5.4+due+10.7

π β α θ ² ³
 * The following symbols can be copied & pasted for your hw problem: √ ±**

2. Done by Kelsey Harmon

For part A, cos390deg is **√3/2** so when you multiply it by sin120deg, you get 3/4 -1/2. The result answer would be positive 1/2. For part B, sin390deg is 1/2, so the result answer would be [(1+ **√3)/2]**. Checked by Morgan Miller


 * 4

Done By: Leslie Baker

For part (b) i got the same answer! Checked by Matthew Milan
 * 1) 4. For part (a) i got (-sqrt6-sqrt2)/4 because cos(2π/3) is -1/2, not 1/2.


 * 1) 6a)


 * 1) 6b) [[image:sc027e75d3.jpg]]

Done by Mady Smith Everything looks right except on 6a i dont think you make the pi/3 negative i think you just subtract. That doesnt make a difference except that sin of pi/3 is positive root 3/2. Other than that i got the same anwsers! Checked By: Alyssa Sturgis
 * #6A correction: **


 * #10 **
 * #18 **

sin(3n/12+2n/12)= sin(n/4+n/6) sin (n/4)cos(n/6)+cos(n/4)sin(n/6) (sqrt2/2)*(root 3/2)+ (root 2/2)(1/2) (root6/4)+(root2/4) SIN=(root6 +root2)/4 cos(n/4+n/6) Cos(n/4)Cos(n/6)-Sin(n/4)sin(n/6) (root2/2)(root3/2)-(root2/2)(1/2) (root6/4)-(root2/4) COS=(Root6-root2)/4 Tan(n/4+n/6) (Tan(n/4)+tan(n/6))/ 1-tan(n/4)*tan(n/6) Tan=(1+(root3/3))/ (1-(root3/3))
 * 1) 20 (N IS PI)
 * #20 tangent (it is often easier to just use the identity than to try sin/cos) **

SIN(3n/4+N/6) Sin(3n/4)cos(n/6)+cos(3n/4)sin(n/6) (-root2/2)(root2/2)+(-root2/2)(1/2) (-1/2)+(-root2/4) SIN=((-2-root2)/4) COS(3n/4+n/6) COS(3n/4)Cos(n/6)-Sin(3n/4)sin(n/6) (-root2/2)(root3/2)- (root2/2)(1/2) (-root6/4)-(root2/4) Cos= ((-root6-root2)/4) Tan(3n/4 + N/6) ((tan(3n/4)+tan(n/6)/1-tan(3n/4)*tan(n/6) TAN= (1+(root3/3)/(1-(root3/3) Done by Daniel Pugliano...work
 * 22

I got the same thing puggles! way to do work Checked by Cameron Johnson
 * #22 sin & tan **

Well the work wasn't here but this is what I got... 24. sin(110 degrees + 80 degrees) ** = sin (190 degrees) ** 26. tan(152 degrees - 47 degrees) ** = tan (105 degrees) ** 30. sin(4pi/9 + pi/8) ** = (sin 41pi/72) ** done by: Rebekah Adair

Done by Andrew Kim
 * 36

That's exactly how I did it! Checked by Clarissa Rodriguez



This is my work, I'm really sorry it's too late, I got home only around 9:30 because we had a marching contest, I didn't realize this when I was signing for the wiki problems, sorry. Done by Jan Dudek

sin(**u** - **v**) sin**u**cos**v** - cos**u**sin**v** (5/13)(-3/5) - (-12/13)(4/5) (-15/65) - (-48/65) (-15/65) + (48/65) Done?/Checked by Jonathan Eichelberger
 * 1) 38 It's getting pretty late and the work still isn't here, so I'll post how I did it...
 * 33/65**

tan u = sin u / cos u = 7/24 tan v = sin v / cos v = 4/3 tan (u+v) = tan u + tan v / 1 - tan u tan v = (7/24) + (4/3) / 1 - (7/24)(4/3) = (39/24) / 1 - (7/18) = (39/24) / (11/18) = (39/24)*(18/11) = 702/264 = 117/44 Done by Sung Yoo
 * 40
 * #40 CORRECTION: **

sin(3π-x)=sinx sin3πcosx-cos3πsinx=sinx 0cosx-(-1sinx)=sinx 0+sinx=sinx sinx=sinx Done/checked by Ellen Barth
 * 48


 * #50: **


 * 52

sin(x+y) + cos(x-y) = 2 cos x cos y

(cos x)(cos y) - (sin x) (sin y) + (cos x)(cos y) + (sin x)(sin y) = 2 cos x cos y

(cos x)(cos y) + (cos x)(cos y) = 2 cos x cos y

2 cos x cos y = 2 cos x cos y

Done by Miles Hennington

I did this problem the same way and got the same result! Checked by kristen hanslik

Done By:Parker Tennet You were right until you got to the last step, here is how i did it sin(x + y) Sin (x - y) = sin x^2 - sin y^2 1. ( sinx cosy + cosx siny) (sinx cosy - cosx siny) 2. sinx^2 cosy^2 - sinx siny cosx cosy + cosx cosy sinx siny - cosx^2 siny^2 (cancel out) 3. sinx^2 cosy^2 - siny^2 cosx^2 4. sinx^2 (1 - siny^2) - siny^2 (1 - sinx^2) 5. (sinx^2 - sin x^2 y^2) - (siny^2 - sinx^2 y^2) (change the signs and they cancel out) 6. sinx^2 - siny^2 = sinx^2 - siny^2 Checked by Freddie Jordan
 * 54

66) True because sin(X-11π/2) = sinXcos(11π/2) - cosXsin(11π/2) (11π/2) = (3π/2). cos of (3π/2) is 0. And sin of (3π/2) is -1. If you plug those numbers in, the first part cancels and the sin cancels out the negative so it leaves you with cosX.