Question

Find the exact value of tan 75 using a half angle identity.

Answer 1

I got as far as \[\frac{ 1-\cos (150) }{ 1+\cos(150) }=\frac{ 1+\frac{ \sqrt{3} }{ 2 } }{ 1-\frac{ \sqrt{3} }{ 2 } }\]

Answer 2

\[\tan(x/2) = \sqrt{\frac{ 1-\cos(x) }{ 1+\cos(x) }} = \frac{ \sin(x) }{ 1+\cos(x)}\]

Answer 3

\(\tan 75^o = \tan \dfrac{150^o}{2} \)



150 degrees has a reference angle of 30 degrees.

Answer 4

In your first reply, a square root is missing on the RHS. The entire {1-cos(150)}/{1+cos(150)} should be inside a square root.

Multiply top and bottom by 2 (that is every term in the numerator and every term in the denominator should be multiplied by 2 to get rid of the fraction).
Then, rationalize the denominator by multiply top and bottom by { 1 + sqrt)3) }.
Simplify.

Answer 5

And when you take the square root, use the positive value because 75 degrees falls in the first quadrant where tan is positive.

Answer 6

@ranga How did you go from 1-cos(x) to sin(x) in your first post?

Answer 7

Inside the square root multiply top and bottom by (1 + cos(x)).

Answer 8

\[\tan(x/2) = \sqrt{\frac{ 1-\cos(x) }{ 1+\cos(x) }} = \sqrt{\frac{ (1-\cos(x))*(1 +\cos(x)) }{ (1+\cos(x)) * (1+\cos(x)) }} = \\\sqrt{\frac{ 1-\cos^2(x) }{ (1+\cos(x))^2 }} = \sqrt{\frac{\sin^2(x) }{ (1+\cos(x))^2 }} = \frac{ \sin(x) }{ 1+\cos(x)}\]

Answer 9

how did you do the last step from sin^2x/1+cosx^2 to sinx/1+cosx ?

Answer 10

\[\tan(x/2) = \frac{ \sin(x) }{ 1+\cos(x)}\\ \tan(150/2) = \frac{ \sin(150) }{ 1+\cos(150)} = \frac{1/2}{1-\sqrt{3}/2} = \frac{1}{2-\sqrt{3}} = \\\frac{1}{2-\sqrt{3}} * \frac{2+\sqrt{3}}{2+\sqrt{3}} = \frac{2+\sqrt{3}}{2^2-\sqrt{3}^2} = \frac{2+\sqrt{3}}{1} = 2 + \sqrt{3}\\\tan(75) = 2 + \sqrt{3}\]

Answer 11

"how did you do the last step from sin^2x/(1+cosx)^2 to sinx/1+cosx ?"

The square root and the square will cancel each other out.
Top is sin(x)^2. Bottom is (1+cos(x))^2. So both top and bottom are squared. When you take the square root, the square root will cancel out the square.\[\sqrt{A^2} = (A^2)^{1/2} = A^{2/2} = A\]

Answer 12

Square root of A-squared is just A.

Answer 13

(actually plus or minus A but you get the point)

Answer 14

@ranga sorry to bug you again, but why did you multiply \[\frac{ 1 }{ 2-\sqrt{3} } by \frac{ 2+\sqrt{3} }{ 2+\sqrt{3} } rather than \frac{ 2-\sqrt{3} }{ 2-\sqrt{3}}\]

Answer 15

It is called rationalizing the denominator. Many textbooks and teachers don't like to have a radical in the denominator. So to get rid of it we need to multiply the denominator by its CONJUGATE. Conjugate of (2 - sqrt(3)) is (2 + sqrt(3)). If we multiply the bottom by (2 + sqrt(3)) we have to multiply the top by (2 + sqrt(3)) also to preserve the ratio. As you can see we got rid of the radical in the denominator.