TRigonometric formulae

Question 1

sin(A+B)=

Answer 1

sinAcosB + cosAsinB

Question 2

Sin(A-B)

Answer 2

SinAcosB - cosAsinB

Question 3

cos(A+B)

Answer 3

cosAcosB - sinAsinB

Question 4

cos(A-B)

Answer 4

cosAcosB + sinAsinB

Question 5

sin2A=

Answer 5

2sinAcosB

Question 6

cos2A

Answer 6

cos²A-sin²A= 2cos²A-1 =1-2sin²A

(using cos²+sin²=1)

Question 7

show cos(x+90)^o = -sinx

Answer 7

cos(A+B)= cosAcosB-sinAsinB

cosxcos90-sinxsin90

(cosx) (0) - (sinx)(1)
- sinx

Question 8

show cos15= 1+sqrt(3)/2sqrt(2)

Answer 8

cos(60-45)= cos60cos45+sin60sin45

=1+sqrt(3)/2sqrt(2)

Question 9

if sinA=1/y

find a)cos2A b) sin2A

Answer 9

a) cos2A=1-2sin²A=1-(2 x 1/16)= 1- 1/8 =7/8
b) sin²2A + cos²2A=1, sin2A =+-sqrt(1-cos²2A)

sin2A= +-sqrt(1- 49/64) =+-sqrt(15/64) = +-sqrt(15)/8

Question 10

tan(A+B) in terms of sin(A+B) and cos(A+B)

Answer 10

sin(A+B)/cos(A+B)

=sinAcosB+cosAsinB/cosAcosB-sinAsinB

divide top and bottom by cosAcosB

=(sinA/cosA + sinBcosB)/1-(2sinAsinB/cosAcosB)

=tanA + tanB/1-tanAtanB

Question 11

tan(A-B)

Answer 11

tanA-tan(-B)/1-tanAtan(-B)

=tanA-tanB/1+tanAtanB

Question 12

tan2A

Answer 12

tan(A+A)

tanA+tanA/1-tanAtanA

2tanA/1-tan²A

Question 13

tan2A=2 what does tanA=

Answer 13

tan2A=2tanA/1-tan²A

2tanA/1-tan2​A=2

2tanA=2(1-tan2​A)

tan²A+tanA-1=0

tanA=-1+-sqrt(1²-(4x1x-1)/2x1

tanA=-1+-sqrt(5)/2

tanA= -1/2+sqrt(5)/2 or -1/2 -sqrt(5)/2

Question 14

find tan3A in terms of tanA

Answer 14

tan3A= tan(A+2A)= tanA+tan2A/1-tanAtan2A

use double angle formula

tan2A=tanA+tanA/1-tan²A

put tan2A values into tan3A

=tanA+(2tanA/1-tan²A)/1-tanA(2tanA/1-tan²A)

replace tanA by t, to make t+(2t/1-t²)/1-t(2t/1-t²)

then times by (1-t²)

=t(1-t²)+2t/1(1-t²-2txt)=3t-t²/1-3t²

=3tanA-tan³A/1-3tan²A

Question 15

3sin2x=cosx give answers between 0 and 2π

Answer 15

3sin2x=cosx

3x2sinxcosx=cosx

6sinxcosx-cosx=o

cosx(6sinx-1)=0

cosx=0, x =π/2 or 3π/2

6sinx-1=0. sinx=1/6, x= arcsin1/6 or π-arcsin1/6

Question 16

prove 1-cos2A/sin2A =tanA

Answer 16

1-cos2A/sin2A= 1-(1-2sin²A)/2sinAcosA =2sin²A/2sinAcosA =sinA/cosA =tanA

Question 17

n € Z means

Answer 17

n is a member of the integers (…, -2,-1,0,1,2,…)

Question 18

A ≠nπ, n€Z means

Answer 18

A cannot be …,-2π,-π,0,π,2π,…

Question 19

show that cosx-sinx-1=-2sinx/2(sinx/2 +cosx/2)

Answer 19

cos2A=1-2sin²A, replace A by x/2 —>cosx=1-2sin²x/2

sin2A=2sinAcosA

sinx=2sin(x/2)cos(x/2)

cosx-sinx-1=(1-2sin²(x/2)-2sin(x/2)cos(x/2)-1

=-2sin²(x/2)-2sin(x/2)cos(x/2)

=-2sin(x/a)(sin(x/2)+cos(x/2))