Pre calculus

Question

1) fin the value of sin(2 theta)

cos(2 theta)=3/4 and 270 degrees< theta <360 degrees

a) square root of 2/2 

b) sqayre root of 2/4

c) nagative square root of 2/2

d) negative square root of 2/4

 

2) Simplidy the expression

3sin^3 (x) csc(x) + cos^2 (x) +2 cos(-x) cos(x)

a)sin^2 (x)

b) tan^2 (x) + 1 

c) 3

d) cos^2 (x)

 

Answer 1

1. theta is from 270 to 360 therefore it's in quadrant IV. However you are dealing with double angles (2 theta) so your interval would actually be 540 to 720 or getting the lowest positive coterminal angles the interval would be from 180 (540-360) to 360 (720-360).

 

So now, it's like solving a normal angle:

 

180 < 2 theta < 360 (from quadrant III to quadrant IV)

 

cos(2 theta) = 3/4   ------------ Cosine is positive therefore it's in quadrant IV.

 

cos = a/h; a = 3, h = 4

 

a² + o² = h²

 

o = ±sqrt(4² - 3²) = ±sqrt(7)

 

sin(2 theta) = -sqrt(7)/4 --------- Sine is negative in quadrant IV

 

*there's no correct answer in the choices because even though you don't do the process above, you can actually check each one of them by adding the squares of 3/4 and one of the choices, to be kinda correct, they should be equal to 1. Nothing passed. So check the question if it's accurate.

 

2. 3sin³xcscx + cos²x + 2cos(-x)cosx

 

First: cscx and sinx are reciprocals so you can just cancel a pair if they appear, so remove the a sin and the csc.

 

Second: cosx = cos(-x)

 

Therefore:

 

3sin²x + cos²x + 2cosxcosx

 

3sin²x + cos²x + 2cos²x

 

3sin²x + 3cos²x

 

3(sin²x + cos²x)

 

3(1)

 

3