/sci/ - Science & Math

[Anonymous No.5437136]

first to finish #s 23 - 29 will receive $30 through paypal or a drawing from yours truly if you're feeling generous. first to post answers with work shown decides. Will deliver.
only #s 23 - 29. E Z P Z amirite

[Anonymous No.5437142]

just did them all in my head, op. give me my money.

[Anonymous No.5437165]

give me proof that youll actually pay up first

[Anonymous No.5437195]

finished

Where's my money?

[Anonymous No.5437212]

i don't believe i see any answers or work, which already makes me think you're a tard who can't read. good luck with the work, and with learning how to read.

[Anonymous No.5437223]

>mfw OP is 16 and no one on here is going to help him

[Anonymous No.5437255]

>>5437136

Just because I'm feeling generous

#23

(cosX - sinX)^2 = (cosX - sinX)(cosX - sinX) = (cosx)^2 - cosX * sinX - cosX * sinX + (sinX)^2

; from pythagorian identity, (cosx)^2 + (sinx)^2 = 1

therefore

(cosX - sinX)^2 = 1 - 2cosX * sinX


#29

sin x (csc x - sin x) = cos^2 x

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

1 - sin^2 x = cos^2 x

The last part is an identity.

Have fun, faggot

[Anonymous No.5437260]

>>5437136
I'm a last semester math major and the only trigonometric functions I know are cos, sin and tan and I have never used anything else. Why do I keep seeing people using these reciprocal trigonometric functions?

[Anonymous No.5437263]

>>5437136
1: Replace sec with 1/cos, csc with 1/sin, ctn with cos/sin, tan with sin/cos
2: multiply both sides by things on the denominator to clear the denominators
3: apply sin^2 + cos^2 = 1

I bet those three steps will solve almost all of them