Question

1+cos 2 theta + cos 6 theta/cos 4 theta = sin 3 theta / sin theta

Answer 1

Answer:

Step-by-step explanation:

LHS:

$1+\frac{cos2\theta+cos6\theta}{cos4\theta}$

We know that

$cosx+cosy=2cos\frac{x+y}{2}cos\frac{x-y}{2}$

Using the formula

$1+\frac{2cos4\theta cos2\theta}{cos4\theta}=1+2cos2\theta$

$cos2\theta=1-2sin^2\theta$

Using the formula

$1+2(1-2sin^2\theta)$

$1+2-4sin^2\theta$

$3-4sin^2\theta$

RHS:

$\frac{sin3\theta}{sin\theta}$

We know that

$sin3\theta=3sin\theta-4sin^3\theta$

Using the formula

$\frac{3sin\theta-4sin^3\theta}{sin\theta}$

$\frac{3sin\theta}{sin\theta}-\frac{4sin^3\theta}{sin\theta}=3-4sin^2\theta$

LHS=RHS

Hence, proved.

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Answer 2

Step-by-step explanation:

tep-by-step explanation:

LHS:

1+\frac{cos2\theta+cos6\theta}{cos4\theta}1+

cos4θ

cos2θ+cos6θ

We know that

cosx+cosy=2cos\frac{x+y}{2}cos\frac{x-y}{2}cosx+cosy=2cos

2

x+y

cos

2

x−y

Using the formula

1+\frac{2cos4\theta cos2\theta}{cos4\theta}=1+2cos2\theta1+

cos4θ

2cos4θcos2θ

=1+2cos2θ

cos2\theta=1-2sin^2\thetacos2θ=1−2sin

2

θ

Using the formula

1+2(1-2sin^2\theta)1+2(1−2sin

2

θ)

1+2-4sin^2\theta1+2−4sin

2

θ

3-4sin^2\theta3−4sin

2

θ

RHS:

\frac{sin3\theta}{sin\theta}

sinθ

sin3θ

We know that

sin3\theta=3sin\theta-4sin^3\thetasin3θ=3sinθ−4sin

3

θ

Using the formula

\frac{3sin\theta-4sin^3\theta}{sin\theta}

sinθ

3sinθ−4sin

3

θ

\frac{3sin\theta}{sin\theta}-\frac{4sin^3\theta}{sin\theta}=3-4sin^2\theta

sinθ

3sinθ

−

sinθ

4sin

3

θ

=3−4sin

2

θ

LHS=RHS

Hence, proved.