# SOLUTION: given tan a = 4/3, cos a<0,cos B = 2/5, tan B<0, find exact values for, cos (a/2), and cos (2B) a = alpha B = beta

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: given tan a = 4/3, cos a<0,cos B = 2/5, tan B<0, find exact values for, cos (a/2), and cos (2B) a = alpha B = beta      Log On

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 Click here to see ALL problems on Trigonometry-basics Question 535130: given tan a = 4/3, cos a<0,cos B = 2/5, tan B<0, find exact values for, cos (a/2), and cos (2B) a = alpha B = betaAnswer by lwsshak3(6463)   (Show Source): You can put this solution on YOUR website!given tan a = 4/3, cos a<0,cos B = 2/5, tan B<0, find exact values for, cos (a/2), and cos (2B) a = alpha B = beta ** tan a = 4/3, cos a<0 If tan>0 and cos<0, then sin must also be<0. This places a in quadrant III where sin and cos are<0, and therefore, tan>0. In this quadrant, you are working with a right triangle with the opposite side=4 and adjacent side=3. By the Pythagorean Theorem, the hypotenuse=5. So cos a=3/5. cos(a/2)=-√[(1+cos a)/2]=-√[(1+(3/5))/2]=√[(8/5)/2]=√(8/10)=√(4/5)=2/√5=2√5/5 cos(a/2)=2√5/5 .. cosB=2/5, tanB<0 If cos>0 and tan<0, then sin must be<0 This places B in quadrant IV where cos>0, sin<0 and tan<0. In this quadrant you are working with a right triangle with the adjacent side=2 and hypotenuse=5. By the Pythagorean Theorem, the opposite side=√21. So sinB=-√21/5 cos 2B=(cosB)^2-(sinB)^2=(2/5)^2-(√21/5)^2=4/25-21/25=-17/25 cos(2B)=-17/25