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Math14-TanB= 34 and 0<B<90Sin12B=±1-cosB2=?We don't have value of cosB so first we have to find cosBAs we know TanB= PerpendicularBase=34, and CosB=BaseHypotenuse So we have to find out Hypotenuse to find value of CosBHypotenuse=Perpendicular2+base2=32+42=9+16= 25=5CosB=45We know that tangent is positive in first quadrant so we will use Sin12B=1-cosB2So Sin12B=1-452=5-452=15×2=110A

15-As we know sin60°=32 , And sinθ is positive because it lies in the first quadrant

So sin12120=1-cos1202=1-(-12)2=2+122=32×2=34=32

16-As we know cos225°=-22 , And cosθ is negative because it lies in the third quadrant

So cos12450=-1+cos4502=-1-02=-12=-22

17-As we know tan135°=-1 , And tanθ is negative because it lies in the second quadrant

So tan12270=-1-cos2701+cos270=-1-01+0=-11=-1

18-As we know cosA=725 , And 0 < A < 90 so cosA will be positive

sin12A=1-cosA2=1-7252=25-7252 =182 ×25=1850=3252 = 35

cos12A=1+cosA2=1+7252 =25+7252=322 ×25=3250=4252 = 45

tan12A=1-cosA1+cosA=1-7251+725=25-72525+725=1832=3242=34

19-As we know cosB=59 , And 0 < B < 90 so cosB will be positive

sin12B=1-cosB2=1-592=9-592 =42 ×9=232

cos12B=1+cosB2=1+592=9+592 =142 ×9=1432

tan12B=1-cosB1+cosB=1-591+59=9-599+59=414=214

20-As we know cosθ=78 , And 0 < θ < 90 so cosθ will be positive

sinθ2=1-cosθ2=1-782=8-782 =12 ×8=123

cosθ2=1+cosθ2=1+782=8+782 =152 ×8=1523

tanθ2=1-cosθ1+cosθ=1-781+78=8-788+78=115=115

21-As we know sinA=0.6=610, And 0 < A < 90 so sinθ will be positive

However first we have to find out cosA to find the half angle identities

sinA = PerpendicularHypotenuse=610, we have to find the Base using the formula H2 = P2 + B2

Base = B = H2-P2=102-62=100-36=64=8

cosA = BaseHypotenuse=810

sin12A=1-cosA2=1-8102=10-8102 =22 ×10 = 225

cos12A=1+cosA2=1+8102 =10+8102=182 ×10 = 3225

tan12A=1-cosA1+cosA=1-8101+810=10-81010+810=218=232=13

22- (4) sin12x= 12sinx is not an identity

23-2 sin40°

24-4 tan250°

25- (1) 1-cosθ

26- (3) -79

27- (2) 2-√3

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