Question

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Cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A​

Answer 1

cos3A + cos5A + cos7A + cos15A = 4 cos4A cos5A cos6A

Step-by-step explanation:

L.H.S. = cos3A + cos5A + cos7A + cos15A

= (cos3A + cos7A) + (cos5A + cos15A)

= 2 cos{(3A + 7A)/2} cos{(3A - 7A)/2} + 2 cos{(5A + 15A)/2} cos{(5A - 15A)/2}

= 2 cos(10A/2) cos(- 4A/2) + 2 cos(20A/2) cos(- 10A/2)

= 2 cos5A cos2A + 2 cos10A cos5A

= 2 cos5A (cos2A + cos10A)

= 2 cos5A [2 cos{(2A + 10A)/2} cos{(2A - 10A)/2}]

= 2 cos5A * 2 cos(12A/2) cos(- 8A/2)

= 4 cos5A cos6A cos4A

= 4 cos4A cos5A cos6A = R.H.S.

Hence proved.

Rules:

• sinC + sinD = 2 sin{(C+D)/2} cos{(C - D)/2}

• sinC - sinD = 2 cos{(C + D)/2} sin{(C - D)/2}

• cosC + cosD = 2 cos{(C + D)/2} cos{(C - D)/2}

• cosC - cosD = 2 sin{(C + D)/2} sin{(D - C)/2}

• cos(- θ) = cosθ

Other trigonometric questions:

• Prove (b - a cosC) sinA = a cosA sinC. - https://brainly.in/question/13188154
• Find the value of (sec⁴θ - sec²θ). - https://brainly.in/question/13320789

Answer 2

The required prove is shown below:

Step-by-step explanation:

Consider the provided function.

cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A​

Use the trigonometry identity: $\cos A+\cos B=2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$

Consider the LHS

$2cos(\frac{3A+5A}{2})\cos(\frac{3A-5A}{2})+2\cos(\frac{7A+15A}{2})\cos(\frac{7A-15A}{2})$

$2cos(4A)\cos(-A)+2cos(11A)\cos(-4A)$

$2cos(4A)\cos(A)+2cos(11A)\cos(4A)$ (∵cos(-θ)=cosθ)

$2cos(4A)(\cos(A)+cos(11A))$

$2cos(4A)(2\cos(\frac{A+11A}{2})\cos(\frac{A-11A}{2}))$

$2cos(4A)(2\cos(6A)\cos(-5A))$

$2cos(4A)(2\cos(6A)\cos(5A))$ (∵cos(-θ)=cosθ)

$4cos(4A)\cos(5A)\cos(6A)$

RHS=LHS

Hence proved.

#Learn More

Cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A

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