3 years ago

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Maurinko [17]3 years ago

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Given sinx=3/5 and x is in quadrant 2, what is the value of tan(x/2)

xxMikexx [17]

X is in the second quadrant means that x/2 is in the first quadrant.

Consider the right triangle drawn in the figure. Let tan(x/2)=a.

Then, let the length of the opposite side to x/2 be a, the adjacent side be 1 and the hypotenuse be square root of a squared +1, as shown in the figure.

sin(x/2)=|opp side|/ |hypotenuse| = $\frac{a}{ \sqrt{ a^{2}+1 } }$

cos (x/2) = |adj side|/ |hypotenuse| = $\frac{1}{ \sqrt{ a^{2}+1 } }$

from the famous identity: sin(2a)=2sin(a)cos(a), we have:

2sin(x/2)cos (x/2)=sin(x)

thus

$2* \frac{a}{ \sqrt{ a^{2}+1 } }*\frac{1}{ \sqrt{ a^{2}+1 } }= \frac{3}{5}$

$2* \frac{a}{ a^{2}+1 }= \frac{3}{5}$

$10a=3 a^{2} +3$

$3 a^{2}-10a +3=0$

(3a-1)(a-3)=0

thus a=1/3 or a=3

thus tan(x/2)=1/3 or tan(x/2)=3

Answer: {1/3, 3}