1樓:匿名使用者
用反三角函式拉
給你定義了一下
'secant
public function sec(byval number as double) as double
sec = 1 / cos(number)
end function
'cosecant
public function csc(byval number as double) as double
csc = 1 / sin(number)
end function
'cotangent
public function ctn(byval number as double) as double
ctn = 1 / tan(number)
end function
'inverse sine
public function asin(byval number as double) as double
asin = atn(number / sqr(-number * number + 1))
end function
'inverse cosine
public function acos(byval number as double) as double
acos = atn(-number / sqr(-number * number + 1)) + 2 * atn(1)
end function
'inverse secant
public function asec(byval number as double) as double
asec = atn(number / sqr(number * number - 1)) + sgn((number) - 1) * (2 *
atn(1))
end function
'inverse cosecant
public function acsc(byval number as double) as double
acsc = atn(number / sqr(number * number - 1)) + (sgn(number) - 1) * (2 *
atn(1))
end function
'inverse cotangent
public function actn(byval number as double) as double
actn = atn(number) + 2 * atn(1)
end function
'hyperbolic sine
public function sinh(byval number as double) as double
sinh = (exp(number) - exp(-number)) / 2
end function
'hyperbolic cosine
public function cosh(byval number as double) as double
cosh = (exp(number) + exp(-number)) / 2
end function
'hyperbolic tangent
public function tanh(byval number as double) as double
tanh = (exp(number) - exp(-number)) / (exp(number) + exp(-number))
end function
'hyperbolic secant
public function sech(byval number as double) as double
sech = 2 / (exp(number) + exp(-number))
end function
'hyperbolic cosecant
public function csch(byval number as double) as double
csch = 2 / (exp(number) - exp(-number))
end function
'hyperbolic cotangent
public function ctnh(byval number as double) as double
ctnh = (exp(number) + exp(-number)) / (exp(number) - exp(-number))
end function
'inverse hyperbolic sine
public function asinh(byval number as double) as double
asinh = log(number + sqr(number * number + 1))
end function
'inverse hyperbolic cosine
public function acosh(byval number as double) as double
acosh = log(number + sqr(number * number - 1))
end function
'inverse hyperbolic tangent
public function atanh(byval number as double) as double
atanh = log((1 + number) / (1 - number)) / 2
end function
'inverse hyperbolic secant
public function asech(byval number as double) as double
asech = log((sqr(-number * number + 1) + 1) / number)
end function
'inverse hyperbolic cosecant
public function acsch(byval number as double) as double
acsch = log((sgn(number) * sqr(number * number + 1) + 1) / number)
end function
'inverse hyperbolic cotangent
public function actnh(byval number as double) as double
actnh = log((number + 1) / (number - 1)) / 2
end function
'logarithm to base n
public function logn(byval base as double, byval number as double)
logn = log(number) / log(base)
end function
'convert degrees to radians
public function degtorad(byval number as double) as double
degtorad = number * 1.74532925199433e-02 'pi/180
end function
'convert radians to degrees
public function radtodeg(byval number as double) as double
radtodeg = number * 57.2957795130824 ' 180/pi
end function
'asin(sin(x)) = x!!62616964757a686964616fe59b9ee7ad9431333236373163!
'好累呀!!
vb程式題,急!謝謝!編寫程式,從文字框輸入x與y的值,在第三個文字框中輸出結果:
2樓:
設text1和text2分別為x和y的輸入框,text3輸出結果,另有***mand1是計算按鈕。
在窗體代版碼「通權用」段編寫函式過程:
private function f(byval x as double, byval y as double) as double
select case x
case is > 0
if y > 0 then
f = log(x) + log(y)
else
f = sin(x) + sin(y)
end if
case else
f = sin(x) + cos(y)
end select
end function
'計算按鈕事件呼叫f:
private sub ***mand1_click()text3.text = f(val(text1.text), val(text2.text))
end sub
3樓:匿名使用者
有圖有真相,望採納。。。。
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VB中label控制項中的文字怎麼「垂直居中「呀
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