the solved ones :
cos(n - x) = - cos x
sin (n +x) = - sin x
-----------------------------
the unsolved ones :
tan (2n - x) = ?
cot (2n +x) = ?
cos(5n +x) = ?
sin (7n - x) = ?
tan (3n/2 - x) = ?
cos (n/2 + x) = ?
sin (n/2 - x) = ?
sin (3n/2 - x) = ?
cos(3n/2 - x) = ?
sin (5n/2 +x) = ?
tan(11n/2 - x) = ?
------------------------------...
please answer JUST IF YOU KNOW THE ANSWER
and cheak my solved answers.
I know it's a long home work but, take your time solving it.
please describe what you write and DON'T WRITE THE ANSWERS ONLY.
and I'll be thankful.
Answers:
I guess you mean n = π
you use the rules
cos(a+b) = cosacosb - sinasinb
cos(a-b) = cosacosb + sinasinb
sin(a+b) = sinacosb + sinbcosa
sin(a-b) = sinacosb - sinbcosa
tan(a+b) = (tana + tanb) / (1 - tanatanb)
tan(a-b) = (tana - tanb) / (1 + tanatanb)
and remember cot is 1/tan
your solved ones are correct
surely you don't expect me to do all of them for you
I'll do one of each type
tan (2π - x)
= (tan2π - tanx) / (1 + tan2πtanx)
= (0 - tanx) / (1 + 0*tanx)
= -tanx
cos (π/2 + x)
= cosπ/2cosx - sinπ/2sinx
= 0*cosx - 1*sinx
= -sinx
sin (3π/2 - x)
= sin3π/2cosx - sinxcos3π/2
= -1*cosx - (sinx)*0
= -cosx
simple as that
just plug and chug
.
x=101010 i know
If your first two solutions are correct, then cos(n-x) = -cos x, and sin(n+x) = -sin x indicates that n is meant to be 180 degrees.
Based on n= 180 degrees, then the following is true:
tan (2n - x) = -tan x This is like tan(360-x) = tan(-x) = -tan x
cot (2n +x) = cot x Same as above
cos(5n +x) = cos(n+x) = -cos x. This is like cos(n+x) = -cos x
sin (7n - x) = sin(n-x) =sin x. This is like sin(n-x) = +sin x
tan (3n/2 - x) = +cot x. This as tan(n-x) = cot x
cos (n/2 + x) = -sin x. Same as cos(90+x) = -sinx
sin (n/2 - x) = +cos x. Same as sin(90-x) =cosx
sin (3n/2 - x) = -cos x. Same as sin(270-x) = -cosx
cos(3n/2 - x) = -sin x. Same as cos(270-x) = -snx
sin (5n/2 +x) = +cos x. Same as sin(90-x) = cosx
tan(11n/2 - x) = +cot x. Do this one yourself
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