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| Product-to-Sum Formulas | Sum-to-Product Formulas | Applications of Transformations in Proving Identities and Solving Problems |
Trigonometric Transformations: Product-to-Sum and Sum-to-Product
Product-to-Sum Formulas
The Product-to-Sum formulas (also known as Product formulas or De Moivre's formulas, though the latter is primarily for complex numbers) are a set of trigonometric identities that allow us to rewrite a product of two sine and/or cosine functions as a sum or difference of sine or cosine functions. These identities are derived directly from the sum and difference identities we discussed earlier.
These formulas are particularly useful in calculus for integration and in other areas of mathematics and physics for simplifying expressions or solving certain types of problems involving oscillations or waves.
Derivation from Sum and Difference Formulas
We begin by recalling the sum and difference identities:
$\sin(A + B) = \sin A \cos B + \cos A \sin B$
... (7.3)
$\sin(A - B) = \sin A \cos B - \cos A \sin B$
... (7.4)
$\cos(A + B) = \cos A \cos B - \sin A \sin B$
... (7.2)
$\cos(A - B) = \cos A \cos B + \sin A \sin B$
... (7.1)
1. Formula for $2 \sin A \cos B$
To obtain a product involving $ \sin A \cos B $, we can add the identities for $ \sin(A + B) $ and $ \sin(A - B) $ (Equations 7.3 and 7.4):
$\sin(A + B) + \sin(A - B) = (\sin A \cos B + \cos A \sin B) + (\sin A \cos B - \cos A \sin B)$
[Adding (7.3) and (7.4)]
Combine the terms on the right side. The $ \cos A \sin B $ terms cancel out:
$ \sin(A + B) + \sin(A - B) = \sin A \cos B + \sin A \cos B $
$ \sin(A + B) + \sin(A - B) = 2 \sin A \cos B $
Rearranging gives the first product-to-sum formula:
$ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) $
$2 \sin A \cos B = \sin(A + B) + \sin(A - B)$
... (8.1)
2. Formula for $2 \cos A \sin B$
To obtain a product involving $ \cos A \sin B $, we subtract the identity for $ \sin(A - B) $ from the identity for $ \sin(A + B) $ (Equation 7.4 from Equation 7.3):
$\sin(A + B) - \sin(A - B) = (\sin A \cos B + \cos A \sin B) - (\sin A \cos B - \cos A \sin B)$
[Subtracting (7.4) from (7.3)]
$ \sin(A + B) - \sin(A - B) = \sin A \cos B + \cos A \sin B - \sin A \cos B + \cos A \sin B $
Combine the terms on the right side. The $ \sin A \cos B $ terms cancel out:
$ \sin(A + B) - \sin(A - B) = \cos A \sin B + \cos A \sin B $
$ \sin(A + B) - \sin(A - B) = 2 \cos A \sin B $
Rearranging gives the second formula:
$ 2 \cos A \sin B = \sin(A + B) - \sin(A - B) $
$2 \cos A \sin B = \sin(A + B) - \sin(A - B)$
... (8.2)
3. Formula for $2 \cos A \cos B$
To obtain a product involving $ \cos A \cos B $, we add the identities for $ \cos(A + B) $ and $ \cos(A - B) $ (Equations 7.2 and 7.1):
$\cos(A + B) + \cos(A - B) = (\cos A \cos B - \sin A \sin B) + (\cos A \cos B + \sin A \sin B)$
[Adding (7.2) and (7.1)]
$ \cos(A + B) + \cos(A - B) = \cos A \cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B $
Combine the terms on the right side. The $ \sin A \sin B $ terms cancel out:
$ \cos(A + B) + \cos(A - B) = \cos A \cos B + \cos A \cos B $
$ \cos(A + B) + \cos(A - B) = 2 \cos A \cos B $
Rearranging gives the third formula:
$ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) $
$2 \cos A \cos B = \cos(A + B) + \cos(A - B)$
... (8.3)
4. Formula for $2 \sin A \sin B$
To obtain a product involving $ \sin A \sin B $, we subtract the identity for $ \cos(A + B) $ from the identity for $ \cos(A - B) $ (Equation 7.2 from Equation 7.1):
$\cos(A - B) - \cos(A + B) = (\cos A \cos B + \sin A \sin B) - (\cos A \cos B - \sin A \sin B)$
[Subtracting (7.2) from (7.1)]
$ \cos(A - B) - \cos(A + B) = \cos A \cos B + \sin A \sin B - \cos A \cos B + \sin A \sin B $
Combine the terms on the right side. The $ \cos A \cos B $ terms cancel out:
$ \cos(A - B) - \cos(A + B) = \sin A \sin B + \sin A \sin B $
$ \cos(A - B) - \cos(A + B) = 2 \sin A \sin B $
Rearranging gives the fourth formula:
$ 2 \sin A \sin B = \cos(A - B) - \cos(A + B) $
$2 \sin A \sin B = \cos(A - B) - \cos(A + B)$
... (8.4)
Note: This formula is sometimes written as $ -2 \sin A \sin B = \cos(A + B) - \cos(A - B) $. Be careful with the order of subtraction on the right side.
Summary of Product-to-Sum Formulas:
Here are the four product-to-sum formulas:
$ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) $
$ 2 \cos A \sin B = \sin(A + B) - \sin(A - B) $
$ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) $
$ 2 \sin A \sin B = \cos(A - B) - \cos(A + B) $
These can also be written by dividing by 2:
$ \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)] $
$ \cos A \sin B = \frac{1}{2}[\sin(A + B) - \sin(A - B)] $
$ \cos A \cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)] $
$ \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] $
Examples
Example 1. Express $ 2 \sin 3x \cos 2x $ as a sum or difference of trigonometric functions.
Answer:
Solution:
We use the product-to-sum formula $ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) $. Here, $ A = 3x $ and $ B = 2x $.
Substitute A and B into the formula:
$ 2 \sin 3x \cos 2x = \sin(3x + 2x) + \sin(3x - 2x) $
Simplify the angles:
$ 2 \sin 3x \cos 2x = \sin(5x) + \sin(x) $
Therefore, $ 2 \sin 3x \cos 2x = \sin 5x + \sin x $.
Example 2. Evaluate $ \cos 75^\circ \cos 15^\circ $.
Answer:
Solution:
We use the product-to-sum formula $ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) $, or its variation $ \cos A \cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)] $.
Here, $ A = 75^\circ $ and $ B = 15^\circ $.
Substitute A and B into the formula:
$ \cos 75^\circ \cos 15^\circ = \frac{1}{2}[\cos(75^\circ + 15^\circ) + \cos(75^\circ - 15^\circ)] $
Simplify the angles:
$ = \frac{1}{2}[\cos(90^\circ) + \cos(60^\circ)] $
Substitute the known values of $ \cos 90^\circ $ and $ \cos 60^\circ $ from the table of specific angles:
$ \cos 90^\circ = 0 $
$ \cos 60^\circ = \frac{1}{2} $
$ = \frac{1}{2}\left[0 + \frac{1}{2}\right] $
$ = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} $
Therefore, $ \cos 75^\circ \cos 15^\circ = \frac{1}{4} $.
Note for Competitive Exams
Product-to-sum formulas are invaluable for transforming products of trigonometric functions into sums or differences, which is often required for integration in calculus or simplifying expressions. Memorising all four forms is recommended. Pay close attention to the coefficient 2 on the left side and the order of terms in the cosine difference for the $ \sin A \sin B $ formula. Practice converting various products into sums/differences. These formulas can also be used in reverse, but the Sum-to-Product formulas are explicitly designed for that transformation.
Sum-to-Product Formulas
The Sum-to-Product formulas (also known as Factorisation formulas) are the reverse of the product-to-sum formulas. They allow us to rewrite a sum or difference of two sine or cosine functions as a product of sine and/or cosine functions. These identities are useful for factoring trigonometric expressions, solving trigonometric equations, and proving other identities, especially those involving sums or differences on one side and products on the other.
Derivation using Substitution
We derive the sum-to-product formulas by making a substitution in the product-to-sum formulas. Let's introduce two new angles, C and D, related to A and B from the product-to-sum formulas:
Let $A + B = C$
... (i)
Let $A - B = D$
... (ii)
Now, we solve this system of linear equations for A and B in terms of C and D.
Adding Equation (i) and Equation (ii):
$ (A + B) + (A - B) = C + D $
$ 2A = C + D $
$ A = \frac{C+D}{2} $
Subtracting Equation (ii) from Equation (i):
$ (A + B) - (A - B) = C - D $
$ A + B - A + B = C - D $
$ 2B = C - D $
$ B = \frac{C-D}{2} $
Now, we substitute $A = \frac{C+D}{2}$, $B = \frac{C-D}{2}$, $A+B=C$, and $A-B=D$ into the product-to-sum formulas (Equations 8.1, 8.2, 8.3, and 8.4).
1. Formula for $ \sin C + \sin D $
Start with the product-to-sum formula: $ 2 \sin A \cos B = \sin(A + B) + \sin(A - B) $ (Equation 8.1)
Substitute A, B, C, D:
$ 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) = \sin C + \sin D $
Rearranging gives the first sum-to-product formula:
$ \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $
$\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$
... (8.5)
2. Formula for $ \sin C - \sin D $
Start with the product-to-sum formula: $ 2 \cos A \sin B = \sin(A + B) - \sin(A - B) $ (Equation 8.2)
Substitute A, B, C, D:
$ 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) = \sin C - \sin D $
Rearranging gives the second formula:
$ \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) $
$\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$
... (8.6)
3. Formula for $ \cos C + \cos D $
Start with the product-to-sum formula: $ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) $ (Equation 8.3)
Substitute A, B, C, D:
$ 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) = \cos C + \cos D $
Rearranging gives the third formula:
$ \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $
$\cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right)$
... (8.7)
4. Formula for $ \cos C - \cos D $
Start with the product-to-sum formula: $ 2 \sin A \sin B = \cos(A - B) - \cos(A + B) $ (Equation 8.4)
Substitute A, B, C, D:
$ 2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) = \cos D - \cos C $
We want the formula for $ \cos C - \cos D $. Multiply both sides by -1:
$ -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) = -(\cos D - \cos C) $
$ \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) $
$\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$
... (8.8)
Alternatively, using the odd property of sine $ \sin(-x) = -\sin x $, we can write $ \sin\left(\frac{C-D}{2}\right) = \sin\left(-\frac{D-C}{2}\right) = -\sin\left(\frac{D-C}{2}\right) $. Substituting this into $ 2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) = \cos D - \cos C $ gives:
$ 2 \sin\left(\frac{C+D}{2}\right) \left[-\sin\left(\frac{D-C}{2}\right)\right] = \cos D - \cos C $
$ -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{D-C}{2}\right) = \cos D - \cos C $
This leads back to the same formula form for $ \cos C - \cos D $ (Equation 8.8), just with the order of C and D swapped inside the second sine term's argument, along with the negative sign.
Summary of Sum-to-Product Formulas:
Here are the four sum-to-product formulas:
$ \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $
$ \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) $
$ \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $
$ \cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) $
Notice the symmetry: when summing like functions (sin+sin, cos+cos), the product involves those functions. When differencing like functions (sin-sin, cos-cos), the product involves the other functions (cos * sin or sin * sin). The $ \cos C - \cos D $ formula is unique with the leading -2 or involving $ \sin(\frac{D-C}{2}) $.
Examples
Example 1. Express $ \sin 5x - \sin 3x $ as a product of trigonometric functions.
Answer:
Solution:
We use the sum-to-product formula for $ \sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right) $. Here, $ C = 5x $ and $ D = 3x $.
Calculate the half sum and half difference of the angles:
$ \frac{C+D}{2} = \frac{5x + 3x}{2} = \frac{8x}{2} = 4x $
$ \frac{C-D}{2} = \frac{5x - 3x}{2} = \frac{2x}{2} = x $
Substitute these into the formula:
$ \sin 5x - \sin 3x = 2 \cos(4x) \sin(x) $
Therefore, $ \sin 5x - \sin 3x = 2 \cos 4x \sin x $.
Example 2. Prove that $ \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x $. (Implicit based on Applications subheading).
Answer:
To Prove:
$ \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x $
Proof:
Start with the Left-Hand Side (LHS):
LHS = $ \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} $
Use the sum-to-product formula for the numerator: $ \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $. Here, $ C = 5x $, $ D = 3x $.
$ \sin 5x + \sin 3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \sin(4x) \cos(x) $
Use the sum-to-product formula for the denominator: $ \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $. Here, $ C = 5x $, $ D = 3x $.
$ \cos 5x + \cos 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \cos(4x) \cos(x) $
Substitute these back into the LHS expression:
LHS = $ \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)} $
Assuming $ \cos x \ne 0 $ and $ \cos 4x \ne 0 $, cancel common factors ($2$ and $ \cos x $):
LHS = $ \frac{\cancel{2} \sin(4x) \cancel{\cos(x)}}{\cancel{2} \cos(4x) \cancel{\cos(x)}} = \frac{\sin(4x)}{\cos(4x)} $
Use the quotient identity $ \tan \theta = \frac{\sin \theta}{\cos \theta} $:
LHS = $ \tan(4x) $
This is equal to the Right-Hand Side (RHS).
Therefore, LHS = RHS. The identity is Proved.
Note for Competitive Exams
Sum-to-product formulas are essential for factorising trigonometric expressions, which is particularly useful when solving trigonometric equations or simplifying complex fractions involving sums/differences. Memorise all four formulas. Pay attention to the $ \cos C - \cos D $ formula, which has a negative sign or involves $ \sin(\frac{D-C}{2}) $. Practice applying these formulas to evaluate sums like $ \sin 75^\circ + \sin 15^\circ $ or prove identities by transforming one side into the other using these formulas. Often, problems require combining sum-to-product formulas in the numerator and denominator, leading to a tangent or cotangent expression, as shown in Example 2.
Applications of Transformations in Proving Identities and Solving Problems
The Product-to-Sum and Sum-to-Product formulas, collectively referred to as transformations or identities, are powerful tools in trigonometry. They allow us to change the form of trigonometric expressions involving sums, differences, or products of sines and cosines. These transformations have significant applications in various areas, including proving trigonometric identities, solving trigonometric equations, and evaluating or simplifying expressions.
1. Proving Trigonometric Identities
Transformations are frequently used to prove other trigonometric identities, especially when one side of the identity involves sums/differences of trigonometric functions with different arguments (angles), and the other side involves products, or vice versa. The strategy often involves converting one side (usually the more complex one) into the form of the other side using these formulas along with other fundamental identities.
Example 1: Proving an Identity using Sum-to-Product Formulas
Example 1. Prove the identity: $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x$.
Answer:
To Prove:
$ \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x $
Proof:
We start with the Left-Hand Side (LHS) of the identity and apply sum-to-product formulas to the numerator and the denominator.
LHS = $ \frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} $
Apply the Sum-to-Product formula for the numerator, which is of the form $ \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $. Here, $ C = 5x $ and $ D = 3x $.
The angles for the formula are:
$ \frac{C+D}{2} = \frac{5x+3x}{2} = \frac{8x}{2} = 4x $
$ \frac{C-D}{2} = \frac{5x-3x}{2} = \frac{2x}{2} = x $
So, the numerator $ \sin 5x + \sin 3x = 2 \sin(4x) \cos(x) $. (Using Equation 8.5)
Apply the Sum-to-Product formula for the denominator, which is of the form $ \cos C + \cos D = 2 \cos\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $. Here, $ C = 5x $ and $ D = 3x $.
The angles for the formula are the same as above:
$ \frac{C+D}{2} = 4x $
$ \frac{C-D}{2} = x $
So, the denominator $ \cos 5x + \cos 3x = 2 \cos(4x) \cos(x) $. (Using Equation 8.7)
Now substitute these expressions back into the LHS:
LHS = $ \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)} $
Assuming $ \cos x \ne 0 $ and $ \cos 4x \ne 0 $, we can cancel the common factors in the numerator and denominator:
LHS = $ \frac{\cancel{2} \sin(4x) \cancel{\cos(x)}}{\cancel{2} \cos(4x) \cancel{\cos(x)}} $
LHS = $ \frac{\sin(4x)}{\cos(4x)} $
Using the quotient identity $ \tan \theta = \frac{\sin \theta}{\cos \theta} $:
LHS = $ \tan(4x) $
This is equal to the Right-Hand Side (RHS).
Therefore, LHS = RHS. The identity is Proved.
2. Solving Trigonometric Equations
Sum-to-Product formulas are particularly useful for solving trigonometric equations where different trigonometric functions or functions with different arguments (angles) are added or subtracted. Converting a sum/difference into a product allows us to use the property that if a product of factors is zero, then at least one of the factors must be zero. This breaks down a complex equation into simpler ones.
Example 2: Solving an Equation using Sum-to-Product Formula
Example 2. Solve the equation $\sin 3\theta + \sin \theta = 0$ for $0 \le \theta < 2\pi$.
Answer:
Given Equation:
$\sin 3\theta + \sin \theta = 0$
Solution:
The given equation is a sum of two sine functions. We can convert this sum into a product using the Sum-to-Product formula: $ \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $. Here, $ C = 3\theta $ and $ D = \theta $.
Calculate the angles for the product formula:
$ \frac{C+D}{2} = \frac{3\theta+\theta}{2} = \frac{4\theta}{2} = 2\theta $
$ \frac{C-D}{2} = \frac{3\theta-\theta}{2} = \frac{2\theta}{2} = \theta $
Substitute these into the formula and the given equation:
$2 \sin(2\theta) \cos(\theta) = 0 $
[Using Sum-to-Product]
For this product to be zero, at least one of the factors must be zero.
This gives us two separate, simpler equations to solve:
Case 1: $ \sin(2\theta) = 0 $
The general solution for $ \sin x = 0 $ is $ x = n\pi $, where $n$ is any integer ($n \in \mathbb{Z}$).
So, $ 2\theta = n\pi $.
Divide by 2: $ \theta = \frac{n\pi}{2} $.
We need to find the values of $ \theta $ in the interval $ 0 \le \theta < 2\pi $. Let's substitute integer values for $n$:
If $n=0$, $ \theta = \frac{0 \times \pi}{2} = 0 $. This is in the interval.
If $n=1$, $ \theta = \frac{1 \times \pi}{2} = \frac{\pi}{2} $. This is in the interval.
If $n=2$, $ \theta = \frac{2 \times \pi}{2} = \pi $. This is in the interval.
If $n=3$, $ \theta = \frac{3 \times \pi}{2} = \frac{3\pi}{2} $. This is in the interval.
If $n=4$, $ \theta = \frac{4 \times \pi}{2} = 2\pi $. This is not included in the interval $ [0, 2\pi) $.
If $n=-1$, $ \theta = \frac{-1 \times \pi}{2} = -\frac{\pi}{2} $. This is outside the interval.
The solutions from Case 1 in the given interval are $ 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} $.
Case 2: $ \cos(\theta) = 0 $
The general solution for $ \cos x = 0 $ is $ x = \frac{\pi}{2} + n\pi $, where $n$ is any integer ($n \in \mathbb{Z}$).
So, $ \theta = \frac{\pi}{2} + n\pi $.
We need to find the values of $ \theta $ in the interval $ 0 \le \theta < 2\pi $. Let's substitute integer values for $n$:
If $n=0$, $ \theta = \frac{\pi}{2} + 0 \times \pi = \frac{\pi}{2} $. This is in the interval.
If $n=1$, $ \theta = \frac{\pi}{2} + 1 \times \pi = \frac{\pi}{2} + \pi = \frac{\pi + 2\pi}{2} = \frac{3\pi}{2} $. This is in the interval.
If $n=2$, $ \theta = \frac{\pi}{2} + 2 \times \pi = \frac{\pi}{2} + 2\pi = \frac{\pi + 4\pi}{2} = \frac{5\pi}{2} $. This is outside the interval.
If $n=-1$, $ \theta = \frac{\pi}{2} - \pi = -\frac{\pi}{2} $. This is outside the interval.
The solutions from Case 2 in the given interval are $ \frac{\pi}{2}, \frac{3\pi}{2} $.
The complete set of distinct solutions in the interval $ [0, 2\pi) $ is the union of the solutions from both cases:
$ \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\} \cup \{\frac{\pi}{2}, \frac{3\pi}{2}\} = \{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}\} $.
Therefore, the solutions for $ \theta $ in the interval $ 0 \le \theta < 2\pi $ are $0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$.
3. Evaluating or Simplifying Expressions
Product-to-Sum formulas are useful for evaluating expressions that involve products of trigonometric functions of angles which might not be standard angles themselves, but whose sums or differences are standard angles.
Example 3: Evaluating a Product using Product-to-Sum Formula
Example 3. Express $2 \cos(75^\circ) \cos(15^\circ)$ as a sum and find its value.
Answer:
Solution:
We are asked to express $ 2 \cos(75^\circ) \cos(15^\circ) $ as a sum. We use the Product-to-Sum formula: $ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) $. Here, $ A = 75^\circ $ and $ B = 15^\circ $.
Calculate the sum and difference of the angles:
$ A + B = 75^\circ + 15^\circ = 90^\circ $
$ A - B = 75^\circ - 15^\circ = 60^\circ $
Substitute these into the formula:
$ 2 \cos(75^\circ) \cos(15^\circ) = \cos(90^\circ) + \cos(60^\circ) $
This is the required expression as a sum.
Now, find the value of this sum. We know the values of $ \cos 90^\circ $ and $ \cos 60^\circ $ from the table of specific angles:
$ \cos 90^\circ = 0 $
$ \cos 60^\circ = \frac{1}{2} $
Substitute these values:
$ 2 \cos(75^\circ) \cos(15^\circ) = 0 + \frac{1}{2} = \frac{1}{2} $
Therefore, the value of $ 2 \cos(75^\circ) \cos(15^\circ) $ is $\frac{1}{2}$.
Similarly, Sum-to-Product formulas can be used to evaluate sums or differences. For example, to find $ \sin 75^\circ + \sin 15^\circ $, we use the formula $ \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right) \cos\left(\frac{C-D}{2}\right) $ with $ C = 75^\circ $ and $ D = 15^\circ $.
$ \sin 75^\circ + \sin 15^\circ = 2 \sin\left(\frac{75^\circ+15^\circ}{2}\right) \cos\left(\frac{75^\circ-15^\circ}{2}\right) $
$ = 2 \sin\left(\frac{90^\circ}{2}\right) \cos\left(\frac{60^\circ}{2}\right) = 2 \sin(45^\circ) \cos(30^\circ) $
$ = 2 \times \left(\frac{1}{\sqrt{2}}\right) \times \left(\frac{\sqrt{3}}{2}\right) = \frac{2\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}} $
Rationalising the denominator: $ \frac{\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{2} $.
So, $ \sin 75^\circ + \sin 15^\circ = \frac{\sqrt{6}}{2} $.
Note for Competitive Exams
Proficiency in using Product-to-Sum and Sum-to-Product formulas is crucial for simplifying expressions and solving equations in competitive exams. Recognising when to apply these formulas based on the structure of the expression (sum/difference vs. product) is key. For equations, converting a sum/difference to a product allows you to set factors to zero. For identities, choose the side that benefits most from the transformation. Practice solving various problems involving these formulas to build speed and confidence. Remember the conditions for which the formulas are valid (e.g., denominators not being zero).