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.

Slide 22:

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

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