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Trigonometry

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nid404:
Ex 5 A

2) a)sec (1/2pi -x)

= 1/ cos (1/2pi -x)

cos(1/2 pi-x)=sin x

so 1/cos(1/2pi-x)=1/sinx = cosec x


f) cosec (pi + x)

=1/sin(pi+x)

sin(pi+x)= -sin x

1/-sin x= -cosec x


3)  c) cot 5/6 pi

=1/tan 5/6pi

= 1/-1/

=-

e) cot(-1/3pi)

1/ tan(-1/3 pi)

= 1/-

5) b) cot A= 1/tan A

tan A= sin A/cos A

1/tan A= cos A/sin A

1/tan A= (4/5)/ (3/5)= 4/3


d) cosecB = 1/sin B

sin= / 2

cosecB= 2/ = 2/3

nid404:
Additional formulae

cos(A-B)=cosAcosB+sinAsinB
cos(A+B)=cosAcosB-sinAsinB
sin(A+B)=sinAcosB+cosAsinB
sin(A-B)= sinAcosB-cosAsinB


tan(A+B)=

tan(A-B)=


Ex 5B

2)

a) cos 105o

cos (60+45) = cos60.cos45- sin60.sin45
                  =1/2. 1/- /2. 1/
                  =1/2- /2
                  = 1-3/ 2
                  = -/ 4

4)  sin(3/2pi + x)

= sin3/2picosx+ cos3/2pisinx
=-1cosx+0sinx
=-1cosx

nid404:
Double angle formulae


sin (A+B)= sinAcosB+cosAsinB

when A=B

sin (A+A)= sinAcosA+cosAsinA
sin 2A= 2cosAsinA


cos(A+B)=cosAcosB-sinAsinB
when A=B
cos 2A= cosAcosA-sinAsinA
cos2A=cos2A- sin2A

cos 2A in terms of sin2A

cos2A+ sin2A=1
cos2A=1-sin2A

cos2A= 1-sin2A-sin2A
        =1-2sin2A

in terms of cos2A

cos2A=  cos2A-(1- cos2A)
        =2cos2A-1

tan(A+B)=

if A=B

tan 2A=

nid404:
5C

3) sin 3A

sin (2A+A)= sin2AcosA+cos2AsinA
sin2A=2cosAsinA and cos2A= 1-2sin2A
sin3A= 2cosAsinAcosA + 1-2sin2AsinA
sin3A=2(1-sin2A)sinA+ sinA- 2sin3A
sin3A= 2sinA-2sin3A+sinA-2sin3A
sin3A=3sinA-4sin3A

8)tan 2A= 12/5
tan 2A= 2tanA/1-tan2A
12/5=2tanA/1-tan2A

12-12tan2A=10tanA
-12tan2A-10tanA+12=0

let tanA=x

-12tan2x-10tanx+12=0

solve quadratic

x=-1.5 and 2/3

thus tan A= -1.5 and 2/3

shedow_21:
 :o Did you learned all that......

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