Mathematics
Double Angle and Half Angle Formulas
Double angle and half angle formulas are trigonometric identities that express the sine, cosine, and tangent of double or half angles in terms of the sine and cosine of the original angle. These formulas are useful for simplifying trigonometric expressions and solving trigonometric equations. They are derived from the sum and difference identities for sine and cosine functions.
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11 Key excerpts on "Double Angle and Half Angle Formulas"
No longer available |Learn more- James Stewart, Lothar Redlin, Saleem Watson(Authors)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
[Hint: Use tan1 s t 2 sin1 s t 2 cos 1 s t 2 and divide the numerator and denominator by cos s cos t.] 7.3 DOUBLE-ANGLE, HALF-ANGLE, AND PRODUCT-SUM FORMULAS ■ Double-Angle Formulas ■ Half-Angle Formulas ■ Evaluating Expressions Involving Inverse Trigonometric Functions ■ Product-Sum Formulas The identities we consider in this section are consequences of the addition formulas. The Double-Angle Formulas allow us to find the values of the trigonometric functions at 2x from their values at x. The Half-Angle Formulas relate the values of the trigono- metric functions at 1 2 x to their values at x. The Product-Sum Formulas relate products of sines and cosines to sums of sines and cosines. ■ Double-Angle Formulas The formulas in the box on the next page are immediate consequences of the addition formulas, which we proved in Section 7.2. Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 554 CHAPTER 7 ■ Analytic Trigonometry DOUBLE-ANGLE FORMULAS Formula for sine: sin 2 x 2 sin x cos x Formulas for cosine: cos 2 x cos 2 x sin 2 x 1 2 sin 2 x 2 cos 2 x 1 Formula for tangent: tan 2 x 2 tan x 1 tan 2 x The proofs for the formulas for cosine are given here. You are asked to prove the remaining formulas in Exercises 35 and 36. Proof of Double-Angle Formulas for Cosine cos 2x cos 1 x x 2 cos x cos x sin x sin x cos 2 x sin 2 x The second and third formulas for cos 2 x are obtained from the formula we just proved and the Pythagorean identity.
eBook - PDF- Cynthia Y. Young(Author)
- 2021(Publication Date)
- Wiley(Publisher)
7.5.1 Conceptual Understand that the half-angle identities are derived from the double-angle identities. We now use the double-angle identities from Section 7.4 to derive the half-angle identities. Like the double-angle identities, the half-angle identities will allow us to find certain exact values of trigonometric functions and to verify other trigonometric identities. We start by rewriting the second and third forms of the cosine double-angle identity to obtain identities for the square of the sine and cosine functions, sin 2 x and cos 2 x, otherwise known as the power reduction formulas. Words Math Power reduction formula for the sine function Write the second form of the cosine double-angle identity. cos(2A) = 1 − 2 sin 2 A Isolate the 2 sin 2 A term on one side of the equation. 2 sin 2 A = 1 − cos(2A) Divide both sides by 2. sin 2 A = 1 − cos(2A) __________ 2 Power reduction formula for the cosine function Write the third form of the cosine double-angle identity. cos(2A) = 2 cos 2 A − 1 Isolate the 2 cos 2 A term on one side of the equation. 2 cos 2 A = cos(2A) + 1 Divide both sides by 2. cos 2 A = 1 + cos(2A) __________ 2 Power reduction formula for the tangent function Taking the quotient of these leads us to another identity. tan 2 A = sin 2 A _ cos 2 A = ( 1 − cos(2A) __________ 2 ) _____________ ( 1 + cos(2A) _ 2 ) = 1 − cos(2A) __________ 1 + cos(2A) These three identities for the squared functions, which are essentially alternative forms of the double-angle identities, are used in calculus as power reduction formulas (identities that allow us to reduce the power of the trigonometric function from 2 to 1): sin 2 A = 1 − cos(2A) __________ 2 cos 2 A = 1 + cos(2A) __________ 2 tan 2 A = 1 − cos(2A) __________ 1 + cos(2A) Derivation of the Half-Angle Identities We can now use these forms of the double-angle identities to derive the half-angle identities.
eBook - PDF- Cynthia Y. Young(Author)
- 2017(Publication Date)
- Wiley(Publisher)
5.4 HALF-ANGLE IDENTITIES 5.4.1 Applying Half-Angle Identities We now use the double-angle identities from Section 5.3 to derive the half-angle identities. Like the double-angle identities, the half-angle identities will allow us to find certain exact values of trigonometric functions and to verify other trigonometric identities. We start by rewriting the second and third forms of the cosine double-angle identity to obtain identities for the square of the sine and cosine functions, sin 2 x and cos 2 x, otherwise known as the power reduction formulas. WORDS MATH Power reduction formula for the sine function Write the second form of the cosine double-angle identity. cos 1 2 A2 5 1 2 2 sin 2 A Isolate the 2 sin 2 A term on one side of the equation. 2 sin 2 A 5 1 2 cos 1 2 A2 Divide both sides by 2. sin 2 A 5 1 2 cos 1 2 A2 2 Power reduction formula for the cosine function Write the third form of the cosine double-angle identity. cos 1 2 A2 5 2 cos 2 A 2 1 Isolate the 2 cos 2 A term on one side of the equation. 2 cos 2 A 5 cos 1 2A2 1 1 Divide both sides by 2. cos 2 A 5 1 1 cos 1 2 A2 2 Power reduction formula for the tangent function Taking the quotient of these leads us to another identity. tan 2 A 5 sin 2 A cos 2 A 5 1 2 cos 1 2 A2 2 1 1 cos 1 2 A2 2 5 1 2 cos 1 2 A2 1 1 cos 1 2 A2 These three identities for the squared functions, which are essentially alternative forms of the double-angle identities, are used in calculus as power reduction formulas (identities that allow us to reduce the power of the trigonometric function from 2 to 1): sin 2 A 5 1 2 cos 1 2 A2 2 cos 2 A 5 1 1 cos 1 2 A2 2 tan 2 A 5 1 2 cos 1 2 A2 1 1 cos 1 2 A2 5.4.1 SKILL Use half-angle identities in simplifying some trigonometric expressions. 5.4.1 CONCEPTUAL Understand that the half-angle identities are derived from the double-angle identities.
eBook - PDF- Cynthia Y. Young(Author)
- 2021(Publication Date)
- Wiley(Publisher)
5.4.1 Conceptual Understand that the half-angle identities are derived from the double-angle identities. We now use the double-angle identities from Section 5.3 to derive the half-angle identities. Like the double-angle identities, the half-angle identities will allow us to find certain exact values of trigonometric functions and to verify other trigonometric identities. We start by rewriting the second and third forms of the cosine double-angle identity to obtain identities for the square of the sine and cosine functions, sin 2 x and cos 2 x, otherwise known as the power reduction formulas. Words Math Power reduction formula for the sine function Write the second form of the cosine double-angle identity. cos(2A) = 1 − 2 sin 2 A Isolate the 2 sin 2 A term on one side of the equation. 2 sin 2 A = 1 − cos(2A) Divide both sides by 2. sin 2 A = 1 − cos(2A) __________ 2 Power reduction formula for the cosine function Write the third form of the cosine double-angle identity. cos(2A) = 2 cos 2 A − 1 Isolate the 2 cos 2 A term on one side of the equation. 2 cos 2 A = cos(2A) + 1 Divide both sides by 2. cos 2 A = 1 + cos(2A) __________ 2 Power reduction formula for the tangent function Taking the quotient of these leads us to another identity. tan 2 A = sin 2 A _ cos 2 A = ( 1 − cos(2A) __________ 2 ) _____________ ( 1 + cos(2A) _ 2 ) = 1 − cos(2A) __________ 1 + cos(2A) These three identities for the squared functions, which are essentially alternative forms of the double-angle identities, are used in calculus as power reduction formulas (identities that allow us to reduce the power of the trigonometric function from 2 to 1): sin 2 A = 1 − cos(2A) __________ 2 cos 2 A = 1 + cos(2A) __________ 2 tan 2 A = 1 − cos(2A) __________ 1 + cos(2A) Derivation of the Half-Angle Identities We can now use these forms of the double-angle identities to derive the half-angle identities.
No longer available |Learn more- Alfred Basta, Stephan DeLong, Nadine Basta, , Alfred Basta, Stephan DeLong, Nadine Basta(Authors)
- 2013(Publication Date)
- Cengage Learning EMEA(Publisher)
The primary use for the formulas lies in their applicability to the solution of trigonometric equations, which are (in turn) useful to solve applications in which harmonic motion or resonance occurs. To this point, we haven’t seen enough to solve trigonometric equations, but we can investigate an application where a double angle occurs. EXAMPLE 8.11 The range R (in feet) of a projectile (an airborne object that is not self-propelled) fired at an angle A with the horizontal and having initial velocity v 0 feet per second is given by R v A H11005 1 32 2 0 2 sin( ) . Suppose a jai alai player throws a ball with an initial velocity of 150 miles per hour at an angle of 2 ° with the horizon-tal. What is the range of the ball? SOLUTION The first thing we notice here is that the initial velocity is given in miles per hour, so we must convert that velocity to feet per second, which we can do through dimensional analysis: 150 3600 5280 miles hour hour seconds feet mile H11003 H11003 feet second H11005 220 Now, inserting that value into the formula, we find R R sin[2(2 sin(4 H11005 H11005 1 32 220 1 32 48 400 2 ( ) )] ( , ) ) ° ° R R 1,512.5 105.5725 105.6 feet H11005 H11005 H11005 ( . ) 0 0698 Thus, the range of the projectile is about 105.6 feet. The Half-Angle Identities The half-angle formulas allow us to find the values of the trigonometric functions of angles having the form H9251 2 for some angle a . They can be derived from the double-angle formulas, provided that we look at things in a clever manner. Note When you see an expression like sin(3 x ), it is a natural (and mistaken!) thought that the “3” can be moved in front of the expression and conclude that sin (3 x ) 5 3 sin x 2 3 sin 3 x . Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
eBook - PDFPrecalculus
A Prelude to Calculus
- Sheldon Axler(Author)
- 2016(Publication Date)
- Wiley(Publisher)
If instead we start out knowing the value of cos(2θ ), then the equation above can be solved for cos θ. The next example illustrates this procedure. Section 5.5 Double-Angle and Half-Angle Formulas 405 Example 3 Find an exact expression for cos 15 ◦ . solution We know that cos 30 ◦ = √ 3 2 . We want to find the cosine of half of 30 ◦ . Example 2 in Section 5.6 uses a different method to find a different exact expression for cos 15 ◦ . Thus we set θ = 15 ◦ in the identity cos(2θ ) = 2 cos 2 θ - 1, getting cos 30 ◦ = 2 cos 2 15 ◦ - 1. In this equation, replace cos 30 ◦ with its value, getting √ 3 2 = 2 cos 2 15 ◦ - 1. Now solve the equation above for cos 15 ◦ (which is positive, so we take the positive square root below), getting This value for cos 15 ◦ (or cos π 12 if using radians) was used in exercises in Sections 4.4 and 4.6. cos 15 ◦ = s 1 + √ 3 2 2 = s ( 1 + √ 3 2 ) · 2 2 · 2 = p 2 + √ 3 2 . To find a general formula for cos θ 2 in terms of cos θ, we will carry out the procedure followed in the example above. The key idea is that we can substitute any value for θ in the identity cos(2θ ) = 2 cos 2 θ - 1, provided that we make the same substitution on both sides of the equation. We want to find a formula for cos θ 2 . Thus we replace θ by θ 2 on both sides of the equation above, getting cos θ = 2 cos 2 θ 2 - 1. Now solve this equation for cos θ 2 , getting the following half-angle formula. Half-angle formula for cosine cos θ 2 = ± r 1 + cos θ 2 The choice of the plus or minus sign in the formula above depends on knowledge of the sign of cos θ 2 . For example, if 0 < θ < π, then 0 < θ 2 < π 2 , which implies that cos θ 2 is positive (thus we would choose the plus sign in the formula above). To find a formula for sin θ 2 , we start with the double-angle formula cos(2θ ) = 1 - 2 sin 2 θ. In the identity above, replace θ by θ 2 on both sides of the equation, getting cos θ = 1 - 2 sin 2 θ 2 .
eBook - PDF- Paul A. Calter, Michael A. Calter(Authors)
- 2011(Publication Date)
- Wiley(Publisher)
cos 2u 2 cos 2 u 1 A/ 2 sin 100° sin 100° 0.9848 100° sin 200° 2 A 1 cos 200° 2 A 1 (0.9397) 2 0.9848 sin 100°. cos 200° 0.9397, Section 3 ◆ Functions of Double Angles and Half-Angles 479 Substituting into (1) gives 136a Another form of this identity is obtained by multiplying numerator and denom- inator by which simplifies to 136b We can obtain a third formula for the tangent by dividing the sine by the cosine. 136c ◆◆◆ Example 21: Prove that Solution: Using the identities for the sine and cosine of half an angle we get Multiplying numerator and denominator by 2 ◆◆◆ 3 cos u 3 cos u 2 1 cos u 2 1 cos u 3 cos u 3 cos u 1 1 cos u 2 1 1 cos u 2 3 cos u 3 cos u 1 sin 2 u 2 1 cos 2 u 2 tan a 2 A 1 cos a 1 cos a Tangent of Half an Angle tan a 2 sin a 2 cos a 2 A 1 cos a 2 A 1 cos a 2 tan a 2 sin a 1 cos a Tangent of Half an Angle 1 cos 2 a sin a(1 cos a) sin 2 a sin a(1 cos a) tan a 2 1 cos a sin a # 1 cos a 1 cos a 1 cos a tan a 2 1 cos a sin a Tangent of Half an Angle v 0 x y FIGURE 16–13 s x s s x 480 Chapter 16 ◆ Trigonometric Identities and Equations FIGURE 16–12 Exercise 3 ◆ Functions of Double Angles and Half-Angles Double Angles Simplify. 1. 2. 3. 4. Prove each identity. 5. 6. 7. 8. 9. 10. 11. 12. 13. Half-Angles Prove each identity. 14. 15. 16. 17. 18. 19. 20. 21. Applications 22. Shear Stress: The shear stress on a cross section of a bar in tension, Fig. 16–12, is related to the axial stress by the formula where is the angle between the axis of the bar and the normal to the cross section. Use the double-angle identities to write this expression with just a sin- gle trigonometric ratio. 23. Trajectories: A projectile is launched on level ground at an angle and with an initial velocity of Fig. 16–13. In time t the x and y displacements are and (a) Write an expression for the time elapsed before the projectile hits the ground.
eBook - PDF- Michael A. Calter, Paul A. Calter, Paul Wraight, Sarah White(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
Nor is the cosine (or tangent) of twice an angle equal to twice the cosine (or tangent) of that angle. Remember to use the formulas from this section for all of the trig functions of double angles. ◆◆◆ Exercise 3 ◆ Functions of Double Angles Simplify. 1. 2 sin 2 x + cos 2x 2. 2 sin 2θ cos 2θ 3. + x x 2 tan 1 tan 2 4. - x x 2 sec sec 2 2 5. - x x x x sin 6 sin 2 cos 6 cos 2 Prove. 6. θ θ θ - = 2 tan 1 tan tan 2 2 7. - + = x x x 1 tan 1 tan cos 2 2 2 8. 2 csc 2θ = tan θ + cot θ 9. 2 cot 2x = cot x - tan x 10. = + - x x x sec 2 1 cot cot 1 2 2 11. - + = + x x x x 1 tan tan 1 cos 2 sin 2 1 12. θ θ θ θ θ + + + = sin 2 sin 1 cos cos 2 tan 13. + = - x x x x 1 cot 2 cos 2 sin 2 2 sin 2 14. 4 cos 3 x - 3 cos x = cos 3x 15. α α α α α + = + + sin cos sin 2 1 cos sin 16. - = x x x cot 1 2 cot cot 2 2 If A and B are the two acute angles in a right triangle, show that: 17. sin 2A = sin 2B. 18. tan(A - B) = -cot 2A. 19. sin 2A = cos(A - B). 18–4 Functions of Half-Angles Sine of α/2 The double-angle formulas derived in Sec. 18–3 can also be regarded as half-angle formulas, because if one angle is double another, the second angle must be half the first. Starting with Eq. 171b, we obtain cos 2θ = 1 - 2 sin 2 θ We solve for sin θ. θ θ θ θ = - = ± - 2 sin 1 cos 2 sin 1 cos 2 2 2 405 Section 18–4 ◆ Functions of Half-Angles For emphasis, we replace θ by α 2 . Sine of Half an Angle α α = ± - sin 2 1 cos 2 173 The ± sign in Eq. 173 (and in Eqs. 174 and 175 c as well) is to be read as plus or minus, but not both. This sign is different from the ± sign in the quadratic formula, for example, where we took both the positive and the negative values. The reason for this difference is clear from Fig. 18-4, which shows a graph of sin α 2 and a graph of α + - (1 cos )/2 . FIGURE 18-4 α (deg) 1 0.5 y 720˚ 540˚ y = sin 180˚ 0 360˚ α 2 α (deg) 0.5 y 0 1 - cos α 2 y = + Note that the two curves are the same only when sin α 2 is positive.
eBook - PDF- Charles P. McKeague(Author)
- 2020(Publication Date)
- XYZ Textbooks(Publisher)
Simplify. 49. √ 1 − 3 __ 5 2 50. 1 − 2 1 _ √ — 5 2 51. 1 + 2 2 _ √ — 5 2 52. √ — 1 − 5 __ 13 2 — 5.3 Double-Angle Formulas 265 5.3 DOUBLE-ANGLE FORMULAS Learning Objectives A Find the value of trigonometric functions using double-angle formulas. B Prove trigonometric identities using double-angle formulas. We will begin this section by deriving the formulas for sin 2 A and cos 2 A using the formulas for sin ( A + B ) and cos ( A + B ). Th e formulas we derive for sin 2 A and cos 2 A are called double-angle formulas . Here is the derivation of the formula for sin 2 A . sin 2 A = sin ( A + A ) Write 2 A as A + A . = sin A cos A + cos A sin A Sum formula = sin A cos A + sin A cos A = 2 sin A cos A Commutative property Th e last line gives us o ur fir st double-angle formula. sin 2 A = 2 sin A cos A Th e fir st thing to notice about this formula is that it indicates that the 2 in sin 2 A cannot be factored out and written as a co efficient. Th at is, sin 2 A ≠ 2 sin A A Solving Trigonometric Functions Here are some examples of how we can apply the double-angle formula sin 2 A = 2 sin A cos A . __ and A terminates in Q II, fin d sin 2 A . Solution In order to apply the formula for sin 2 A , we mus t fir s t fin d cos A . 5.3 Videos ________ ___ Note that we chose the negative radical since cos A < 0 when A is in QII. Now we can apply the formula for sin 2 A : cos A = − √ — 1 − sin 2 A = − √ 1 − 3 __ 5 2 = − √ 16 ___ 25 = − 4 __ 5 sin 2 A = 2 sin A cos A = 2 3 __ 5 − 4 __ 5 = − 24 ___ 25 Example 1 If sin A = 3 5 266 Chapter 5 Identities and Formulas We can also use our new formula to expand the work we did previously with identities. Example 2 Prove: (sin θ + cos θ ) 2 = 1 + sin 2 θ . Proof (sin θ + cos θ ) 2 = sin 2 θ + 2 sin θ cos θ + cos 2 θ Expand = 1 + 2 sin θ cos θ Pythagorean identity = 1 + sin 2 θ Double-angle identity Th ere are three forms of the double-angle formula for cos 2 A .
eBook - PDF- David Cohen, Theodore Lee, David Sklar, , David Cohen, Theodore Lee, David Sklar(Authors)
- 2016(Publication Date)
- Cengage Learning EMEA(Publisher)
SOLUTION using the formula for cos 2 u using the formula for cos 2 u with u 2 t An easy way to simplify this last expression is to multiply both the numerator and the denominator by 2. As you should check for yourself, the final result is The last three formulas we are going to prove in this section are the half-angle formulas, which follow. 1. 2. 3. Note: In the half-angle formulas the symbol is intended to mean either positive or negative, but not both; the sign before the radical is determined by the quadrant in which the angle (or arc) s 2 terminates. To derive the formula for cos( s 2), we begin with one of the alternative forms of the cosine double-angle formula: Since this identity holds for all values of u , we may replace u by s 2 to obtain This is the required formula for cos( s 2). To derive the formula for sin( s 2), we follow exactly the same procedure, except that we begin with the identity sin 2 . [Exercise 34(c) asks you to complete the proof.] In both for-mulas the sign before the radical is determined by the quadrant in which the angle or arc s 2 terminates. u 1 2 (1 cos 2 u ) cos s 2 B 1 cos 2( s 2) 2 B 1 cos s 2 cos 2 u 1 cos 2 u 2 or cos u B 1 cos 2 u 2 tan s 2 sin s 1 cos s cos s 2 B 1 cos s 2 sin s 2 B 1 cos s 2 The Half-Angle Formulas cos 4 t 3 4 cos 2 t cos 4 t 8 1 2 cos 2 t 1 2 (1 cos 4 t ) 4 1 2 cos 2 t cos 2 2 t 4 cos 4 t (cos 2 t ) 2 a 1 cos 2 t 2 b 2 Copyright 201 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.2 The Double-Angle Formulas 615 EXAMPLE 5 Using a Half-Angle Formula Evaluate cos 105 using a half-angle formula.
eBook - PDF- Bernard Kolman, Arnold Shapiro(Authors)
- 2014(Publication Date)
- Academic Press(Publisher)
7.3 DOUBLE AND HALF-ANGLE FORMULAS 291 19. sin 7TT/6 20. cos 5TT/6 21. tan 15° 22. tan 165° In Exercises 23-28 write the given expression in terms of cofunctions of com-plementary angles. 23. sin 47° 24. cos 78° 25. tan TT/6 26. tan 84° 27. cos TT/3 28. sin 72° 30' 29. If sin / = -3 / 5 with W(t) in quadrant III, find sin (TT/2 -t). 30. If cos t = -5/13 with W(t) in quadrant II, find sin(f -TT). 31. If tan 6 = 4/3 and angle 0 lies in quadrant III, find tan(0 + TT/4). 32. If sec 0 = 5/3 and angle 6 lies in quadrant I, find sin(0 + TT/6). 33. If cos t = 0.4 with W(t) in quadrant IV, find tan(r + TT). 34. If sec a = 1.2 and angle a lies in quadrant IV, find tan (a - 77-). 35. If sin s = 3/5 and cos t = -12/13, with W(s) in quadrant II and W(t) in quadrant III, find sin (s + t). 36. If sin s = -4 / 5 and csc t = 13/5, with W(s) in quadrant IV and W(t) in quadrant II, find cos(s -t). 37. If cos a = 5/13 and tan /3 = -2 , with angle a in quadrant I and angle /3 in quadrant II, find tan(a + /3). 38. If sec a = 5/3 and cot (3 = 15/8, with angle a in quadrant IV and angle (3 in quadrant III, find tan(a - /3). Prove each of the following identities by transforming the left-hand side of the equation into the expression on the right-hand side. 39. sin 2 a = 2 sin a cos a 40. cos It = cos 2 1 - sin 2 1 A 1 4r ~> 2 tan a 41. tan 2 a = 1 :—5— 1 - tan 2 a 42. sin(jc + y) sin(x -y) = sin 2 JC - sin 2 y 43. cos(x -y) cos(x + y) = cos 2 JC - sin 2 x sin (s + /) _ tan s + tan f sin (s - 0 tan s - tan f 45. csc(7 + 7r/2) = sec / 1 + tan x 1 - tan JC 1 + tan s tan t 47. tan(;c + TT/4) = 49. cot{s - t) = tan s - tan t 51. sin(s + f) + sin(s - f) = 2 sins cos / 52. cos(s + f) + cos(s -0 = 2 cos s cos / sin(x + h) -sinx . /cos h - /sin/i 53. : = sin x : + cos x 46. 48. 50. tan (a + 90°) = -cot a csc(/ -TT) = -csc / w . v cot w cot i; - 1 COt(l4 + V) = : ; z COt Li + COt V h h ) h cos(x + h) -cos x /cos h — .
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