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Double- and half-angle formulae Trigonometry. Sine double-angle formulae Recall from the last section, the sine of the sum of two angles; sin(α + β) =

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Presentation on theme: "Double- and half-angle formulae Trigonometry. Sine double-angle formulae Recall from the last section, the sine of the sum of two angles; sin(α + β) ="— Presentation transcript:

1 Double- and half-angle formulae Trigonometry

2 Sine double-angle formulae Recall from the last section, the sine of the sum of two angles; sin(α + β) = sin α cos β + cos α sin β We will use this to obtain the sine of a double angle.

3 Sine of double angle: substitution of α for β If we take the left-hand side—sin(α + β) —and replace β with α, we get: sin(α + β) = sin(α + α) = sin 2α Now consider the right-hand side of the equation: sin α cos β + cos α sin β If we replace β with α, we obtain: sin α cos α + cos α sin α = 2 sin α cos α

4 Sine of double angle: conclusion Putting our results for the LHS and RHS together, we obtain the important result: sin 2α = 2 sin α cos α This result is called the sine of a double angle. It is useful for simplifying expressions later.

5 Cosine of a double angle Using a similar process, we can start with the cosine of the sum of two angles: cos(α + β) = cos α cos β − sin α sin β Again substituting α for β, we get: cos(α + α) = cos α cos α − sin α sin α = cos 2 α − sin 2 α.

6 Cosine of double angle: alternative forms(1) Just as the Pythagorean identities have alternate forms, we can rewrite the double-angle formula for cosine in two alternate ways. Starting with cos(2α) = cos 2 α − sin 2 α, we can substitute (1− sin 2 α) for cos 2 α, and get cos(2α) = (1− sin 2 α) − sin 2 α = 1 − 2sin 2 α.

7 Cosine of double angle: alternative forms(2) Likewise, we can substitute (1 − cos 2 α) for sin 2 α into the right-hand side of the equation and obtain: cos 2 α − sin 2 α = cos 2 α − (1 − cos 2 α) = 2cos 2 α − 1

8 Summary - Cosine of a Double Angle The following have equivalent value, and we can use whichever one we like, depending on the situation: cos 2α = cos 2 α − sin 2 α cos 2α = 1− 2sin 2 α cos 2α = 2cos 2 α − 1

9 But what about tan 2x?

10 But what about tan 2α?

11 Example 1 Find cos 60° by using the functions of 30°. We can only use the sine and cosine functions of 30°, so we need to start with 60° = 2 × 30°. We will use the result cos 2α = cos 2 α − sin 2 α, and the triangle:

12 Example 1 (continued) Now we proceed to find the exact value of cos60° using the ratios of 30°:

13 Example 2 Find cos 2x if sin x = -12/13 (in Quadrant III). We chose the following form of the cosine of a double angle formula, cos 2α = 1− 2sin 2 α, because it has a squared form of sin α which we can use.

14 Example 2 (calculated)

15 Note: We didn't find the value of x using calculator first, and then find the required value. If we had done that, we would not have found the exact value, and we would have missed the pleasure of seeing the double angle formula in action.

16 Exercises Exercise 1: Without finding x, find sin 2x if cos x = 4/5 (in Quadrant I). We that we need to use the triangle (because of the 4 and 5 in the question).

17 Exercise 1 (calculated) We can use our formula for the sine of a double angle to find the required value: Simple and straightforward, and no need for arcsin!

18 Exercise 2 Prove that When we need to prove an identity, we start on one side (usually the most complicated side) and work on it until it is equivalent to the other side. In this example, we start on the left hand side and use our various identities from earlier sections to simplify it. Here, “LHS” = “left-hand side” of the equation, and “RHS” = “right-hand side” of the equation

19 Exercise 2 (calculated)

20 Example 3 Prove that 2 csc 2x tan x = sec 2 x A good approach with these proofs is to reduce everything to sine and cosine only. Then you find the steps are easier to simplify, and it is easier to recognize the various formulas we have learned.

21 Exercise 3 (calculated)


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