Product-to-Sum Identities

 

By adding the sum and difference identities for cosine:

\[ \cos(a+b) = \cos a \cdot \cos b \; – \; \sin a \cdot \sin b \]
\[ \underline{+ \quad \cos(a-b) = \cos a \cdot \cos b + \sin a \cdot \sin b} \]
\[ \cos(a+b) + \cos(a-b) = 2 \cos a \cdot \cos b \]

We obtain our first product-to-sum identity:

1) \( \displaystyle \cos a \cdot \cos b = \frac{1}{2} [\cos(a+b) + \cos(a-b)] \)

Similarly, by subtracting the two equations, we get:

\[ \cos(a+b) – \cos(a-b) = -2 \sin a \sin b \]

2) \( \displaystyle \sin a \cdot \sin b = -\frac{1}{2} [\cos(a+b) \; – \; \cos(a-b)] \)

By adding the sum and difference identities for sine:
\[ \sin(a+b) = \sin a \cos b + \sin b \cos a \]
\[ \underline{+ \quad \sin(a-b) = \sin a \cos b – \sin b \cos a} \]
\[ \sin(a+b) + \sin(a-b) = 2 \sin a \cos b \]

We get the product-to-sum identity for mixed terms:

3) \( \displaystyle \sin a \cdot \cos b = \frac{1}{2} [\sin(a+b) + \sin(a \; – \; b)] \)

 

QUESTION 48

 

Evaluate the following statement:

\[ \cos 80^\circ \cdot \cos 40^\circ + \frac{1}{2} \cos 140^\circ \]

 

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

 

Solution:

 

Apply the product-to-sum identity to the grouped product term:

\[ \underbrace{\cos 80^\circ \cdot \cos 40^\circ} + \frac{1}{2} \cos 140^\circ \]
\[ = \frac{1}{2} [\cos (80^\circ + 40^\circ) + \cos (80^\circ – 40^\circ)] + \frac{1}{2} \cos 140^\circ \]
\[ = \frac{1}{2} \cos 120^\circ + \frac{1}{2} \cos 40^\circ + \frac{1}{2} \cos 140^\circ \]

Using the supplementary angle relation \(\cos 140^\circ = -\cos 40^\circ\):

\[ = \frac{1}{2} \cos 120^\circ + \frac{1}{2} \cos 40^\circ – \frac{1}{2} \cos 40^\circ \]
\[ = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) = -\frac{1}{4} \]

 

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

 

QUESTION 49

 

Find the value of the following continuous product:

\[ \cos 70^\circ \cdot \cos 50^\circ \cdot \cos 10^\circ \]

 

\[ A) \frac{1}{2} \quad B) \frac{1}{4} \quad C) \frac{\sqrt{3}}{2} \quad D) \frac{\sqrt{3}}{4} \quad E) \frac{\sqrt{3}}{8} \]

 

Solution:

 

Group the first two factors and apply the product-to-sum identity:

\[ \underbrace{\cos 70^\circ \cdot \cos 50^\circ} \cdot \cos 10^\circ \]

\[ = \frac{1}{2} [\cos (70^\circ + 50^\circ) + \cos (70^\circ – 50^\circ)] \cdot \cos 10^\circ \]

\[ = \frac{1}{2} \cdot \left( \cos 120^\circ + \cos 20^\circ \right) \cdot \cos 10^\circ \]

\[ = \frac{1}{2} \cdot \left( -\frac{1}{2} + \cos 20^\circ \right) \cdot \cos 10^\circ \]

Distribute \(\cos 10^\circ\):

\[ = -\frac{1}{4} \cos 10^\circ + \frac{1}{2} \cos 20^\circ \cdot \cos 10^\circ \]

Apply the identity a second time to the product component:

\[ = -\frac{1}{4} \cos 10^\circ + \frac{1}{2} \cdot \frac{1}{2} (\cos 30^\circ + \cos 10^\circ) \]

\[ = -\frac{1}{4} \cos 10^\circ + \frac{1}{4} \left( \frac{\sqrt{3}}{2} + \cos 10^\circ \right) \]

\[ = -\frac{1}{4} \cos 10^\circ + \frac{\sqrt{3}}{8} + \frac{1}{4} \cos 10^\circ = \frac{\sqrt{3}}{8} \]

 

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

 

QUESTION 50

 

Simplify the following fraction:

\[ \frac{\cos 25^\circ \cdot \cos 10^\circ – \cos 20^\circ \cdot \cos 15^\circ}{\sin^2 5^\circ} \]

 

\[ A) \ 1 \quad B) \ 2 \quad C) -2 \quad D) \ 2 \cos 5^\circ \quad E) -2 \cos 5^\circ \]

 

Solution:

 

Apply product-to-sum identities to both parts of the numerator:

\[ = \frac{\frac{1}{2} (\cos 35^\circ + \cos 15^\circ) – \frac{1}{2} (\cos 35^\circ + \cos 5^\circ)}{\sin^2 5^\circ} \]

\[ = \frac{\cos 15^\circ – \cos 5^\circ}{2 \sin^2 5^\circ} \]

Now apply the sum-to-product identity to the remaining terms in the numerator:

\[ = \frac{-2 \sin 10^\circ \cdot \sin 5^\circ}{2 \sin^2 5^\circ} = \frac{- \sin 10^\circ}{\sin 5^\circ} \]

Expand the numerator using the double-angle formula (\(\sin 10^\circ = 2\sin 5^\circ\cos 5^\circ\)):

\[ = \frac{-2 \sin 5^\circ \cdot \cos 5^\circ}{\sin 5^\circ} = -2 \cos 5^\circ \]

 

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

 

QUESTION 51

 

Find the simplified form of the following rational expression:

\[ \frac{\tan (x + \frac{\pi}{4}) + \tan (x – \frac{\pi}{4})}{\cot (x + \frac{\pi}{4}) + \cot (x – \frac{\pi}{4})} \]

 

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

 

Solution:

 

Let us use substitutions to simplify handling the arguments:

\[ \left. \begin{array}{l} x + \frac{\pi}{4} = a \\ x – \frac{\pi}{4} = b \end{array} \right\} \Rightarrow a + b = 2x \text{ and } a – b = \frac{\pi}{2} \]

Rewrite the original expression with variables \(a\) and \(b\):

\[ \frac{\tan a + \tan b}{\cot a + \cot b} \]

Convert tangents and cotangents into sine and cosine fractions to combine them:

\[ = \frac{\displaystyle \frac{\sin (a + b)}{\cos a \cdot \cos b}}{\displaystyle \frac{\sin (a + b)}{\sin a \cdot \sin b}} = \frac{\sin a \cdot \sin b}{\cos a \cdot \cos b} \]

Now, transform both the numerator and the denominator back into sums using product-to-sum identities:

\[ = \frac{\displaystyle -\frac{1}{2} [\cos (a + b) – \cos (a – b)]}{\displaystyle \frac{1}{2} [\cos (a + b) + \cos (a – b)]} = \frac{- (\cos 2x – \cos \frac{\pi}{2})}{\cos 2x + \cos \frac{\pi}{2}} \]

Since \(\cos \frac{\pi}{2} = 0\):

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

 

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

 

QUESTION 52

 

Evaluate the following trigonometric subtraction:

\[ \frac{1}{\sin 40^\circ} \; – \; 4 \sin 80^\circ \]

 

\[ A) -2 \quad B) \ 2 \quad C) -2 \cot 40^\circ \quad D) \ 2 \cot 40^\circ \quad E) -\cot 40^\circ \]

 

Solution:

 

Find a common denominator:

\[ \frac{1}{\sin 40^\circ} \; – \; 4 \sin 80^\circ = \frac{1 – 4 \sin 80^\circ \cdot \sin 40^\circ}{\sin 40^\circ} \]

Apply the product-to-sum identity to the numerator component:

\[ = \frac{\displaystyle 1 – 4 \cdot \left( -\frac{1}{2} \right) [\cos 120^\circ \; – \; \cos 40^\circ]}{\sin 40^\circ} \]

\[ = \frac{\displaystyle 1 + 2 \left( -\frac{1}{2} – \cos 40^\circ \right)}{\sin 40^\circ} \]

\[ = \frac{1 \; – \; 1 \; – \; 2 \cos 40^\circ}{\sin 40^\circ} = \frac{-2 \cos 40^\circ}{\sin 40^\circ} = -2 \cot 40^\circ \]

 

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

 

QUESTION 53

 

Given \(x \ – \ y = 30^\circ \), find an equivalent evaluation for the product below:

\[ \cot(x \ – \ 2y) \cdot \cot(2x \ – \ y) \]

 

\[ A) -1 \quad B) \ 1 \quad C) -2 \quad D) \ \cot 3x \quad E) \ \cot(x+y) \]

 

Solution:

 

Convert the cotangents to cosine/sine fractions:

\[ = \frac{\displaystyle \cos(x \ – \ 2y)}{\displaystyle \sin(x \ – \ 2y)} \cdot \frac{\displaystyle \cos(2x \ – \ y)}{\displaystyle \sin(2x \ – \ y)} \]

Apply product-to-sum identities to both the entire numerator and denominator products:

\[ = \frac{\displaystyle \frac{1}{2} [\cos(3x \ – \ 3y) + \cos(-x \ – \ y)]}{\displaystyle -\frac{1}{2} [\cos(3x \ – \ 3y) – \cos(-x \ – \ y)]} \]

Since \(x – y = 30^\circ \implies 3x – 3y = 90^\circ\), we substitute this value into the expression:

\[ = \frac{\displaystyle \cos 90^\circ + \cos(-x \ – \ y)}{\displaystyle -(\cos 90^\circ \ – \ \cos(-x \ – \ y))} \]

Knowing that \(\cos 90^\circ = 0\):

\[ = \frac{\cos(-x – y)}{\cos(-x – y)} = 1 \]

 

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