sin[arcsin(1/5)+ arccos x] =1 find the value of x?
This question is best solved in smaller parts. For instance,
sin (theta) = 1
theta = pi/2 or 90 degrees. I'm going to assume all answers for angles will be in radians.
Then, arcsin (1/5) + arccos x equals pi/2, since sin (pi/2)=1
arccos x = pi/2 - arcsin (1/5)
x = cos (pi/2 - arcsin (1/5))
Using the identity: cos (a - b) = cos a cos b + sin a sin b
x = cos (pi/2) cos (arcsin (1/5)) + sin (pi/2) sin (arcsin (1/5))
x = 0 * cos (arcsin (1/5)) + 1 * 1/5
x = 1/5
To solve this problem, you use that sin (pi/2) = 1
Therefore the entire expression on the inside of the sine brackets is equal to pi/2.
Then solve for x by taking the cos of both sides.
Use the difference identity for cosine to evaluate x.