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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.

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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