Skip to content
Exercise 3.2 · Q1

Q.Find the values of other five trigonometric functions if cos⁡x=−12\cos x = -\frac{1}{2}, xx lies in third quadrant.

Chhattisgarh CgbseTextbookSubjective· 3mImportance★★★★★est
9% · 13/150 Questions
✓ Free question

In the third quadrant, both sine and cosine are negative. Using the Pythagorean identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 with cos⁡x=−12\cos x = -\frac{1}{2}, we find sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}, then compute the remaining four functions from these two.

The key to this problem is understanding how the signs of trigonometric functions change across quadrants. In the third quadrant (where both xx and yy coordinates are negative on the unit circle), sine and cosine are both negative, while tangent is positive (negative divided by negative). Once we know sine and cosine, the other four functions follow from their definitions.

The Pythagorean identity connects sine and cosine for any angle. Since we're given cos⁡x=−12\cos x = -\frac{1}{2} and told xx is in the third quadrant, we can find sin⁡x\sin x and then build everything else.

Finding the remaining five functions

1. Find sin⁡x\sin x using the Pythagorean identity

The fundamental identity is:

sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1

Substituting cos⁡x=−12\cos x = -\frac{1}{2}:

sin⁡2x+(−12)2=1\sin^2 x + \left(-\frac{1}{2}\right)^2 = 1

sin⁡2x+14=1\sin^2 x + \frac{1}{4} = 1

sin⁡2x=34\sin^2 x = \frac{3}{4}

sin⁡x=±32\sin x = \pm \frac{\sqrt{3}}{2}

Since xx lies in the third quadrant where sine is negative:

sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}

2. Find tan⁡x\tan x from the ratio definition

Tangent is the ratio of sine to cosine:

tan⁡x=sin⁡xcos⁡x=−32−12=32×21=3\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}

Note

Notice that tan⁡x\tan x is positive in the third quadrant, which confirms our signs are correct — both sine and cosine are negative, so their ratio is positive.

3. Find csc⁡x\csc x as the reciprocal of sine

csc⁡x=1sin⁡x=1−32=−23=−233\csc x = \frac{1}{\sin x} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}

4. Find sec⁡x\sec x as the reciprocal of cosine

sec⁡x=1cos⁡x=1−12=−2\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{1}{2}} = -2

5. Find cot⁡x\cot x as the reciprocal of tangent

cot⁡x=1tan⁡x=13=33\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}

Alternatively, cot⁡x=cos⁡xsin⁡x=−12−32=13=33\cot x = \frac{\cos x}{\sin x} = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}.

Watch out

A common mistake is forgetting to apply the quadrant restriction. The equation sin⁡2x=34\sin^2 x = \frac{3}{4} gives two possible values, but only sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2} is valid in the third quadrant.

✓Final answer

The five trigonometric functions are: sin⁡x=−32\sin x = -\frac{\sqrt{3}}{2}, tan⁡x=3\tan x = \sqrt{3}, csc⁡x=−233\csc x = -\frac{2\sqrt{3}}{3}, sec⁡x=−2\sec x = -2, and cot⁡x=33\cot x = \frac{\sqrt{3}}{3}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous question · 3.3.2Example 9 — Find the value of cos(-1710^°).→Next question · 3.3.2Q2 — Find the values of other five trigonometric functions if sin x = (3)/(…