prove that cos theta * cos 2 theta * cos 4 theta * cos 8 theta = sin 16 theta / 16 sin theta
Answers
(instead of θ, I use A)
Answer:
Step-by-step explanation:
Given :
To prove that :
Proof :
We know that,
____
LHS =
(Multiplying & Dividing by 2 sin A)
⇒
⇒
⇒
(Multiplying & Dividing by 2 )
⇒
⇒
⇒
(Multiplying & Dividing by 2 )
⇒
⇒
⇒
(Multiplying & Dividing by 2 )
⇒
⇒
⇒ = RHS
Hence, proved,.
Answer:
- cos A × cos 2A× cos 4A× cos 8A=16 sin Asin 16A
Proof :
We know that,
sin \ 2A \ = 2 \ sin \ A \ cos \ Asin 2A =2 sin A cos A
____
LHS = cos \ A \ \times \ cos \ 2A\times \ cos \ 4A\times \ cos \ 8Acos A × cos 2A× cos 4A× cos 8A
(Multiplying & Dividing by 2 sin A)
⇒ \frac{2 \ sin \ A \ \times \ cos \ A \ \times \ cos \ 2A\times \ cos \ 4A\times \ cos \ 8A}{2 \ sin \ A}2 sin A2 sin A × cos A × cos 2A× cos 4A× cos 8A
⇒ \frac{(2 \ sin \ A \ \times \ cos \ A) \ \times \ cos \ 2A\times \ cos \ 4A\times \ cos \ 8A}{2 \ sin \ A}2 sin A(2 sin A × cos A) × cos 2A× cos 4A× cos 8A
⇒ \frac{sin \ 2A \ \times \ cos \ 2A\times \ cos \ 4A\times \ cos \ 8A}{2 \ sin \ A}2 sin Asin 2A × cos 2A× cos 4A× cos 8A
(Multiplying & Dividing by 2 )
⇒ \frac{2 \times \ sin \ 2A \ \times \ cos \ 2A \ \times \ cos \ 4A\times \ cos \ 8A}{2 \times \ 2 \ sin \ A}2× 2 sin A2× sin 2A × cos 2A × cos 4A× cos 8A
⇒ \frac{(2 \times \ sin \ 2A \ \times \ cos \ 2A) \ \times \ cos \ 4A\times \ cos \ 8A}{2 \times \ 2 \ sin \ A}2× 2 sin A(2× sin 2A × cos 2A) × cos 4A× cos 8A
⇒ \frac{ sin \ 4A \ \times \ cos \ 4A\times \ cos \ 8A}{4 \ sin \ A}4 sin Asin 4A × cos 4A× cos 8A
(Multiplying & Dividing by 2 )
⇒ \frac{2 \times \ sin \