Please help. Calculus question.

Question

. If a is a Quadrant IV angle with cos(a)=√5/5 and sin(ß)=√10/10 where pi/2 ‹ß‹pi, find

a. cos(a+B)

b. sin(a+B)

c. Tan(a+B)

Answer 1

use the sum of trig formulas cosa=sqr5/5 sinb=sqr10/10. sina=-2sqr5/5, cosb= -3sqr10/10

tana = sina/cosa = -2, tanb = sinb/cosb = -1/3

substitute into the trig sum formulas:

.

cos(a+b) = cosacosb - sinasinb = -sqr50/50 = approximately -0.14

sin(a+b) = sinacosb - sinbcosa = 7sqr50/50 = approximately +0.99

tan(a+b) = (tana+tanb)/(1-tanatanb) = 7sqr50/50)/(-sqr50/50) = exactly -7

a+b = about 98.13 degrees

tan98.13 = -7

use an inverse trig caluculator to find a, and b

b= 161.565 degrees

a = - 63.435 degrees

a+b = 161.565 + (-63.435) = 98.13

Answer 2

Just need to use the angle summation formulas. But first, need to find sin(a) and cos(B)

cos(a) = √5/5 --> sin(a) = -2√5/5 (Q IV makes sine negative)

sin(B) = √10/10 --> cos(B) = -3√10/10 (Q II makes cosine negative)

a) cos(a+B) = cos(a)cos(B)-sin(a)sin(B)

=(√5/5)(-3√10/10) -(-2√5/5)(√10/10) can be simplified

b)sin(a+B) = sin(a)cos(B)+sin(B)cos(a)

=(-2√5/5)( -3√10/10)+(√10/10)(√5/5)

c) tan(a) = sin(a)/cos(a) = (-2√5/5)(√5/5)

tan(B) = sin(B)/cos(B) = (√10/10)(-3√10/10)

tan(a+B) = (tan(a)+tan(B))/(1-tan(a)tan(B))

Just substitute in the values for tan(a) and tan(B)