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Maths-

General

Easy

Question

If f(α) = $open square brackets table row cell cos invisible function application alpha end cell cell sin invisible function application alpha end cell row cell negative sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close square brackets$ and $alpha comma beta comma gamma$ are angles of triangle then f( $alpha$ ). f( $beta$ ).f ( $gamma$ ) =

I₂
– I₂
0
None

The correct answer is: – I₂

Here
f ( $alpha$ ) . f ( $beta$ ) . f( $gamma$ ) = $open square brackets table row cell cos invisible function application left parenthesis alpha plus beta plus gamma right parenthesis end cell cell sin invisible function application left parenthesis alpha plus beta plus gamma right parenthesis end cell row cell negative sin invisible function application left parenthesis alpha plus beta plus gamma right parenthesis end cell cell cos invisible function application left parenthesis alpha plus beta plus gamma right parenthesis end cell end table close square brackets$
= $open square brackets table row cell cos invisible function application pi end cell cell sin invisible function application pi end cell row cell negative sin invisible function application pi end cell cell cos invisible function application pi end cell end table close square brackets$ = $open square brackets table row cell negative 1 end cell 0 row 0 cell negative 1 end cell end table close square brackets$ = –I₂

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chemistry-General

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