i-precalculus.com

I-Precalculus

Sin theta = 2 n sin frac{ theta}{2 n} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 n} $. Clearly true if $n=1$ because $ sin theta = 2 sin frac{ theta}{2} cos frac{ theta}{2} $. Assume its true for $n=k$. So we have. Sin theta = 2 k sin frac{ theta}{2 k} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $. 2 k left(2 sin frac{ theta}{2 {k 1} cos frac{ theta}{2 {k 1} right) cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $.

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I-Precalculus | i-precalculus.com Reviews

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Sin theta = 2 n sin frac{ theta}{2 n} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 n} $. Clearly true if $n=1$ because $ sin theta = 2 sin frac{ theta}{2} cos frac{ theta}{2} $. Assume its true for $n=k$. So we have. Sin theta = 2 k sin frac{ theta}{2 k} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $. 2 k left(2 sin frac{ theta}{2 {k 1} cos frac{ theta}{2 {k 1} right) cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $.

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I-Precalculus | i-precalculus.com Reviews

https://i-precalculus.com

Sin theta = 2 n sin frac{ theta}{2 n} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 n} $. Clearly true if $n=1$ because $ sin theta = 2 sin frac{ theta}{2} cos frac{ theta}{2} $. Assume its true for $n=k$. So we have. Sin theta = 2 k sin frac{ theta}{2 k} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $. 2 k left(2 sin frac{ theta}{2 {k 1} cos frac{ theta}{2 {k 1} right) cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $.

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I-Precalculus: Prove by Induction

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Sin theta = 2 n sin frac{ theta}{2 n} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 n} $. Clearly true if $n=1$ because $ sin theta = 2 sin frac{ theta}{2} cos frac{ theta}{2} $. Assume its true for $n=k$. So we have. Sin theta = 2 k sin frac{ theta}{2 k} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $. 2 k left(2 sin frac{ theta}{2 {k 1} cos frac{ theta}{2 {k 1} right) cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $.

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Sin theta = 2 n sin frac{ theta}{2 n} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 n} $. Clearly true if $n=1$ because $ sin theta = 2 sin frac{ theta}{2} cos frac{ theta}{2} $. Assume its true for $n=k$. So we have. Sin theta = 2 k sin frac{ theta}{2 k} cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $. 2 k left(2 sin frac{ theta}{2 {k 1} cos frac{ theta}{2 {k 1} right) cos frac{ theta}{2} cos frac{ theta}{4} cdots cos frac{ theta}{2 k} $.

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