If sina = 4/5 calculate tan a/2. a is inth first quadrant. Deduct a relation for tan a/2

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

sin a = 4/5

We know that sin^2 a + cos^2 a = 1

==> cosa = sqrt(1- sin^2 a)

                  = sqrt(1- 16/25)

                     = 3/5

==> cosa = 3/5

==> tana = sina/cosa = 4/3

Now we know that:

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

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

==> 4/3 = 2 tan(a/2)/(1-tan^2 (a/2)

==> 2/3 = tan(a/2) / (1-tan^2 (a/2)

Cross multiply:

==> 3*tan(a/2) = 2(1-tan^2 (a/2)

==> 3tan(a/2) = 2 - 2 tan^2 (a/2)

==> 2tan^2 (a/2) +3tan(a/2) - 2 = 0

Let tan(a/2) = x

==> 2x^2 + 3x -2 = 0

==> (2x -1)(x+2) = 0

==> x1= 1/2 ==> tan(a/2) = 1/2 ==> a/2 = arctan(1/2)

==> x2= -2 ==> tan(a/2) = -2 ==> a/2 = arctan(-2)

But a/2 is in the 1st quadrant, then the only solution is:

a/2 = arctan (1/2) = 27 degrees (approx.)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

sina = 4/5.

Therefore tana = sina/cosa = sina/sqrt(1-sin^2a) = (4/5)/{sqrt{1-(4/5)^2} = 5(4/5)/sqrt{25-16}

=4/3.

tana = 4/3 = 2t/(1-t^2), where t = tana/2.

4(1-t^2) = 3*2t

4-4t^2 = 6t

2-2t^2 = 3t

2t^2 +3t -2 = 0

(2t-1)(t +2) = 0

t = 1/2 or t = -2.

Therefore inthe 1st quadrant t = tana/2 = 1/2.

tan a/2 - a = 1/2  - 2arc tan (1/2) = -0.4273 rad apprx .

 

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We are given that sin a = 4 / 5. Now sin [2*a/2] = 2* sin a/2* cos a/2 .

sin [2*a/2] = 2* sin a/2* cos a/2 = 4/5

now as (sin a )^2 + (cos a)^2 =1

=> (cos a)^2 = 1 - (sin a )^2

=> (cos a)^2 = 1 - 16/25

=> cos a = sqrt [ 9/ 25]

=> cos a = 3/5

cos (2*a/2) = (cos a/2 )^2 - (sin a/2)^2

=> 3/5 = 1- (sin a/2)^2 - (sin a/2)^2 = 1- 2*(sin a/2)^2

=>2*(sin a/2)^2= 1 - 3/5

dividing by 2* sin a/2* cos a/2

=> (sin a/2) / (cos a/2) = (2/5) / (4/5)

=> tan a/2 = 2/4 = 1/2

Therefore tan a/2 = 1/2

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Before calculating tan a/2, we'll infer the formula for tan a/2.

We know that the function tangent is a ratio of sine and cosine functions.

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

We'll write the formula for the sine of the half-angle.

sin (a/2) = sqrt [(1-cos a)/2]

We'll write the formula for the cosine of the half-angle.

cos (a/2) = sqrt [(1+cos a)/2]

We'll make the ratio:

sin (a/2)/cos(a/2) = sqrt [(1-cos a)/2]/sqrt [(1+cos a)/2]

We'll simplify and we'll get:

tan (a/2) = sqrt [(1-cos a)]/sqrt [(1+cos a)]

We'll multiply by sqrt [(1+cos a)] to eliminate the square root from denominator:

tan (a/2) = sqrt [(1-cos a)]*sqrt [(1+cos a)]/(1+cos a)

tan (a/2) = sqrt [1-(cos a)^2]/(1+cos a)

But 1-(cos a)^2 = (sin a)^2

Since a is in the 1st quadrant, sqrt (sin a)^2 = sin a

Now, we'll calculate tan (a/2) = sin a/(1+cos a)

cos a = sqrt[1 - (sin a)^2]

cos a = sqrt (1-16/25)

cos a = 3/5

tan(a/2) = (4/5)/(1 + 3/5)

tan(a/2) = 4/8

tan(a/2) = 1/2

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