Asked by Jam
I am having problems with proving identities that has a number and an exponent with it.
Like:
*2cosxcsc2x = cscx
*sin^2 2x = 4sin^2 x cos^2 x
*cot2(theta)=csc(theta)-2sin(theta)/2cos(theta)
Like:
*2cosxcsc2x = cscx
*sin^2 2x = 4sin^2 x cos^2 x
*cot2(theta)=csc(theta)-2sin(theta)/2cos(theta)
Answers
Answered by
Reiny
2cosxcsc2x = cscx
LS = 2cosx(1/sin 2x)
= 2cosx(1/(2sinxcosx)
= 1/sinx
= csc x
= RS
sin^2 2x = 4 sin^2 x cos^2 x
will use the same identity as in the first one:
namely, sin 2x = 2sinxcosx
LS = sin^2 2x
= (sin 2x)(sin 2x)
= (2sinxcosx)(2sinxcosx)
= 4 sin^2 x cos^2 x
= RS
cot 2Ø = cscØ - 2sinØ/2cosØ
I think it should be
cot 2Ø = (cscØ - 2sinØ)/2cosØ
RS = (1/sinØ - 2sinØ)/2cosØ
multiply top and bottom by sinØ
= (1 - 2sin^2 Ø)/(2sinØcosØ ----> recall cos 2x - 1 - 2sin^2 x
= cos 2Ø / sin 2Ø
= cot 2Ø
= LS
LS = 2cosx(1/sin 2x)
= 2cosx(1/(2sinxcosx)
= 1/sinx
= csc x
= RS
sin^2 2x = 4 sin^2 x cos^2 x
will use the same identity as in the first one:
namely, sin 2x = 2sinxcosx
LS = sin^2 2x
= (sin 2x)(sin 2x)
= (2sinxcosx)(2sinxcosx)
= 4 sin^2 x cos^2 x
= RS
cot 2Ø = cscØ - 2sinØ/2cosØ
I think it should be
cot 2Ø = (cscØ - 2sinØ)/2cosØ
RS = (1/sinØ - 2sinØ)/2cosØ
multiply top and bottom by sinØ
= (1 - 2sin^2 Ø)/(2sinØcosØ ----> recall cos 2x - 1 - 2sin^2 x
= cos 2Ø / sin 2Ø
= cot 2Ø
= LS
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