If \sin t = 3/5 and the terminal point ofis in quadrant II, then \cos t = \tan t = \cot t = \sec t = \csc t =

/ Exams & Certification / By

If \sin t = 3/5 and the terminal point of
is in quadrant II, then \cos t = \tan t = \cot t = \sec t = \csc t =

The Correct Answer and Explanation is:

Given:
sin⁡t=35\sin t = \dfrac{3}{5}sint=53​, and the terminal point of ttt is in Quadrant II.

Step 1: Use the Pythagorean identity to find cos⁡t\cos tcost

The identity is:sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1sin2t+cos2t=1

Substitute sin⁡t=35\sin t = \dfrac{3}{5}sint=53​:(35)2+cos⁡2t=1925+cos⁡2t=1cos⁡2t=1−925=1625\left(\dfrac{3}{5}\right)^2 + \cos^2 t = 1 \\ \dfrac{9}{25} + \cos^2 t = 1 \\ \cos^2 t = 1 – \dfrac{9}{25} = \dfrac{16}{25}(53​)2+cos2t=1259​+cos2t=1cos2t=1−259​=2516​

So,cos⁡t=±45\cos t = \pm \dfrac{4}{5}cost=±54​

But since the angle is in Quadrant II, and cosine is negative in Quadrant II:cos⁡t=−45\cos t = -\dfrac{4}{5}cost=−54​

Step 2: Find the remaining trig functions

Now that we have:

  • sin⁡t=35\sin t = \dfrac{3}{5}sint=53​
  • cos⁡t=−45\cos t = -\dfrac{4}{5}cost=−54​

We can compute:

  1. tan⁡t=sin⁡tcos⁡t\tan t = \dfrac{\sin t}{\cos t}tant=costsint​ tan⁡t=3/5−4/5=−34\tan t = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}tant=−4/53/5​=−43​
  2. cot⁡t=1tan⁡t=1−3/4=−43\cot t = \dfrac{1}{\tan t} = \dfrac{1}{-3/4} = -\dfrac{4}{3}cott=tant1​=−3/41​=−34​
  3. sec⁡t=1cos⁡t=1−4/5=−54\sec t = \dfrac{1}{\cos t} = \dfrac{1}{-4/5} = -\dfrac{5}{4}sect=cost1​=−4/51​=−45​
  4. csc⁡t=1sin⁡t=13/5=53\csc t = \dfrac{1}{\sin t} = \dfrac{1}{3/5} = \dfrac{5}{3}csct=sint1​=3/51​=35​

Final Answers:

  • sin⁡t=35\sin t = \dfrac{3}{5}sint=53​
  • cos⁡t=−45\cos t = -\dfrac{4}{5}cost=−54​
  • tan⁡t=−34\tan t = -\dfrac{3}{4}tant=−43​
  • cot⁡t=−43\cot t = -\dfrac{4}{3}cott=−34​
  • sec⁡t=−54\sec t = -\dfrac{5}{4}sect=−45​
  • csc⁡t=53\csc t = \dfrac{5}{3}csct=35​

Explanation

To solve this trigonometric problem, we start with the given value of sin⁡t=35\sin t = \dfrac{3}{5}sint=53​, meaning the opposite side of a right triangle is 3 and the hypotenuse is 5. We need to find all other trigonometric functions based on this.

Using the Pythagorean identity, sin⁡2t+cos⁡2t=1\sin^2 t + \cos^2 t = 1sin2t+cos2t=1, we can find cos⁡t\cos tcost. Substituting sin⁡t=35\sin t = \dfrac{3}{5}sint=53​, we calculate:cos⁡2t=1−(35)2=1−925=1625\cos^2 t = 1 – \left(\dfrac{3}{5}\right)^2 = 1 – \dfrac{9}{25} = \dfrac{16}{25}cos2t=1−(53​)2=1−259​=2516​

Taking the square root gives cos⁡t=±45\cos t = \pm \dfrac{4}{5}cost=±54​. However, the quadrant of the angle matters. In Quadrant II, sine is positive but cosine is negative. So, cos⁡t=−45\cos t = -\dfrac{4}{5}cost=−54​.

Now that both sine and cosine are known, we can derive the remaining trigonometric functions:

  • tan⁡t=sin⁡tcos⁡t=3/5−4/5=−34\tan t = \dfrac{\sin t}{\cos t} = \dfrac{3/5}{-4/5} = -\dfrac{3}{4}tant=costsint​=−4/53/5​=−43​
  • cot⁡t=1tan⁡t=−43\cot t = \dfrac{1}{\tan t} = -\dfrac{4}{3}cott=tant1​=−34​
  • sec⁡t=1cos⁡t=−54\sec t = \dfrac{1}{\cos t} = -\dfrac{5}{4}sect=cost1​=−45​
  • csc⁡t=1sin⁡t=53\csc t = \dfrac{1}{\sin t} = \dfrac{5}{3}csct=sint1​=35​

Understanding quadrant signs is crucial: in Quadrant II, only sine and cosecant are positive. The other functions are negative due to the signs of x and y coordinates in that quadrant.

Related Posts

Scroll to Top