# What is product `sin 7^o cos 14^o cos 28^o cos 56^o` ?

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### 1 Answer

You could use the double angle formula, such that:

`2 sin alpha cos alpha = sin 2alpha`

Starting from left, you should notice that you just have the factor `sin 7^o` , hence, you may multiplicate by `2 cos 7^o` , such that:

`P = sin 7^o*cos 14^o*cos 28^o*cos 56^o`

`2 cos 7^o*P = 2 cos 7^o*sin 7^o*cos 14^o*cos 28^o*cos 56^o`

Converting `2 cos 7^o*sin 7^o` in `sin 14^o` yields:

`2 cos 7^o*P = sin 14^o*cos 14^o*cos 28^o*cos 56^o `

`4cos 7^o*P = 2sin 14^o*cos 14^o*cos 28^o*cos 56^o`

Converting `2 sin 14^o*cos 14^o` in `sin 28^o` yields:

`4cos 7^o*P = sin 28^o*cos 28^o*cos 56^o`

`8cos 7^o*P = 2 sin 28^o*cos 28^o*cos 56^o`

Converting `2 sin 28^o*cos 28^o` in `sin 56^o` yields:

`8cos 7^o*P = sin 56^o*cos 56^o`

`16cos 7^o*P = 2 sin 56^o*cos 56^o`

Converting `2 sin 56^o*cos 56^o` in `sin 112^o` yields:

`16cos 7^o*P = sin 112^o => P = (sin 112^o)/(16cos 7^o)`

`P = 0.927/(16*0.992) => P ~~ 0.05`

**Hence, evaluating the given product, under the given conditions, yields **`P ~~ 0.05.`