Precalculus question?

Question

can someone explain to me how to figure out a roblem like

cos(Sin^-1 1/2)

or

Sin^-1(sin7pi/6)

breaking it down for me would be amaaaaaazing :)

Answer 1

don't listen to the moron above me. work inside out.

1) cos(sin-1 1.2)

first find sin-1 of 1/2. it's 30 or pi/6

now you have cos(pi/6) which is rt3/2

2) same thing. inside out. sin 7pi/6 = -1/2

then do Sin-1(-1/2). that equals 150 or 5pi/6, the reason i use 150 and not 210 is because of the range of Sin-1 x

do you mean how I know sin-1 1/2 is 30? that's memorizing all the "nice angles" (30, 45, 60 and any multiple of those) It also helps to know ALL - S -T -C corresponding to quadrants I, II, III, IV respectively. ALL the trig functions are positive in quadrant I, Sine is + in II, Tangent is + III and Cosine is + in IV

Answer 2

Alright so check it out.

The protocol remainder of sin^ appears to be 1 1/2 yet it is able to be converted to pi4/sin due to its association with a right beta wing bonzai call, you follow?

so due to the right beta wing bonzai, the pi4 would be read as sin^ 1 1/3.

because beta wings are always increased by ones, as a whole, so from two it increases into a 3.

now following that calculation it is easy to decipher the proto wings of the left cilouite.

Now the left cilouite in this problem is sin^1 1/3, so by understanding the protocal of the right criolike, you can see that it may easily be converted into 3^1/2, as simple as that.

Answer 3

yes