# LarPCalcLim2_05_05

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### What You Should Learn:

Use multiple-angle formulas to rewrite and evaluate trigonometric functions. Use power-reducing formulas to rewrite and evaluate trigonometric functions. Use half-angle formulas to rewrite and evaluate trigonometric functions. What You Should Learn

### What You Should Learn:

Use product-to-sum and sum-to-product formulas to rewrite and evaluate trigonometric functions. Use trigonometric formulas to rewrite real-life models. What You Should Learn

### Slide 5:

Multiple-Angle Formulas

### Multiple-Angle Formulas:

Multiple-Angle Formulas In this section, you will study four other categories of trigonometric identities. 1. The first category involves functions of multiple angles such as sin ku and cos ku . 2. The second category involves squares of trigonometric functions such as sin 2 u . 3. The third category involves functions of half-angles such as sin( u / 2). 4. The fourth category involves products of trigonometric functions such as sin u cos v .

### Multiple-Angle Formulas:

Multiple-Angle Formulas You should learn the double-angle formulas because they are used often in trigonometry and calculus.

### Example 1 – Solving a Multiple-Angle Equation:

Example 1 – Solving a Multiple-Angle Equation Solve 2 cos x + sin 2 x = 0. Solution: Begin by rewriting the equation so that it involves functions of x (rather than 2 x ). Then factor and solve. 2 cos x + sin 2 x = 0 2 cos x + 2 sin x cos x = 0 2 cos x (1 + sin x ) = 0 2 cos x = 0 and 1 + sin x = 0 Write original equation. Double-angle formula Factor. Set factors equal to zero.

### Example 1 – Solution:

Example 1 – Solution So, the general solution is and where n is an integer. Try verifying these solutions graphically. Solutions in [0, 2  ) cont’d

### Slide 10:

Power-Reducing Formulas

### Power-Reducing Formulas:

Power-Reducing Formulas The double-angle formulas can be used to obtain the following power-reducing formulas. Example 5 shows a typical power reduction that is used in calculus.

### Example 5 – Reducing a Power:

Example 5 – Reducing a Power Rewrite sin 4 x as a sum of first powers of the cosines of multiple angles. Solution: Note the repeated use of power-reducing formulas. Property of exponents Power-reducing formula Expand.

### Example 5 – Solution:

Example 5 – Solution Power-reducing formula Distributive Property Factor out common factor. cont’d

### Slide 14:

Half-Angle Formulas

### Half-Angle Formulas:

Half-Angle Formulas You can derive some useful alternative forms of the power-reducing formulas by replacing u with u / 2. The results are called half-angle formulas.

### Example 6 – Using a Half-Angle Formula:

Example 6 – Using a Half-Angle Formula Find the exact value of sin 105  . Solution: Begin by noting that 105  is half of 210  . Then, using the half-angle formula for sin( u / 2) and the fact that 105  lies in Quadrant II, you have

### Example 6 – Solution:

Example 6 – Solution . The positive square root is chosen because sin  is positive in Quadrant II. cont’d

### Slide 18:

Product-to-Sum Formulas

### Product-to-Sum Formulas:

Product-to-Sum Formulas Each of the following product-to-sum formulas can be verified using the sum and difference formulas. Product-to-sum formulas are used in calculus to evaluate integrals involving the products of sines and cosines of two different angles.

### Example 8 – Writing Products as Sums:

Example 8 – Writing Products as Sums Rewrite the product cos 5 x sin 4 x as a sum or difference. Solution: Using the appropriate product-to-sum formula, you obtain cos 5 x sin 4 x = [sin(5 x + 4 x ) – sin(5 x – 4 x )] = sin 9 x – sin x .

### Product-to-Sum Formulas:

Product-to-Sum Formulas Occasionally, it is useful to reverse the procedure and write a sum of trigonometric functions as a product. This can be accomplished with the following sum-to-product formulas.

Application

### Example 12 – Projectile Motion:

Example 12 – Projectile Motion Ignoring air resistance, the range of a projectile fired at an angle  with the horizontal and with an initial velocity of v 0 feet per second is given by where r is the horizontal distance (in feet) that the projectile will travel.

### Example 12 – Projectile Motion:

Example 12 – Projectile Motion A place kicker for a football team can kick a football from ground level with an initial velocity of 80 feet per second (see Figure 5.18). a. Write the projectile motion model in a simpler form. b. At what angle must the player kick the football so that the football travels 200 feet? c. For what angle is the horizontal distance the football travels a maximum? cont’d Figure 5.18

### Example 12 – Solution:

Example 12 – Solution a. You can use a double-angle formula to rewrite the projectile motion model as r = v 0 2 (2 sin  cos  ) = v 0 2 sin 2  . b. r = v 0 2 sin 2  200 = ( 80 ) 2 sin 2  Rewrite original projectile motion model. Rewrite model using a double-angle formula. Write projectile motion model. Substitute 200 for r and 80 for v 0 .

### Example 12 – Solution:

Example 12 – Solution 200 = 200 sin 2  1 = sin 2  You know that 2  =  / 2, so dividing this result by 2 produces  =  / 4. Because  / 4 = 45  , you can conclude that the player must kick the football at an angle of 45  so that the football will travel 200 feet. Simplify. Divide each side by 200. cont’d

### Example 12 – Solution:

Example 12 – Solution c. From the model r = 200 sin 2  you can see that the amplitude is 200. So the maximum range is r = 200 feet. From part (b), you know that this corresponds to an angle of 45  . Therefore, kicking the football at an angle of 45  will produce a maximum horizontal distance of 200 feet. cont’d