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Federal (FBISE) · Chapter test · Free

1st Year Mathematics Chapter 8 Test8. Fundamentals of Trigonometry

30 marks · 40 minutes · 21 questions. Covers: 8. Fundamentals of Trigonometry.

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1st Year Mathematics Chapter 8 Test

1st Year · Mathematics · Federal (FBISE)

Name: ____________Roll No: ________Class/Section: ________Date: ________Total marks: 30Time: 40 minutes

Q1. Multiple choice questions (Attempt all · 10 × 1 = 10)

  1. If cos⁡θ=−310 \cos \theta = -\frac{3}{\sqrt{10}} and 90∘<θ<180∘ 90^\circ < \theta < 180^\circ than tan⁡2θ= \tan 2\theta =
    1. 34\frac{3}{4}
    2. ±34\pm\frac{3}{4}
    3. −43\frac{-4}{3}
    4. −34\frac{-3}{4}
  2. The exact value of sin⁡7.5∘cos⁡7.5∘cos⁡15∘ \sin 7.5^\circ \cos 7.5^\circ \cos 15^\circ is ?
    1. 14\frac{1}{4}
    2. 12\frac{1}{2}
    3. 18\frac{1}{8}
    4. 116\frac{1}{16}
  3. sin⁡(π2−θ)sin⁡(π2+θ)=\sin\left(\frac{\pi}{2} - \theta\right) \sin\left(\frac{\pi}{2} + \theta\right) =
    1. 1+cos⁡2θ1 +\cos 2\theta
    2. sin⁡2θ\sin 2\theta
    3. 1+cos⁡2θ2\frac{1 +\cos 2\theta}{2}
    4. 12sin⁡2θ\frac{1}{ 2}\sin2\theta
  4. If sin⁡A=15 \sin A = \frac{1}{5} and sin⁡B=13 \sin B = \frac{1}{3} and 0<A,B<π2 0 < A, B < \frac{\pi}{2} , then cos⁡(A+B)= \cos(A + B) =
    1. 83−115\frac{8\sqrt{3} - 1}{15}
    2. −83+115-\frac{8\sqrt{3} + 1}{15}
    3. 83+115\frac{8\sqrt{3} + 1}{15}
    4. −83+115\frac{-8\sqrt{3} + 1}{15}
  5. Using product to sum identities, the exact value of −cos⁡195∘sin⁡45∘ -\cos 195^\circ \sin 45^\circ is :
    1. 33−34\frac{3\sqrt{3} - 3}{4}
    2. 33+34\frac{3\sqrt{3} + 3}{4}
    3. 3+14\frac{\sqrt{3} + 1}{4}
    4. None of these
  6. tan⁡(π2+θ)=…\tan\left(\frac{\pi}{2} + \theta\right) = \ldots
    1. cot θ
    2. −cot⁡θ-\cot\theta
    3. tan θ
    4. −tan⁡θ-\tan\theta
  7. 2cos⁡αsin⁡β= 2 \cos \alpha \sin \beta =
    1. sin⁡(α+β)+sin⁡(α−β){\sin(\alpha + \beta)}+{\sin(\alpha - \beta)}
    2. sin⁡(α+β)−sin⁡(α−β){\sin(\alpha + \beta)}-{\sin(\alpha - \beta)}
    3. cos⁡(α+β)+cos⁡(α−β){\cos(\alpha + \beta)}+{\cos(\alpha - \beta)}
    4. cos⁡(α−β)−cos⁡(α+β){\cos(\alpha - \beta)}-{\cos(\alpha +\beta)}
  8. cos⁡23∘sin⁡17∘=\cos 23^\circ \sin 17^\circ =
    1. sin⁡40∘−sin⁡6∘\sin 40^\circ - \sin 6^\circ
    2. sin⁡40∘+sin⁡6∘\sin 40^\circ + \sin 6^\circ
    3. 12[sin⁡40∘−sin⁡6∘]\frac{1}{2}[\sin 40^\circ - \sin 6^\circ ]
    4. 12[sin⁡40∘+sin⁡6∘]\frac{1}{2}[\sin 40^\circ + \sin 6^\circ ]
  9. sin⁡40∘cos⁡20∘+cos⁡40∘sin⁡20∘=\sin 40^\circ \cos 20^\circ + \cos 40^\circ \sin 20^\circ =
    1. 32\frac{\sqrt{3}}{2}
    2. sin60°
    3. sin30°
    4. All of these
  10. cos⁡u+cos⁡3u+cos⁡5u+cos⁡7u=\cos u + \cos 3u + \cos 5u + \cos 7u =
    1. cos⁡ucos⁡2ucos⁡4u\cos u \cos 2u \cos 4u
    2. sin⁡usin⁡2usin⁡4u\sin u \sin 2u \sin 4u
    3. −4sin⁡usin⁡2usin⁡4u-4\sin u \sin 2u \sin 4u
    4. 4cos⁡ucos⁡2ucos⁡4u4\cos u \cos 2u \cos 4u

Q2. Short questions (Attempt any 6 of 8 · 6 × 2 = 12)

  1. Use the product-to-sum formula to change the sum or difference. sin (- u) sin 5u
  2. Verify:sin⁡(π2−α)=cos⁡α\sin\left(\frac{\pi}{2} - \alpha\right) = \cos\alpha
  3. Rewrite as a single expression.tan⁡75∘−tan⁡45∘1+tan⁡75∘tan⁡45∘\frac{\tan 75^\circ - \tan 45^\circ}{1 + \tan 75^\circ \tan 45^\circ}
  4. Verify the following identity. 8 sin⁴ θ = 3 + cos 4θ - 4 cos 2θ
  5. Express the following as single trigonometric ratio.12sin⁡75∘+12cos⁡75∘\frac{1}{\sqrt{2}} \sin 75^\circ + \frac{1}{\sqrt{2}} \cos 75^\circ
  6. Find the exact value of sin 22.5⁰ × cos 22.5⁰.
  7. Given that , sin⁡θ=35,sin⁡ϕ=513,\sin \theta = \frac{3}{5}, \quad \sin \phi = \frac{5}{13}, where θ is obtuse and Φ is acute. Find the value of. tan ( θ - Φ )
  8. Prove that :tan⁡(θ2)=±(1−cos⁡θsin⁡θ)\tan\left(\frac{\theta}{2}\right) = \pm\left(\frac{1 - \cos\theta}{\sin\theta}\right)

Q3. Long questions (Attempt any 2 of 3 · 2 × 4 = 8)

  1. Verify the following identity.1−tan⁡2x21+tan⁡2x2=cos⁡x\frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \cos x
  2. Verify the following identity. cos⁴ x - sin⁴ x = cos 2x
  3. Find (a) sin⁡2θ\sin 2\theta (b) cos⁡2θ\cos 2\theta (c) tan⁡2θ\tan 2\theta (d) sin⁡θ2\sin \frac{\theta}{2} (e) cos⁡θ2\cos \frac{\theta}{2}(f) tan⁡θ2\tan \frac{\theta}{2} when: cot⁡θ=−12\cot \theta =-\frac{1}{2} where π2<θ<π\frac{\pi}{2} < \theta < {\pi}.