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JEECUP (UP Polytechnic) Trigonometric — practice questions

Explore 25 free Trigonometric questions in JEECUP (UP Polytechnic) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Trigonometric questions and exercises with answers

Showing questions 1–25 of 25.

$\int_{\pi/6}^{\pi/3} \cos x\,dx$ equals:Read the answer and worked solution →$\int_0^{\pi/2} \cos^3 x\,dx$ equals:Read the answer and worked solution →$\int_0^{\pi/2} \sin x\,dx$ equals:Read the answer and worked solution →$\int_0^{\pi/2} \sin^2 x\,dx$ equals:Read the answer and worked solution →$\dfrac{d}{dx}(\sin x)$ equals:Read the answer and worked solution →$\dfrac{d}{dx}(\tan x)$ equals:Read the answer and worked solution →SAT · Mathematics · Trig Ratios (SOHCAHTOA)Read the answer and worked solution →SAT · Mathematics · Trig Identities — T/FRead the answer and worked solution →Match the trig identity to its resultRead the answer and worked solution →Match the trig valueRead the answer and worked solution →Fill the gaps in the triangle area ruleRead the answer and worked solution →Positive or negative angleRead the answer and worked solution →Which ratios are positiveRead the answer and worked solution →Type the trig functionRead the answer and worked solution →Pop what is defined everywhereRead the answer and worked solution →Which unit of angleRead the answer and worked solution →One word is wrongRead the answer and worked solution →Order the build upRead the answer and worked solution →Sine and cosineRead the answer and worked solution →Which one does not belongRead the answer and worked solution →Which way does it turnRead the answer and worked solution →Spot the false statementRead the answer and worked solution →Which ones belongRead the answer and worked solution →Knock out the true statementsRead the answer and worked solution →Trigonometry bingoRead the answer and worked solution →

Worked practice questions

$\int_{\pi/6}^{\pi/3} \cos x\,dx$ equals:

  1. $(\sqrt 3 - 1)/2$
  2. $(\sqrt 3 + 1)/2$
  3. $1$
  4. $0$
Show answer and solution

Correct answer: A. $(\sqrt 3 - 1)/2$

$\int \cos x\,dx = \sin x$, so $[\sin x]_{\pi/6}^{\pi/3} = \sin(\pi/3) - \sin(\pi/6) = \sqrt 3/2 - 1/2 = (\sqrt 3 - 1)/2$.

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$\int_0^{\pi/2} \cos^3 x\,dx$ equals:

  1. $1/3$
  2. $2/3$
  3. $3/4$
  4. $\pi/4$
Show answer and solution

Correct answer: B. $2/3$

Use $\cos^3 x = (1/4)(\cos 3x + 3\cos x)$. **Integrating:** $(1/4)[\sin 3x/3 + 3\sin x]_0^{\pi/2} = (1/4)[(-1/3 + 3) - 0] = (1/4)(8/3) = 2/3$.

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$\int_0^{\pi/2} \sin x\,dx$ equals:

  1. $0$
  2. $1$
  3. $2$
  4. $\pi/2$
Show answer and solution

Correct answer: B. $1$

$\int_0^{\pi/2} \sin x\,dx = [-\cos x]_0^{\pi/2} = -\cos(\pi/2) + \cos 0 = 0 + 1 = 1$.

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