Question

Sin2a+sin4a+sin6a+sin8a/cos2a+cos4a+cos6a+cos8a=tan5a

Answer 1

Step-by-step explanation:

2sin(2A+4A/2)cos(2A-4A/2) + 2sin(6A+8A/2)cos(6A-8A/2)/2cos(2A+4A/2)cos(2A-4A/2)+2cos(6A+8A/2)cos(6A-8A/2)

------using identity sinA + sinB=2sin(A+B/2)cos(A-B/2)

cosA+cosB= 2cosA+B/2 cos A-B/2

2 sin3Acos(-A)+2sin7Acos(-A) /2cis3Acis(-A) +2cos7Acos(-A)

2cos(-A) (sin3A+sin7A)/2cos(-A) (cos3A+cos7A)

sin3a+sin7a/cos3a+cos7a

2sin3a+7a/2 cos3a-7a/2/2cos3a+7a/2cos3a-7a/2

sin5a cos -4a/cos5a cos-4a

sin5a/cos5a=tan5a

Answer 2

Therefore ( Sin 2a + Sin 4a + Sin 6a + Sin 8a ) / ( Cos 2a + Cos 4a + Cos 6a + Cos 8a ) = Tan 5a has been proved successfully.

Given:

( Sin 2a + Sin 4a + Sin 6a + Sin 8a ) / ( Cos 2a + Cos 4a + Cos 6a + Cos 8a ) = Tan 5a

To Find:

Prove that, ( Sin 2a + Sin 4a + Sin 6a + Sin 8a ) / ( Cos 2a + Cos 4a + Cos 6a + Cos 8a ) = Tan 5a

Solution:

The given question can be answered as shown below.

Given:

LHS: ( Sin 2a + Sin 4a + Sin 6a + Sin 8a ) / ( Cos 2a + Cos 4a + Cos 6a + Cos 8a )

RHS: Tan 5a

Some trigonometric identities,

Sin A + Sin B = 2sin [ ( A + B ) / 2 ]. Cos [ ( A - B ) / 2 ]

Cos A + Cos B = 2 Cos [ ( A + B ) / 2 ]. Cos [ ( A - B ) / 2 ]

Solving LHS:

⇒ ( Sin 2a + Sin 4a + Sin 6a + Sin 8a ) / ( Cos 2a + Cos 4a + Cos 6a + Cos 8a )

⇒ [ ( Sin 2a + Sin 8a ) + ( Sin 4a + Sin 6a ) ] / [ ( Cos 2a + Cos 8a ) + ( Cos 4a + Cos 6a ) ]

Applying trigonometric identities,

⇒ ( Sin 2a + Sin 8a ) = 2 Sin [ ( 2a + 8a ) / 2 ]. Cos [ ( 2a - 8a ) / 2] = 2 Sin 5a. Cos 3a

⇒ ( Sin 4a + Sin 6a ) = 2 Sin [ ( 4a + 6a ) / 2 ]. Cos [ ( 4a - 6a ) / 2] = 2 Sin 5a. Cos a

⇒ ( Cos 2a + Cos 8a ) = 2 Cos [ ( 2a + 8a ) / 2 ]. Cos [ ( 2a - 8a ) / 2] = 2 Cos 5a. Cos 3a

⇒ ( Cos 4a + Cos 6a ) = 2 Cos [ ( 4a + 6a ) / 2 ]. Cos [ ( 4a - 6a ) / 2] = 2 Cos 5a. Cos a

Now in the question,

⇒ [ ( Sin 2a + Sin 8a ) + ( Sin 4a + Sin 6a ) ] / [ ( Cos 2a + Cos 8a ) + ( Cos 4a + Cos 6a ) ] = [ 2 Sin 5a. Cos 3a + 2 Sin 5a. Cos a ] / [ 2 Cos 5a. Cos 3a + 2 Cos 5a. Cos a ]

⇒ 2 Sin 5a ( Cos 3a + Cos a ) / 2 Cos 5a ( Cos 3a + Cos a )

⇒ Sin 5a / Cos 5a = Tan 5a

RHS: Tan 5a

⇒ LHS = RHS

Hence proved.

Therefore ( Sin 2a + Sin 4a + Sin 6a + Sin 8a ) / ( Cos 2a + Cos 4a + Cos 6a + Cos 8a ) = Tan 5a has been proved successfully.

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