If #tana=21/20# and #a# lies in #Q3#, find #tan(a/2),sin(a/2)# and #cos(a/2)#?

Question

Violet Anderson · Accepted Answer

We know

#tana=(2tan(a/2))/(1-tan^2(a/2))#

#=>21/20=(2x)/(1-x^2)" where "x=tan(a/2)#

#=>21-21x^2=40x#

#=>21x^2+40x-21=0#

#=>21x^2+49x-9x-21=0#

#=>7x(3x+7)-3(3x+7)=0#

#=>(3x+7)(7x-3)=0#

So

#x=-7/3 or x=3/7#

Now it is given

#180^@ < a < 270^@#

#=>90^@ < a/2 < 135^@#

So #a/2 in "2nd quadrant"#

Hence #x=tan(a/2)->"negative"#

So #x=tan(a/2)=-7/3#

#sin(a/2)->+ve#

So #sin(a/2)=1/csc(a/2)=1/sqrt(1+cot^2(a/2)#

#=1/sqrt(1+1/tan^2(a/2)#

#=1/sqrt(1+9/49#

#=7/sqrt58#

Now

#cos(a/2)=sin(a/2)/tan(a/2)#

#=(7/sqrt58)/(-7/3)=-3/sqrt58#

Abigail Armstrong · Answer

undefined Given that $ \tan{a} = \frac{21}{20} $ and angle $ a $ lies in Quadrant 3 (Q3), we can use trigonometric identities and the properties of angles in Q3 to find $ \tan{\frac{a}{2}} $, $ \sin{\frac{a}{2}} $, and $ \cos{\frac{a}{2}} $.

First, let's find $ \tan{\frac{a}{2}} $. We can use the half-angle formula for tangent:
$$ \tan{\frac{a}{2}} = \frac{\sqrt{\frac{1 - \cos{a}}{1 + \cos{a}}}}{1} $$

Next, we can find $ \sin{\frac{a}{2}} $ and $ \cos{\frac{a}{2}} $ using the relationships between tangent, sine, and cosine:
$$ \sin{\frac{a}{2}} = \frac{\sin{a}}{1 + \sqrt{1 + \tan^2{\frac{a}{2}}}} $$
$$ \cos{\frac{a}{2}} = \frac{\cos{a}}{1 + \sqrt{1 + \tan^2{\frac{a}{2}}}} $$

Given that $ \tan{a} = \frac{21}{20} $, we can find $ \sin{a} $ and $ \cos{a} $ using the Pythagorean identity:
$$ \sin^2{a} = 1 - \cos^2{a} $$
$$ \sin{a} = -\frac{21}{29} $$
$$ \cos{a} = -\frac{20}{29} $$

Substitute these values into the formulas for $ \tan{\frac{a}{2}} $, $ \sin{\frac{a}{2}} $, and $ \cos{\frac{a}{2}} $ to get the final results.