Various math problems

[zschnopz]

Hi there. Just got back a math test and wanted someone explain my wrong answers to me, if any of you would.

1) tan(42/6) = 0, pi ...I got marked down for the pi
2) arcsin(1/2) = pi/6, 5pi/6 ...I got marked down for the 5pi/6
3)csc[arcsin(1/2) = pi/6, 5pi/6 ...I got marked down for the 5pi/6

4) tan(x) + 2cos(x) = sec(x)
I solved this for (2sin -1)(sin-1) so sin = -1/2, 1 = pi/2, 11pi/2, 7pi/2 ...i got marked down for the pi/2

5) Establish identity:
sec(x) - tan(x) = cos(x)/(1+sin(x))
1/cos - sin/cos + sin/cos - sin^2/cos
(1-sin^2)/cos = cos
1-sin^2 = cos^2
(i got no credit for this)

Did I just make a ton of little mistakes?

thank you. and, obviously, I'm open to reworking any problem that you want me to.

 

[Deleted member 4993]

Hi there. Just got back a math test and wanted someone explain my wrong answers to me, if any of you would.

1) tan(42/6) = 0, pi ...I got marked down for the pi .... That problem and anf answer does not make any sense to me
2) arcsin(1/2) = pi/6, 5pi/6 ...I got marked down for the 5pi/6 ............. In general your response is correct
3)csc[arcsin(1/2) = pi/6, 5pi/6 ...I got marked down for the 5pi/6 ............... The answer should be 2 - I don't know why you were marked "correct" for pi/6.

4) tan(x) + 2cos(x) = sec(x)
I solved this for (2sin -1)(sin-1) so sin = -1/2, 1 = pi/2, 11pi/2, 7pi/2 ...i got marked down for the pi/2

The given function is DNE at x = π/2

5) Establish identity:
sec(x) - tan(x) = cos(x)/(1+sin(x))
1/cos - sin/cos + sin/cos - sin^2/cos

What are you doing in the line above? [You should always write sin(x) or cos(x) - instead of sin or cos]

sec(x) - tan(x)

= 1/cos(x) - sin(x)/cos(x)

= [1-sin(x)]/cos(x)

= [1-sin(x)]/cos(x) * [1+sin(x)]/[1+sin(x)]

= [1-sin²(x)]/{cos(x) * [1+sin(x)]}

= cos²(x)/{cos(x) * [1+sin(x)]}

= cos(x)/ [1+sin(x)]


(1-sin^2)/cos = cos
1-sin^2 = cos^2
(i got no credit for this)

Did I just make a ton of little mistakes?

thank you. and, obviously, I'm open to reworking any problem that you want me to.

.