trig problem (1 Viewer)

Aznmichael92

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hey all i just encountered another trig problem. Can someone help me?

Questions are.

1.If sin a = 3/5, 0* < a < 90* and sin b = 5/13, 90* < b < 180*, find exact values of

i) sin 2a
ii) cos (a-b)

2. Simplify sinAcosAcos2A.

My first instinct for this problem is 1/2 sin2Acos2A. Is that correct?
 

vds700

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Aznmichael92 said:
hey all i just encountered another trig problem. Can someone help me?

Questions are.

1.If sin a = 3/5, 0* < a < 90* and sin b = 5/13, 90* < b < 180*, find exact values of

i) sin 2a
ii) cos (a-b)

2. Simplify sinAcosAcos2A.

My first instinct for this problem is 1/2 sin2Acos2A. Is that correct?
i)sin2A = 2sinAcosA
=2(3/5)(4/5) = 24/25 draw a right angles triangle with sinA = 3/5, use pythagoras' theorem to find other side

ii)cos(A-B) = cosAcosB+sinAsinB same thiung, just draw up another triangle with B, and sub in the values

sinAcosAcos2A = (1/2)sin2Acos2A = (1/4)sin4A
 

Timothy.Siu

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Aznmichael92 said:
hey all i just encountered another trig problem. Can someone help me?

Questions are.

1.If sin a = 3/5, 0* < a < 90* and sin b = 5/13, 90* < b < 180*, find exact values of

i) sin 2a
ii) cos (a-b)

2. Simplify sinAcosAcos2A.

My first instinct for this problem is 1/2 sin2Acos2A. Is that correct?
is this 2 unit?
i)sin 2a=2sin acos a= 2 x 3/5 x 4/5 = 24/25
ii) cos (a-b)= cos a cos b + sin a sin b = 4/5 x -5/12 + 3/4x5/13 = -7/156

2. if its not linked to question 1. then
sin Acos Acos 2A = 1/2sin 2A cos 2A = 1/4sin 4A?
 

trailblazer

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Timothy.Siu said:
is this 2 unit?
i)sin 2a=2sin acos a= 2 x 3/5 x 4/5 = 24/25
ii) cos (a-b)= cos a cos b + sin a sin b = 4/5 x -5/12 + 3/4x5/13 = -7/156

2. if its not linked to question 1. then
sin Acos Acos 2A = 1/2sin 2A cos 2A = 1/4sin 4A?
Yeh is it really 2-unit? And what rule was used in part 2?
 

bored of sc

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Aznmichael92 said:
hey all i just encountered another trig problem. Can someone help me?

Questions are.

1.If sin a = 3/5, 0* < a < 90* and sin b = 5/13, 90* < b < 180*, find exact values of

i) sin 2a
ii) cos (a-b)

2. Simplify sinAcosAcos2A.

My first instinct for this problem is 1/2 sin2Acos2A. Is that correct?
This is extension 1.
 

Aznmichael92

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yes apparently its 2 unit. :( I am currently in year 10 and my maths teacher say this is 2 unit work. I think he took it out from some excel book. Thanks for the help.

Um just another question I need help with.

Prove cos 2a = 1 - tan^2 a/ 1 + tan^2 a

edit woops is it?

Guess from now on have to post in ext 1 thread
 

Timothy.Siu

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Aznmichael92 said:
yes apparently its 2 unit. :( I am currently in year 10 and my maths teacher say this is 2 unit work. I think he took it out from some excel book. Thanks for the help.

Um just another question I need help with.

Prove cos 2a = 1 - tan^2 a/ 1 + tan^2 a

edit woops is it?

Guess from now on have to post in ext 1 thread
Prove cos 2a = 1 - tan^2 a/ 1 + tan^2 a

RHS= (1 - tan^2 a)/ (1 + tan^2 a)

= 1-(sec^2 a -1)/sec^2 a
=2cos^2 a - 1 = 2cos^2 a - (sin^2 a+cos^2 a)
= cos^2 a - sin^2 a = cos 2a
LHS= cos 2a
RHS=LHS

the key is to change all of one trig thing into another, since theres cos on one side and tan on the other.
 

Aznmichael92

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Timothy.Siu said:
Prove cos 2a = 1 - tan^2 a/ 1 + tan^2 a

RHS= (1 - tan^2 a)/ (1 + tan^2 a)

= 1-(sec^2 a -1)/sec^2 a
=2cos^2 a - 1 = 2cos^2 a - (sin^2 a+cos^2 a)
= cos^2 a - sin^2 a = cos 2a
LHS= cos 2a
RHS=LHS

the key is to change all of one trig thing into another, since theres cos on one side and tan on the other.
ty but there is also part ii to the question

hence show tan67*30' = 1 + root 2
 

Timothy.Siu

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tan 135= sin 135/cos 135

-1= 1/root2 x (1+tan^2 a) /(1-tan^2 a) where a is 67.5 degrees
multiply root2(1-tan^2 a) to both sides
-root 2 + root2tan^2 a= 1+tan^a
tan^2 a(root2 -1)=root 2 +1
tan^2 a= (root2+1)/(root 2 -1)
= (root2+1)2
therefore tan 67.5= root 2 +1
 

Aznmichael92

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Timothy.Siu said:
is this 2 unit?
i)sin 2a=2sin acos a= 2 x 3/5 x 4/5 = 24/25
ii) cos (a-b)= cos a cos b + sin a sin b = 4/5 x -5/12 + 3/4x5/13 = -7/156

2. if its not linked to question 1. then
sin Acos Acos 2A = 1/2sin 2A cos 2A = 1/4sin 4A?
y negative 5 on 12? isnt it 12/13?
 

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