Question

Please convert this into trigonometrical form: -
\[\sin 32^o+i \cos 32^o.\]

Answer 1

assuming you meant exponential form :
we have this :
-i^2 sin32 + icos32
take i common :
i(cos32 - isin32)
=ie^(-i32)

Answer 2

\[ \huge a + ib = e^{\ln |\sqrt{a^2 + b^2 }| + \tan^{-1}\left ( b \over a\right )}\]

Answer 3

and covert that degree into radians ..

Answer 4

But In my book answer is given as
wait a minute

Answer 5

OH ... what do you mean by trigonometric form??
if it's in the from of cos \theta + i sin \theta
then just subtract angle from 90 ... on both cos and sin

Answer 6

\[\cos(360^ok+58^o)+i \sin (360^ok+58^o), k \in z\]

Answer 7

why should i do that @experimentX Plz tell

Answer 8

Hmm ... trigonometric functions are periodic in 360 ... adding or subtracting 360 or it's multiple does not have any effect. so ti's general solution.
plus 90 - 32 = 58
and sin(90 - \theta) = cos \theta

Answer 9

ahh i see..
the book wants you to answer like this :
substitute cos 32 = sin 58 and sin 42 = cos58
also note that since periodicity of both these func is 2pi,,you may easily write
cos(58) = cos(2npi + 58) and similar with sin(58)
then you'll get the ans which book says..

Answer 10

k! thanx @shubhamsrg & @experimentX :)

Answer 11

yw

Answer 12

glad to help ^_^

Answer 13

^_^