Question

Solve: cos2 θ + cos8 θ = cos5 θ

Answer 1

||✪✪ QUESTION ✪✪||

Solve: cos2 θ + cos8 θ = cos5θ

|| ✰✰ ANSWER ✰✰ ||

Solving LHS,

→ cos2θ + cos8θ

using cosC + cosD = 2 cos(C+D/2) * cos(C-D/2) , we get,

→ 2 * cos(2θ + 8θ/2) * cos(2θ - 8θ/2)

→ 2 * cos5θ * cos(-3θ)

Now, we know that, cos(-θ) = cosθ

→ 2*cos5θ*cos3θ -------- Equation (1)

_________________________

Now, Lets Try to Solve RHS part :-

→ cos5θ

→ cos(3θ + 2θ)

Now using cos(C + D) = cosC*cosD - sinC*sinD

→ cos3θ*cos2θ - sin3θ*sin2θ -------- Equation (2)

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Putting Equation (1) and Equation (2) Equal now,

→ 2*cos5θ*cos3θ = cos3θ*cos2θ - sin3θ*sin2θ

→ sin3θ*sin2θ = cos3θ*cos2θ - 2*cos5θ*cos3θ

→ sin3θ*sin2θ = cos3θ [ cos2θ - 2*cos5θ ]

→ (sin3θ/cos3θ) * sin2θ = [ cos2θ - 2*cos5θ ]

→ Tan3θ * sin2θ = [ cos2θ - 2*cos5θ ]

I think Cant solve Further Now.

Answer 2

Answer:

Step-by-step explanation:

|✪✪ QUESTION ✪✪||

Solve: cos2 θ + cos8 θ = cos5θ

|| ✰✰ ANSWER ✰✰ ||

Solving LHS,

→ cos2θ + cos8θ

using cosC + cosD = 2 cos(C+D/2) * cos(C-D/2) , we get,

→ 2 * cos(2θ + 8θ/2) * cos(2θ - 8θ/2)

→ 2 * cos5θ * cos(-3θ)

Now, we know that, cos(-θ) = cosθ

→ 2*cos5θ*cos3θ -------- Equation (1)

_________________________

Now, Lets Try to Solve RHS part :-

→ cos5θ

→ cos(3θ + 2θ)

Now using cos(C + D) = cosC*cosD - sinC*sinD

→ cos3θ*cos2θ - sin3θ*sin2θ -------- Equation (2)

__________________________________

Putting Equation (1) and Equation (2) Equal now,

→ 2*cos5θ*cos3θ = cos3θ*cos2θ - sin3θ*sin2θ

→ sin3θ*sin2θ = cos3θ*cos2θ - 2*cos5θ*cos3θ

→ sin3θ*sin2θ = cos3θ [ cos2θ - 2*cos5θ ]

→ (sin3θ/cos3θ) * sin2θ = [ cos2θ - 2*cos5θ ]

→ Tan3θ * sin2θ = [ cos2θ - 2*cos5θ ]

I think Cant solve Further Now.

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