Precalculus 12 Addition, subtraction, and double angle identities 6.1 and 6.2 Part A – proving some identities sin  A B 
1. If you want to calculate tan(A+B), it’s pretty easy to calculate cos A B  . But there is a cleaner way to write the identity. Show that tan( A  B) 
tan A  tan B
1  tan A tan B
a) write tan( A  B ) 
sin  A B 
cos A B 
b) use the addition formulae for both sin( ) and cos( ). A cos B
c) multiply by a specially chosen 1, in this case cos
cos A cos B
d) simplify 2. Repeat the above process for tan(A‐B) 3. Add these two new identities to your identities sheet And the double angle identities… 4. We can also write an identity for cos(2A). We just need to use the existing identity for cos(A+A) and then simplify: . Cos(2A) = cos(A+A) = cosAcosA ‐ sinAsinA = cos2A ‐ sin2A (Note: this LOOKS like the pythagorean identity sin2A + cos2A = 1, but it’s not) Write this identity onto your sheet. 5. Repeat this process for sin(2A). Write it into your sheet. 6. There are actually two other ways to write the cos(2A). a) start with cos(2A) = cos2A – sin2A b) use the pythagorean identity sin2A + cos2A = 1, rearrange and let sin2A = 1 – cos2A, and make a substitution. c) Simplify. What did you get? d) repeat that process, but instead use a different re‐arrangement of the pythag identity: cos2A – 1 = sin2A e) Write these two other forms of cos(2A) onto your sheet. In all, you should have three ways to write cos(2A) Part B ‐ solving some problems 1. Use the addition and subtraction identities to write each expression as on trigonometric expression of one angle a) cos(18o)cos(24o) – sin(18o)sin(24o) b) sin  76  cos  3   cos  76  sin  3  c) sin(5)cos(3) – cos(5)sin(3) 2. Without the use of calculators, use your knowledge of special triangles to find the exact values of each 
a) Tan( 12
) b) Sec(15o) c) sin  1712  3. If A is a first quadrant angle and cos( A) 
3
5
while B is a second quadrant angle such that sin( B ) 
, find tan(A+B) 5
13
4. If P is a third quadrant angle and sec( P )   5 and Q is a fourth quadrant angle where csc(Q )   2 , find a) Sin(P + Q) 

5. Show that 2 cos   
b) Cos(P ‐ Q)  , expanded and simplified, is cos   sin  4



c) Tan(P + Q) 
6. Solve for x on the interval  0, 2  : sin  x 
7. If sin K

  cos x 3
4
3
and cos K  then find the exact values of sin(2 K ) and cos(2 K ) 5
5
8. Use an appropriate formula to re‐write as a single trig function a) 4cos(x)sin(x) b) 2sin2(x) – 1 c) 1 – 2cos2(3x) d) cos
( )  sin 2 ( 2x ) 2 x
2
9. Solve on the interval 0  x  2 a) sin(2x) + sin(x) = 0 b) cos(2x) + 3cos(x) = 1 Answers for part B. Some could be wrong. Pretty sure they’re all there. There are also some extra answers that don’t apply. sin 60  
cos 42 6 2 3  2 
15 ,195 ( or  
 13
,
12 12
 3
sin 
 2
sin  2  6 2
2 6
2  3 4
4
3
4
12
33  10
10
1
2 6
3 
 4
2
5
13
56
10
5
3
24
25
)  cos(2 x)  cos(6 x ) cos( x ) 
 

7
25
1/9 root(3) 
0, 23 ,  , 43

3
,  , 53
4 5
9
 18 
3 7
8
1
4
(why two answers?) 13 15
4
10
10

3 10
10
2 2
3
2sin(2 x) Part C – proving more identities (don’t worry about non‐permissible values) a) (2sin x  3cos x) 2  (3sin x  2 cos x) 2  13 b) tan x  cot y 
c) sin( x  y )
sin x sin y
a) (sin x  cos x)  1  sin(2 x) b) cos
b) cot y  cot x 
cos( x  y )
cos x sin y
1  sin x
 2sec 2 x  2sec x tan x  1 1  sin x
2
4
x  sin 4 x  cos 2 x