Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Interesting trig/series question

Challenging question for all! Want to see all your different approaches in solving this.

\sum _{ k=0 }^{ 90 }{ \sin ^{ k }{ x } } \cos ^{ 99-k }{ x } =0

Original post by morgzy24

Challenging question for all! Want to see all your different approaches in solving this.

\sum _{ k=0 }^{ 90 }{ \sin ^{ k }{ x } } \cos ^{ 99-k }{ x } =0

It's a bit late for me to do maths, but it looks like

\sin x = -\cos x

solves this by symmetry, so one solution is:

\tan x = -1 \Rightarrow x = -\frac{\pi}{4}+n\pi

(edited 8 years ago)

Original post by morgzy24

Challenging question for all! Want to see all your different approaches in solving this.

\sum _{ k=0 }^{ 90 }{ \sin ^{ k }{ x } } \cos ^{ 99-k }{ x } =0

is it meant to be 90 the top of the sum but 99 - k in the "argument" of the sum?

another solution is

\frac{\pi}{2} + n\pi

OP

Original post by TeeEm

is it meant to be 90 the top of the sum but 99 - k in the "argument" of the sum?

I don't believe so but maybe!

username1560589

cos x = 0 solves this trivially, so if cos x doesn't equal 0, we can factorise cos x out 99 times, which gives a geometric series. Doing some algebra gives

\tan x = \pm 1

.

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