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)

Answer 3

The whole problem really rests on the two values we are not given, sin(α) and cos(β), and the quadrants are what tell us their signs. So let us find those first, before touching any of the sum identities.

We know cos(α) = √5/5 and that α sits in Quadrant IV, where sine is negative. Using sin²α + cos²α = 1:

sin²α = 1 - 1/5 = 4/5

so sin(α) = -2√5/5, and it is the quadrant, not the algebra, that tells us to take the negative root.

For β we are given sin(β) = √10/10 with π/2 < β < π, which places β in Quadrant II where cosine is negative. Same idea:

cos²β = 1 - 1/10 = 9/10

so cos(β) = -3√10/10.

Now that we have all four pieces, the identities are just careful substitution.

cos(α + β) = cos(α)cos(β) - sin(α)sin(β) = (√5/5)(-3√10/10) - (-2√5/5)(√10/10) = -3√50/50 + 2√50/50 = -√50/50

and since √50 = 5√2, that cleans up to -√2/10.

sin(α + β) = sin(α)cos(β) + cos(α)sin(β) = (-2√5/5)(-3√10/10) + (√5/5)(√10/10) = 6√50/50 + √50/50 = 7√50/50 = 7√2/10

For part c you do not need a third identity. Tangent is just sine over cosine, and the tenths cancel:

tan(α + β) = (7√2/10) / (-√2/10) = -7

So our final answers are cos(α + β) = -√2/10, sin(α + β) = 7√2/10, and tan(α + β) = -7.