Question

if tanx = 3/4 and x lies in third quadrant , then find values of sin x/2 cos x/2 and tan x/2

Answer 1

check the attachment

Answer 2

Given:

• Tan(x) = 3/4

To Find:

• Values of sin(x/2), cos(x/2), tan(x/2).

Solution:

• We know that there are many standard trigonometry formulas.
• We should choose the formula in such a way that we get the required answer.
• consider, $2x = \frac{2tan(x)}{1-tan^2(x)}$  
• Applying the above mentioned formula for tan(x),
• $tan(x) = \frac{2tan(x/2)}{1-tan^2(x/2)}$  
• $\frac{3}{4}= \frac{2tan(x/2)}{1-tan^2(x/2)}$
• ⇒ $3-3tan^2(x/2) = 8tan(x/2)$  
• ⇒ $3tan^2(x/2)+8tan^2(x/2)-3=0$
• ⇒ $3tan^(x/2)+9tan(x/2)-tan(x/2)-3=0$
• ⇒ $3tan(x/2)(tan(x/2)+3)-(tan(x/2)+3)=0$
• ⇒ $(3tan(x/2)-1)(tan(x/2)+3) = 0$
• ⇒ $tan(x/2) = -3$  and $tan(x/2) = 1/3$
• It is already mentioned that x lies in the third quadrant.
• We can predict the range where it lies, 180°≤ x ≤ 270°
• ⇒ 90° ≤ x ≤ 135°
• tan(x/2) lies in the second quadrant.
• ∴ tan(x/2)  = - 3
• Now, cos(x/2) = $\frac{1}{sec(x/2)}$  = $-\frac{1}{\sqrt{1+tan^2(x/2)} }$  = $-\frac{1}{\sqrt{1+(-3)^2} } = -\frac{1}{\sqrt{10} }$  
• next, sin(x/2) = $tan(x/2)*cos(x/2)$ = $-3*-\frac{1}{\sqrt{10}} = \frac{3}{\sqrt{10}}$
• ∴ sin(x/2) = $\frac{3}{\sqrt{10} }$ and cos(x/2) = $-\frac{1}{\sqrt{10} }$

∴ tan(x/2)  = - 3, sin(x/2) = $\frac{3}{\sqrt{10} }$,  and  cos(x/2) = $-\frac{1}{\sqrt{10} }$