These form fundamental identities that are defined for acute angles. The extension of these ratios to any angle in terms of radian measure is called the trigonometric function. Sin is positive in the first and second quadrant and cos is positive in the first and fourth quadrant. The range of the sine and cosine functions is [-1,1] under the real number domain. 

What Are Sin Cos Formulas?

If (x,y) is a point on the unit circle, and if a ray from the origin (0, 0) to (x, y) makes an angle θ from the positive axis, then x and y satisfy the Pythagorean theorem x² + y² = 1, where x and y form the lengths of the legs of the right-angled-triangle. Thus the basic sin cos formula becomes cos²θ + sin²θ = 1.

There are many identities related to the sine and cosine that are applied in the trigonometric functions. All the trigonometric expressions are simpler to evaluate using these trigonometric formulas. Let us discuss them in detail.

Sin Cos Formulas

For any acute angle of θ, the functions of negative angles are: 

• sin(-θ) = – sinθ 
• cos (-θ) = cosθ

Identities expressing trig functions in terms of their complements:

• cosθ = sin(90° - θ)
• sinθ = cos(90° - θ)

Sum and Difference of Sin Cos Formulas

An angle that is made up of the sum or difference of two or more angles is called the compound angle. Let us denote the compound angles as α and β. There are Sin Cos Formulas with respect to compound angles for expanding or simplifying trigonometric expressions. Let us investigate them.

• sin (α + β) = sin α cos β + cos α sin β
• sin (α – β) = sin α cos β – cos α sin β
• cos (α + β) = cos α cos β – sin α sin β
• cos (α – β) = cos α cos β + sin α sin β

Transformation of Sin and Cos Formulas

There are a few identities that we pick from one side to work and make substitutions until the side is transformed to the other side. To verify an identity, we rewrite any side of the equation and transform it to the other side. From the above-mentioned sum and difference identities, we derive the product-to-sum and the sum-to-product formulas.

Product-to-sum formulas are applied when given a product of cosines, We express the product as a sum or difference, write the formula, substitute the given angles and finally simplify. 

• 2 sin α cos β = sin (α +β) + sin (α – β)
• 2 cos α sin β = sin (α + β) – sin (α – β)
• 2 cos α cos β = cos (α + β) + cos (α – β)
• 2 sin α sin β = cos (α – β) – cos (α + β)

The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas are as given below,

• sin α + sin β = 2 sin((α+β)/2) cos((α−β)/2)
• sin α – sin β = 2 cos((α+β)/2) sin((α−β)/2)
• cos α + cos β= 2 cos((α+β)/2) cos((α−β)/2)
• cos α – cos β = -2 sin((α+β)/2) sin((α-β)/2)

Derivation of Product to Sum Formulas

Here we express products of cosine and sine as a sum. We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get:

   cosα cosβ + sinα sinβ = cos(α − β)
+ cosα cosβ − sinα sinβ = cos(α + β)
––----------------------------––-------------
  2cosα cosβ = cos(α−β) + cos(α + β)
––----------------------------––-------------

Then, divide by 2 and isolate the product of cosines: cosα cosβ = (1/2)[cos(α−β) + cos(α+β)]

Similarly, we can derive the other formulas by expressing the products as the sum/difference.

Derivation of Sum to Product Formulas 

There are a few problems that require the reverse of the product to sum. Let us see the derivation of these sum to product formulas. For this let us use a few substitutions as (u+v)/2 = α,  (u- v)/2 = β

Then α + β = [(u+v)/2] + [(u- v)/2] = u 

α - β = [(u + v)/2] - [(u- v)/2] = v

Let us derive the sum-to-product formula. We replace α and β in the product-to-sum formula.

Consider (sinα cosβ) = (1/2)[sin(α + β) + sin(α - β)]

Substituting for (α + β) and αβ, we get

sin((u+v)/2) cos ((u-v)/2) = 1/2[sinu + sin v] 

2sin((u+v)/2)) cos ((u-v)/2) = sinu + sin v

Similarly, we can derive the other sum-to-product identities.

Sin Cos Formulas of Multiple Angles

We have double and triple angles formula and the half-angle formulas as follows:

• sin 2θ = 2 sinθ cosθ
• sin 3θ = 3 sinθ - 4 sin³θ
• cos 2θ = cos² θ - sin² θ
• cos 2θ = 2cos²θ - 1
• cos 2θ = 1- 2sin² θ
• cos 3θ = 4 cos³θ - 3cosθ
• sin (θ/2) = ± √((1- cosθ)/2)
• cos (θ/2) = ± √((1+ cosθ)/2)
• sin θ = 2tan (θ/2) /(1 + tan² (θ/2))
• cos θ = (1-tan² (θ/2))/(1 + tan² (θ/2))

Examples Using Sin Cos Formulas

Example 1: When, sin X = 1/2 and cos Y = 3/4 then find cos(X+Y)

Solution: We know cos(X + Y) = cos X cos Y – sin X sin Y

Given sin X = 1/2

We know that, cos X = √(1 - sin²X) = √(1 - (1/4)) = √3/2

Thus, cos X = √3/2

Given cos Y = 3/4

We know that, sin Y = √(1 - cos²Y) = √(1 - (9/16)) = √7/4

Thus, sin Y = √7/4

cos X = √3/2, and sinY = √7/4

Applying the sum of cos formula, we have cos(X+Y) = (√3/2) × (3/4) –  1/2 × (√7/4)

= (3√3 - √7)/8

Answer: cos(X+Y) = (3√3 - √7)/8

Example 2: If sin θ = 3/5, find sin2θ.

Solution: We know that sin2θ = 2 sin θ cos θ

We need to determine cos θ.

Let us use the sin cos formula cos²θ + sin²θ = 1.

Rewriting, we get cos²θ = 1 -  sin²θ

= 1-(9/25)

cos²θ = 16/25

cos θ = 4/5

sin2θ = 2 sin θ cos θ

= 2 × (3/5) × (4/5) = 24/25

Answer: sin2θ = 24/25

Example 3: Prove (cos 4a - cos 2a)/ (sin 4a + sin 2a) = -tan a.

Solution: Using the sin cos formula, let us rewrite the LHS and transform it to the RHS

\(=\dfrac{-2\sin(\dfrac{4a+2a}{2})\sin(\dfrac{4a-2a}{2})}{2\sin(\dfrac{4a+2a}{2})\cos(\dfrac{4a-2a}{2})}\)

\(=\dfrac{-2\sin(3a) sina}{2\sin(3a) cosa}\)

= - sina /cosa

= −tan a

Thus proved.

Answer: (cos 4a - cos 2a)/ (sin 4a + sin 2a) = -tan a.