Sum-to-Product and Product-to-Sum Identities

 

1. Sum-to-Product Identities:

 

Let

\[ \left. \begin{array}{l} a + b = x \\ a \ – \ b = y \end{array} \right\} \Rightarrow a = \frac{x+y}{2} \ \text{ and} \ b = \frac{x-y}{2}\]

Then, by manipulating the sum and difference identities:

\[ \begin{array} \cos(a + b) & = \cos a \cos b \ – \ \sin a \sin b \\ \pm \ \cos(a – b) & = \cos a \cos b + \sin a \sin b \\ \hline \cos(a + b) & + \cos(a – b) = 2 \cos a \cos b \end{array} \]

We derive:

\[ 1) \ \cos x + \cos y = 2 \cos \frac{x+y}{2} \cos \frac{x-y}{2} \]

Similarly:

\[ \cos(a + b) – \cos(a – b) = -2 \sin a \sin b \]
\[ 2) \ \cos x – \cos y = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \]

For sine:

\[ \begin{array} \sin(a + b) & = \sin a \cos b + \sin b \cos a \\ \pm \ \sin(a – b) & = \sin a \cos b – \sin b \cos a \\ \hline \sin(a + b) & + \sin(a – b) = 2 \sin a \cos b \end{array} \]
\[ 3) \ \sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2} \]

And:

\[ \sin(a + b) – \sin(a – b) = 2 \sin b \cos a \]
\[ 4) \ \sin x – \sin y = 2 \sin \frac{x-y}{2} \cos \frac{x+y}{2} \]

For tangent, by finding a common denominator:

\[\tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y } \]
\[ = \frac{\sin x \cos y + \sin y \cos x}{\cos x \ \cos y} \]
\[ 5) \ \tan x + \tan y = \frac{\sin(x+y)}{\cos x \ \cos y} \]

Substituting \(-y\) for \(y\) in the previous identity yields:

\[ 6) \ \tan x \ – \tan y = \frac{\sin(x-y)}{\cos x \ \cos y} \]

Similarly, for cotangent:

\[\cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y}\]
\[ = \frac{\cos x \sin y + \sin x \cos y} {\sin x \ \sin y} \]
\[ 7) \ \cot x + \cot y = \frac{\sin (x + y)} {\sin x \ \sin y} \]

Substituting \(-y\) for \(y\) in the cotangent identity yields:

\[ 8) \ \cot x \ – \ \cot y = \frac{-\sin (x – y)}{\sin x \ \sin y} \]

 

Example:

 

Evaluate the expression \(\cos 72^\circ – \cos 36^\circ\), and use the result to determine the exact value of \(\sin 18^\circ\).

\[ = \cos 72^\circ – \cos 36^\circ\]
\[ = -2 \sin \frac{72^\circ + 36^\circ}{2} \cdot \sin \frac{72^\circ – 36^\circ}{2} \]
\[ = -2 \sin 54^\circ \cdot \sin 18^\circ \]

Multiply the numerator and the denominator by \(\cos 18^\circ\) to apply double-angle identities:

\[ = \frac{-2 \sin 54^\circ \cdot \sin 18^\circ \cdot \cos 18^\circ}{\cos 18^\circ} \]

\[ = -\frac{\cos 36^\circ \cdot \sin 36^\circ}{\cos 18^\circ} \]

\[ = -\frac{\frac{1}{2} \sin 72^\circ}{\sin 72^\circ} = -\frac{1}{2} \]

Now, let’s solve for \(\sin 18^\circ\):

\[ \cos 72^\circ – \cos 36^\circ = -\frac{1}{2} \]

Using cofunction identities, substitute \(\cos 72^\circ = \sin 18^\circ\):

\[ \Rightarrow \sin 18^\circ – \cos 36^\circ = -\frac{1}{2} \]

Apply the double-angle identity \(\cos 2\theta = 1 – 2\sin^2\theta\):

\[ \Rightarrow \sin 18^\circ \ – \ (1 \ -\ 2 \sin^2 18^\circ) = -\frac{1}{2} \]
\[ \Rightarrow 4 \sin^2 18^\circ + 2 \sin 18^\circ – 1 = 0 \]

Let \( \sin 18^\circ = t \):

\[ \Rightarrow 4t^2 + 2t -1 = 0 \]

Applying the quadratic formula and keeping the positive root since \(18^\circ\) is in Quadrant I:

\[ \Rightarrow t =\sin 18^\circ = \frac{-1 + \sqrt{ 5} }{4} \]

 

Example:

 

In \(\triangle ABC\), let us find an alternative expression for \[\frac{b+c}{b-c}\]
Applying the Law of Sines to \(\triangle ABC\):
\[ \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\]
\[\begin{array}{l} b = 2R \sin B \\ c = 2R \sin C \end{array} \]

Substituting these values into the ratio:
\[ \frac{b + c}{b – c} = \frac{2R \sin B + 2R \sin C}{2R \sin B \ – \ 2R \sin C} \]
\[ = \frac{\sin B + \sin C}{\sin B \ -\ \sin C} \]

Applying the sum-to-product identities to both the numerator and the denominator:
\[ = \frac{2 \sin \frac{B + C}{2} \cdot \cos \frac{B – C}{2}}{2 \sin \frac{B – C}{2} \cdot \cos \frac{B + C}{2}} \]
\[ = \tan \frac{B + C}{2} \cdot \cot \frac{B – C}{2} \]
\[ = \frac{\tan \frac{B + C}{2}}{\tan \frac{B – C}{2}} \]

This result is famously known as the Law of Tangents.

 

Question 35

 

Evaluate the following trigonometric statement: \(\cos 80^\circ + \cos 20^\circ + \sqrt{3} \cos 130^\circ\)

 

\[ A) -\cos 10^\circ \quad B) -\sin 10^\circ \quad C) \cos 10^\circ \quad D) \sin 10^\circ \quad E) 0 \]

 

Solution:

 

\[ \cos 80^\circ + \cos 20^\circ + \sqrt{3} \cos 130^\circ \]
Applying the sum-to-product identity to the first two terms:
\[ = 2 \cos \frac{80^\circ + 20^\circ}{2} \cdot \cos \frac{80^\circ – 20^\circ}{2} \ – \ \sqrt{3} \cos 50^\circ \]
\[ = 2 \cos 50^\circ \cdot \cos 30^\circ \ – \ \sqrt{3} \cos 50^\circ \]
Since \(\cos 30^\circ = \frac{\sqrt{3}}{2}\):
\[ = 2 \cdot \frac{\sqrt{3}}{2} \cos 50^\circ \ – \ \sqrt{3} \cos 50^\circ = 0 \]

 

\( \text{Correct Answer: E} \)

 

Question 36

 

Given \(10x = \pi\), find the value of the rational expression:
\[ \frac{cos 7x + \cos 5x}{\cos 5x + \cos 3x} \]

 

\[ A) \ 1 \quad B) \ -1 \quad C) \ 2 \quad D) \ -2 \quad E) \ \frac{1}{2} \]

 

Solution:

 

Apply the sum-to-product identities:
\[ \frac{\cos 7x + \cos 5x}{\cos 5x + \cos 3x} = \frac{2 \cos \frac{7x+5x}{2} \cdot \cos \frac{7x-5x}{2}}{2 \cos \frac{5x+3x}{2} \cdot \cos \frac{5x-3x}{2}} \]
\[ = \frac{\cos 6x \cdot \cos x}{\cos 4x \cdot \cos x} = \frac{\cos 6x}{\cos 4x} \]

Since \(10x = \pi \implies 6x + 4x = \pi \implies 6x = \pi – 4x\):
\[ = \frac{\cos (\pi – 4x)}{\cos 4x} = \frac{-\cos 4x}{\cos 4x} = -1 \]

 

\( \textbf{Correct Answer: B} \)

 

Question 37

 

Evaluate the following statement:
\[ \frac{\cos 75^\circ + \cos 15^\circ}{\cos 75^\circ – \cos 15^\circ} \]

 

\[ A) \ -\sqrt{3} \quad B) \ \sqrt{3} \quad C) \ -\frac{1}{2} \quad D) \frac{1}{2} \quad E) \ -\sqrt{3} \]

 

Solution:

 

Applying sum-to-product formulas:
\[ \frac{\cos 75^\circ + \cos 15^\circ}{\cos 75^\circ – \cos 15^\circ} \]
\[ = \frac{2 \cos \frac{75^\circ + 15^\circ}{2} \cdot \cos \frac{75^\circ – 15^\circ}{2}}{-2 \sin \frac{75^\circ + 15^\circ}{2} \cdot \sin \frac{75^\circ – 15^\circ}{2}} \]
\[= -\frac{\cos 45^\circ \cdot \cos 30^\circ }{\sin 45^\circ \cdot \sin 30^\circ } = -\cot 45^\circ \cdot \cot 30^\circ \]
\[ = -1 \cdot \sqrt{3} = -\sqrt{3} \]

*(Note: The options provided in the source text contain an error or misprint in the answer key tag; the correct value evaluates to \(-\sqrt{3}\).)*

 

\( \textbf{Correct Answer: A} \)

 

Question 38

 

Given \(20x = \pi\), find the value of:
\[ \frac{\cos 4x \ – \ \cos 8x}{\cos 4x \cdot \cos 8x} \]

 

\[ A) \ \frac{1}{2} \quad B) \ -1 \quad C) \ 1 \quad D) \ -2 \quad E) \ 2 \]

 

Solution:

 

Apply sum-to-product identities to the numerator:
\[ \frac{\cos 4x – \cos 8x}{\cos 4x \cdot \cos 8x} = \frac{-2 \sin \frac{4x + 8x}{2} \cdot \sin \frac{4x – 8x}{2}}{\cos 4x \cdot \cos 8x} \]
\[ = \frac{-2 \sin 6x \cdot \sin (-2x)}{\cos 4x \cdot \cos 8x} = \frac{2 \sin 6x \cdot \sin 2x}{\cos 4x \cdot \cos 8x} \]

Since \(20x = \pi \implies 10x = \frac{\pi}{2}\), we use cofunction transitions \(\cos \theta = \sin(\frac{\pi}{2} – \theta)\):
\(\cos 4x = \cos(\frac{\pi}{2} – 6x) = \sin 6x\) and \(\cos 8x = \cos(\frac{\pi}{2} – 2x) = \sin 2x\).
Substituting these back:
\[ = \frac{2 \sin 6x \cdot \sin 2x}{\sin 6x \cdot \sin 2x} = 2 \]

 

\( \textbf{Correct Answer: E} \)
 

Question 39

 

Which of the following is equivalent to the expression below?
\[ \frac{\sin (a + b \,- \, c) + \sin (a \, – \, b + c)}{\cos (a + b\, – \,c) + \cos (a \,- \,b + c)} \]

 

\[ A) \ 1 \quad B) \ \tan (b – c) \quad C) \ \cot (b – c) \quad D) \ \tan a \quad E) \ \cot a \]

 

Solution:

 

Let \[ \left. \begin{array} a + b \,-\, c = x \\ a \,-\, b + c = y \end{array} \right \} \Rightarrow \frac{x+y}{2} = a \] and \[ \Rightarrow \frac{x-y}{2} = b \;- c \]

Substituting \(x\) and \(y\) into our expression:
\[ \frac{\sin x + \sin y}{\cos x + \cos y} = \frac{2 \sin \frac{x+y}{2} \cdot \cos \frac{x\;-y}{2}}{2 \cos \frac{x+y}{2} \cdot \cos \frac{x\;-y}{2}} \]
\[ = \frac{\sin a}{\cos a} = \tan a \]

 

\( \textbf{Correct Answer: D} \)

 

Question 40

 

Evaluate the following trigonometric operation:
\[ \sin 50^\circ \; – \; \frac{\cos^2 10^\circ}{\sin 50^\circ} \]

 

\[ A) \ 3 \quad B) \ 2 \quad C) \ \frac{1}{2} \quad D) \ -2 \quad E) \ -\frac{1}{2} \]

 

Solution:

 

Find a common denominator:
\[ \sin 50^\circ – \frac{\cos^2 10^\circ}{\sin 50^\circ} = \frac{\sin^2 50^\circ – \cos^2 10^\circ}{\sin 50^\circ} \]
Rewrite \(\sin^2 50^\circ\) as \(\cos^2 40^\circ\) using cofunctions, then factor using a difference of squares:
\[ = \frac{\cos^2 40^\circ – \cos^2 10^\circ}{\sin 50^\circ} = \frac{(\cos 40^\circ – \cos 10^\circ)(\cos 40^\circ + \cos 10^\circ)}{\sin 50^\circ} \]

Apply sum-to-product identities to both factors:
\[ = \frac{(-2 \sin 25^\circ \cdot \sin 15^\circ) \cdot (2 \cos 25^\circ \cdot \cos 15^\circ)}{\sin 50^\circ} \]
Regroup to form double-angle components \(2\sin 25^\circ\cos 25^\circ = \sin 50^\circ\):
\[ = \frac{-2 \sin 15^\circ \cdot \cos 15^\circ \cdot (\sin 50^\circ)}{\sin 50^\circ} \]
\[ = -2 \sin 15^\circ \cdot \cos 15^\circ = -\sin 30^\circ = -\frac{1}{2} \]

 

\( \textbf{Correct Answer: E} \)

 

Question 41

 

Evaluate the following expression:
\[ \frac{1}{\sin 54^\circ} \; – \; \frac{1}{\cos 72^\circ} \]

 

\[ A) \ 0 \quad B) \ -1 \quad C) \ 1 \quad D) \ -2 \quad E) \ 2 \]

 

Solution:

 

Find a common denominator:

\[ \frac{1}{\sin 54^\circ} \; – \frac{1}{\cos 72^\circ} = \frac{\cos 72^\circ – \sin 54^\circ}{\sin 54^\circ \cdot \cos 72^\circ} \]
Using cofunctions, substitute \(\cos 72^\circ = \sin 18^\circ\) and \(\sin 54^\circ = \cos 36^\circ\):
\[ = \frac{\sin 18^\circ – \sin 54^\circ}{\cos 36^\circ \cdot \sin 18^\circ} \]

Apply sum-to-product identity to the numerator:

\[ = \frac{2 \cos \left( \frac{18^\circ + 54^\circ}{2} \right) \cdot \sin \left( \frac{18^\circ – 54^\circ}{2} \right)}{\cos 36^\circ \cdot \sin 18^\circ} \]
\[ = \frac{2 \cos 36^\circ \cdot \sin (-18^\circ)}{\cos 36^\circ \cdot \sin 18^\circ} \]
Since \(\sin(-18^\circ) = -\sin 18^\circ\):
\[ = \frac{-2 \cos 36^\circ \cdot \sin 18^\circ}{\cos 36^\circ \cdot \sin 18^\circ} = -2 \]

 

\( \textbf{Correct Answer: D} \)

 

Question 42

 

Simplify the following expression:
\[ \frac{\sin 6^\circ + \sin 12^\circ + \sin 18^\circ}{1 + \cos 6^\circ + \cos 12^\circ} \]

 

\[ A) \ 2 \sin 6^\circ \quad B) \ -2 \sin 6^\circ \quad C) \ -1 \quad D) \ 1 \quad E) \ 2 \]

 

Solution:

 

Rearrange the numerator and expand the double-angle in the denominator (\(\cos 12^\circ = 2\cos^2 6^\circ – 1\)):

\[ = \frac{(\sin 18^\circ + \sin 6^\circ) + \sin 12^\circ}{1 + \cos 6^\circ + (2 \cos^2 6^\circ – 1)} \]

Apply sum-to-product to the grouped term in the numerator:

\[ = \frac{2 \sin 12^\circ \cdot \cos 6^\circ + \sin 12^\circ}{\cos 6^\circ + 2 \cos^2 6^\circ} \]

Factor out common terms:

\[ = \frac{\sin 12^\circ (2 \cos 6^\circ + 1)}{\cos 6^\circ (1 + 2 \cos 6^\circ)} = \frac{\sin 12^\circ}{\cos 6^\circ} \]

Apply the double-angle formula \(\sin 12^\circ = 2\sin 6^\circ\cos 6^\circ\):

\[ = \frac{2 \sin 6^\circ \cdot \cos 6^\circ}{\cos 6^\circ} = 2 \sin 6^\circ \]

 

\( \textbf{Correct Answer: A} \)

 

Question 43

 

Let \(A = \cos 9x + \cos 7x + \cos 3x + \cos x\). Evaluate the expression below:

\[ \frac{A}{\cos x \cdot \cos 3x \cdot \cos 5x} \]

 

\[ A) \ 1 \quad B) \ 2 \quad C) \ 3 \quad D) \ 4 \quad E) \ 5 \]

 

Solution:

 

Group the terms in \(A\) and apply sum-to-product identities:

\[ A = (\cos 9x + \cos 7x) + (\cos 3x + \cos x) \]
\[ = 2 \cos 8x \cdot \cos x + 2 \cos 2x \cdot \cos x \]
\[ = 2 \cos x \cdot (\cos 8x + \cos 2x) \]

Apply the identity once more inside the parenthesis:

\[ = 2 \cos x \cdot (2 \cos 5x \cdot \cos 3x) = 4 \cos x \cdot \cos 3x \cdot \cos 5x \]

Substituting this back into our rational statement:

\[ \frac{4 \cos x \cdot \cos 3x \cdot \cos 5x}{\cos x \cdot \cos 3x \cdot \cos 5x} = 4 \]

 

\( \textbf{Correct Answer: D} \)

 

Question 44

 

Simplify the following trigonometric fractional expression:

\[ \frac{\cos 12x + \cos 8x + \cos 4x}{\sin 12x + \sin 8x + \sin 4x} \]

 

\[ A) 1 \quad B) \tan 10x \quad C) \cot 10x \quad D) \tan 8x \quad E) \cot 8x \]

 

Solution:

 

Rearrange the terms by grouping the outer angles:

\[ = \frac{(\cos 12x + \cos 4x) + \cos 8x}{(\sin 12x + \sin 4x) + \sin 8x} \]

Apply sum-to-product identities to the grouped pairs:

\[ = \frac{2 \cos 8x \cdot \cos 4x + \cos 8x}{2 \sin 8x \cdot \cos 4x + \sin 8x} \]

Factor out the common terms from both numerator and denominator:

\[ = \frac{\cos 8x (2 \cos 4x + 1)}{\sin 8x (2 \cos 4x + 1)} = \frac{\cos 8x}{\sin 8x} = \cot 8x \]

 

\(\textbf{Correct Answer: E} \)

 

Useful Theorem:

 

In expressions of the form

\[ \frac{\cos A + \cos B + \cos C}{\sin A + \sin B + \sin C} \]

If the angles form an arithmetic progression such that

\[ B = \frac{A + C}{2} \]

then the expression strictly simplifies to \(\displaystyle \frac{\cos B}{\sin B} = \cot B \).

 

Question 45

 

Simplify the following trigonometric fractional expression:

\[ \frac{\sin 50^\circ + \cos 55^\circ + \cos 70^\circ}{\cos 50^\circ + \sin 55^\circ + \sin 70^\circ} \]

 

\[ A) \cot 50^\circ \quad B) \cot 55^\circ \quad C) \tan 50^\circ \quad D) \tan 55^\circ \quad E) 1 \]

 

Solution:

 

Convert the first term using cofunctions (\(\sin 50^\circ = \cos 40^\circ\) and \(\cos 50^\circ = \sin 40^\circ\)):

\[ = \frac{\cos 40^\circ + \cos 55^\circ + \cos 70^\circ}{\sin 40^\circ + \sin 55^\circ + \sin 70^\circ} \]

Since the angles form an arithmetic progression with a mean of \( 55^\circ = \displaystyle\frac{40^\circ + 70^\circ}{2} \), we can apply the theorem directly:

\[ = \frac{\cos 55^\circ}{\sin 55^\circ} = \cot 55^\circ \]

 

\(\textbf{Correct Answer: B} \)

 

Question 46

 

Given \( 14x = \pi \), determine the value of:

\[ \frac{\tan 7x + \tan 3x}{\tan 7x \; – \; \tan 3x} \]

 

\[ A) \ 1 \quad B) \ -1 \quad C) -2 \quad D) \ 2 \quad E) \frac{1}{2} \]

 

Solution:

 

Apply the identity for sum and difference of tangents (identities 5 and 6):

\[ \frac{\tan 7x + \tan 3x}{\tan 7x \; – \; \tan 3x} = \frac{\displaystyle\frac{\sin (7x + 3x)}{\cos 7x \cdot \cos 3x}}{\displaystyle\frac{\sin (7x \; – \; 3x)}{\cos 7x \cdot \cos 3x}} \]

Canceling out the common denominators:

\[ = \frac{\sin 10x}{\sin 4x} \]

Since \(14x = \pi \implies 10x = \pi – 4x\), and \(\sin(\pi – \theta) = \sin\theta\):

\[ = \frac{\sin (\pi \; – \; 4x)}{\sin 4x} = \frac{\sin 4x}{\sin 4x} = 1 \]

 

\(\textbf{Correct Answer: A} \)

 

Question 47

 

Given \(16x = \pi\), find the value of the expression: \(\tan 6x \; – \; \tan 2x\)

\[ A) \ \frac{1}{2} \quad B) -\frac{1}{2} \quad C) \ 2 \quad D) -2 \quad E) \ 4 \]

 

Solution:

 

Apply the difference identity for tangents:

\[ \tan 6x \; – \; \tan 2x = \frac{\sin (6x – 2x)}{\cos 6x \cdot \cos 2x} = \frac{\sin 4x}{\cos 6x \cdot \cos 2x} \]

Expand the numerator using the double-angle formula \(\sin 4x = 2\sin 2x\cos 2x\):

\[ = \frac{2 \sin 2x \cdot \cos 2x}{\cos 6x \cdot \cos 2x} = \frac{2 \sin 2x}{\cos 6x} \]

Since \(16x = \pi \implies 8x = \displaystyle\frac{\pi}{2}\), it follows that \(2x = \frac{\pi}{2} – 6x\). Applying cofunction relations:

\[ = \displaystyle\frac{2 \sin \left( \frac{\pi}{2} \; – \; 6x \right)}{\cos 6x} = \frac{2 \cos 6x}{\cos 6x} = 2 \]

 

\(\textbf{Correct Answer: C} \)

 

 

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