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Grade 11Trigonometry

Cos(alpha+beta)=4/5, Sin(alpha-beta)=5/13 then find the value of tan2alpha.

Profile image of Avijit Sen
7 Years agoGrade 11
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2 Answers

Profile image of kkbisht
7 Years ago
we know the T-identities:
Cos(A+B)= CosA. CosB – Sin A.SinB = 4/5  -----(1)
Sin( A- B) = Sin ACosB- CosA. SinB= 5/13   ------(2)
Now multiplying (1) and (2) we get
Cos2B(SinA.CosA) + (SinA.CosA)= 4/13 or  SinA.CosA= 4/13 as( Sin2B +Cos2B =1)
=>2 SinA.CosA = 8/13 or Sin2A= 8/13 
But Tan2A= Sin2A/Cos2A = Sin2A/ (\sqrt{}(1-Sin2 2A)         as you know ( Sin2A= 1-  Cos2A)
=> Tan2A= 8/13/(\sqrt{}(1-64/169) =8/13 /\sqrt{}105/13  = 8/\sqrt{}105
that is Tan 2\alpha }= 8/\sqrt{}105
kkbisht
Profile image of kkbisht
7 Years ago
we know the T-identities:
Cos(A+B)= CosA. CosB – Sin A.SinB = 4/5  -----(1)
Sin( A- B) = Sin ACosB- CosA. SinB= 5/13   ------(2)
Now multiplying (1) and (2) we get
(Cos2B + Sin2B)(SinA.CosA)  – ( (sin2A+ Cos2A) ( SinB.CosB)= 4/13 or  SinA.CosA= 4/13 as( Sin2B +Cos2B =1)
=>SinA.CosA – SinB.CosB= 4/13   ------------------------------  (3)    as ( Sin2A + Cos2A=1)
Also Sin(A+B)=\sqrt{}{1-cos2(A+B)}= b\sqrt{}(1- 16/25) =\sqrt{}(9/25) = 3/5       ( as given : cos(A+B)=4/5 )
and Cos(A-B) =\sqrt{}{1-sin2(A-B)} = \sqrt{}{1- (25/169)}  = \sqrt{}(144/169)= 12/13   ( as  given:Sin(A-B)= 5/13 )
Now Sin(A+B) = 
                          Sin ACosB+ CosA. SinB= 3/5    ---- (4)   and
         Cos(A-B)=
                          CosA. CosB + Sin A.SinB = 12/13 -----(5)
Now multiply (4) & (5) we obtain
SinA.CosA(Cos2B+ Sin2B) + SinB.CosB( Sin2A+Cos2A)= 36/65
SinA.CosA x1    + SinB.CosB x1=36/65
or SinACosA + SinB.CosB= 36/65  ---------------(6)
Now Adding  (3) & (6) we get  2 SinA.CosA=(4/13)  + 36/65  = (20+36) /65=56/65
 
Sin2A=56/65 -----------------(7)  
Hence Tan2A= Sin2A/Cos2A  = Sin2A/\sqrt{}(1-Sin2 2A) =(56/65) / \sqrt{}{1- (56/65)2 = (56/65) / \sqrt{}{(9/65) . (121/65)}
= (56/65) /(33/65)
=56/33
This is correct Answer 
 
kkbisht

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