Given sin x = -4/5 and x is in quadrant III, what is the value of tan x/2?
Trigonometry is the branch of mathematics that deals with the measurement of angles and helps us study the relationship between the sides and angles of a right-angled triangle. Let's solve a question related to trigonometric concepts.
Answer: When sin x = -4/5 and x is in quadrant III, the value of tan x/2 is -2.
Let's solve the problem in detail.
We can write sin x in terms of tan x/2 using the formula:
⇒ sin x = (2 tan (x/2)) / (1 + tan2(x/2))
Therefore, using the above formula, we can find the values of tan x/2 by putting the value of sin x.
⇒ -4/5 = (2 tan (x/2)) / (1 + tan2(x/2))
Now, if we replace tan (x/2) by y, we get a quadratic equation:
⇒ 0.8y2 + 2y + 0.8 = 0
⇒ 2y2 + 5y + 2 = 0
By using the quadratic formula, we get y = -0.5, -2
Hence, the value of tan (x/2) = -0.5, -2
Now, we have two solutions of tan (x/2).
Now, let's check for the ideal solution using the formula tan x = (2 tan (x/2)) / (1 - tan2(x/2)).
For tan (x/2) = -0.5:
⇒ tan x = 2(-0.5) / 1 - (-0.5)2 = -4/3
It is also given that x lies in the third quadrant. We know that tan is positive in the third quadrant, and here we get tan x = -4/3 which is negative.
Hence, we can say that tan (x/2) = -0.5 is not a correct solution. Hence it is rejected.
Now let's check for tan (x/2) = -2.
⇒ tan x = 2(-2) / 1 - (-2)2 = 4/3
Here, we get tan x = 4/3 which is positive.
Hence, we can say that tan (x/2) = -2 is a correct solution.