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What does the following rule mean? arcsin(x) + arccos(x) = pi/2 for every x in the interval <-1, 1>.When is it useful to use this rule?
The rule is an identity. It"s saying that if I pick an $x$ between $-1$ and $1$ inclusive, then I"m guaranteed that $arcsin x + arccos x = fracpi2$.For example, $arcsin frac12 = fracpi6$ and $arccos frac12 = fracpi3$. Thus$$arcsin frac12 + arccos frac12 = fracpi6 + fracpi3 = fracpi2.$$Why $x in <-1, 1>$? Well that"s because $arcsin x$ & $arccos x$ are only defined for those $x$.You would use this rule just like any other trig identity. When is it useful to lớn use $sin^2 x + cos^2 x = 1$? Whenever you can.
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The rule is an identity. It"s saying that if I pick an $x$ between $-1$ and $1$ inclusive, then I"m guaranteed that $arcsin x + arccos x = fracpi2$.For example, $arcsin frac12 = fracpi6$ and $arccos frac12 = fracpi3$. Thus$$arcsin frac12 + arccos frac12 = fracpi6 + fracpi3 = fracpi2.$$Why $x in <-1, 1>$? Well that"s because $arcsin x$ and $arccos x$ are only defined for those $x$.You would use this rule just like any other trig identity. When is it useful to use $sin^2 x + cos^2 x = 1$? Whenever you can.
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How about if I let arcsin (x) = 0?Then x = sin (x) = cos(pi/2 - x)arccos (x) = pi/2 - x = pi/2 - arcsin (x)arccos (x) = pi/2 - arcsin (x)arcsin (x) + arccos (x) = pi/2
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That proves that $arcsin 0 + arccos 0 = dfracpi2$ since you"ve started with the assumption that $arcsin x = 0$. However, the aim would be to lớn prove that $arcsin x + arccos x = dfracpi2$ for all $x$ between $-1$ & $1$.If you want a full proof, here"s one:We start with the identity$$sin heta = cosleft(fracpi2 - heta ight).$$ We can substitute $ heta = arcsin x$ lớn get$$sin(arcsin x) = cosleft(fracpi2 - arcsin x ight)\x = cosleft(fracpi2 - arcsin x ight).$$ Then we take the $arccos$ of both sides lớn get$$arccos x = fracpi2 - arcsin x\arccos x + arcsin x = fracpi2.$$
That proves that $arcsin 0 + arccos 0 = dfracpi2$ since you"ve started with the assumption that $arcsin x = 0$. However, the aim would be khổng lồ prove that $arcsin x + arccos x = dfracpi2$ for all $x$ between $-1$ & $1$.If you want a full proof, here"s one:We start with the identity$$sin heta = cosleft(fracpi2 - heta ight).$$ We can substitute $ heta = arcsin x$ khổng lồ get$$sin(arcsin x) = cosleft(fracpi2 - arcsin x ight)\x = cosleft(fracpi2 - arcsin x ight).$$ Then we take the $arccos$ of both sides to get$$arccos x = fracpi2 - arcsin x\arccos x + arcsin x = fracpi2.$$
That proves that $arcsin 0 + arccos 0 = dfracpi2$ since you"ve started with the assumption that $arcsin x = 0$. However, the aim would be khổng lồ prove that $arcsin x + arccos x = dfracpi2$ for all $x$ between $-1$ and $1$.If you want a full proof, here"s one:We start with the identity$$sin heta = cosleft(fracpi2 - heta ight).$$ We can substitute $ heta = arcsin x$ to lớn get$$sin(arcsin x) = cosleft(fracpi2 - arcsin x ight)\x = cosleft(fracpi2 - arcsin x ight).$$ Then we take the $arccos$ of both sides lớn get$$arccos x = fracpi2 - arcsin x\arccos x + arcsin x = fracpi2.$$
Another way is lớn differentiate arcsin(x) + arccos(x) & show that it"s zero. This means the sum is a constant for all valid x, và we already know the x=0 case which is enough lớn conclude the result.
(displaystyle left(fracpi2 ight)"=0=frac1sqrt1-x^2 -frac1sqrt1-x^2 .)this shows arcsinx+arccosx=constant.but not the value of the constant.
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