Show this is identity sin 10o cos20o cos 40o=1/8

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You need to prove the following trigonometric identity, such that:

`sin10^o*cos 20^o*cos 40^o = 1/8`

Multiplying by 8 both sides yields:

`8sin10^o*cos 20^o*cos 40^o = 1`

You need to multiply by `cos 10^o` both sides, such that:

`8sin10^o*cos 10^o*cos 20^o*cos 40^o = cos 10^o`

You need to use the following double angle identity, such that:

`2 sin alpha*cos alpha = sin 2*alpha`

Reasoning by analogy yields:

`2 sin 10^o*cos 10^o = sin 2*10^o`

Replacing `sin 20^o` for `2 sin 10^o*cos 10^o` yields:

`4*sin 20^o*cos 20^o*cos 40^o = cos 10^o`

`2*2sin 20^o*cos 20^o*cos 40^o = cos 10^o`

Replacing `sin 40^o` for `2sin 20^o*cos 20^` o yields:

`2*sin 40^o*cos 40^o = cos 10^o`

Replacing `sin 80^o` for `2sin 40^o*cos 40^o` yields:

`sin 80^o = cos 10^o`

You need to use the following trigonometric identity, such that:

`sin alpha = cos (90^o - alpha)`

Reasoning by analogy yields:

`sin 80^o = cos(90^o - 80^o) = cos 10^o` valid

**Hence, testing if `sin10^o*cos 20^o*cos 40^o = 1/8` is identity yields the valid identity `sin 80^o = cos 10^o` , hence` sin10^o*cos 20^o*cos 40^o = 1/8` holds.**

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