Find coss t given that cos s 13 with s in quadrant I and s

Find cos(s + t) given that cos s = 1/3, with s in quadrant I, and sin t = - 1/2, with in quadrant IV. 2 squareroot 6 - 1/6 2 squareroot 6 + 1/6 Find tan (s + t) given that cos s = 1/3, with s in quadrant I, and sin t = -1/2, with t in quadrant IV. 9 squareroot 3 - 8 squareroot 2/3 9 squareroot 3 + 8 squareroot 2/3 9 squareroot 3 - 8 squareroot 2/5 Find cos (s - t) given that cos s = - 1/6, with s in quadrant III, and cost t = - 3/5, with t in quadrant III 3 - 4 squarerot 37/30 3 + 4 squareroot 35/30 3 - 4 squareroot 35/30

Solution

7) coss = 1/3 and s is in Ist quadrant

So, sins = 2sqrt2/3

sint = -1/2 and t is in IV quadrant

cost = sqrt(3)/2

cos(s+t) = cosscost -sinssint

= (1/3)(sqrt3/2) - (2sqrt2/3)(-1/2)

= ( sqrt(3) + 2sqrt(2) )/6

Option D

7) coss = 1/3 and s is in Ist quadrant

So, sins = 2sqrt2/3

sint = -1/2 and t is in IV quadrant

cost = sqrt(3)/2

tan(s+t) = sin(s+t)/cos(s+t)

Lets find sin(s+t) = sins*cost +coss*sint

=(2sqrt2/3)(sqrt3/2) +(1/3)((-1/2)

= (2sqrt(6) - 1)/6

tan(s+t) = (2sqrt(6) - 1)/( sqrt(3) + 2sqrt(2) )

rationalise denominator: tan(s+t) = (2sqrt(6) - 1)(sqrt3 -2sqrt2)/(-5)

= (2sqrt18 - 4sqrt12 -sqrt3 +2sqrt2)/-5

= (6sqrt2 - 8sqrt3 -sqrt3 +2sqrt2)/-5

=(8sqrt(2) -9sqrt(3)/-5

={ (9sqrt(3)- 8sqrt(2) }/5

Option C

8) cos s = -1/6 ; s is in IIIrd quadrant

sin s = -sqrt(35)/6

cost = -3/5 t is in quadrant IIIrd

sint = -4/5

cos(s-t) = coss cost +sinssint

=(-1/6)(-3/5) + (-sqrt(35)/6)(-4/5)

={ 3 +4sqrt(35) }/30

Option B