উচ্চতর গণিত ১ম পত্র • Combined Board • ঢাকা, যশোর, সিলেট, দিনাজপুর বোর্ড ২০১৮

উচ্চতর গণিত ১ম পত্র প্রশ্ন

p=sin⁡2α,q=sin⁡2β,r=cos⁡2α,s=cos⁡2β,t=sin⁡2γ p=\sin 2 \alpha, q=\sin 2 \beta, r=\cos 2 \alpha, s=\cos 2 \beta, t=\sin 2 \gamma .

  1. ক)

    প্রমাণ কর যে, sec⁡3x2=224+8+8cos⁡6x \sec \frac{3 x}{2}=\frac{2 \sqrt{2}}{\sqrt{4+\sqrt{8+8 \cos 6 x}}}

    (1)
  2. খ)

    যদি p+q=c,r+s=d p+q=c, r+s=d হয়, তবে দেখাও যে, cos⁡(2α+2β)=d2−c2d2+c2 \cos (2 \alpha+2 \beta)=\frac{d^{2}-c^{2}}{d^{2}+c^{2}}

    (2)
  3. গ)

    যদি α+β+γ=π \alpha+\beta+\gamma=\pi হয়, তবে দেখাও যে, p2+q2+t2=2−2cos⁡2αcos⁡2βcos⁡2γ p^{2}+q^{2}+t^{2}=2-2 \cos 2 \alpha \cos 2 \beta \cos 2 \gamma .

    (3)

উত্তর ও ব্যাখ্যা

  1. ক)

    AI-তৈরি নমুনা উত্তর

    ডানপক্ষ =224+8+8cos⁡6x= \frac{2\sqrt{2}}{\sqrt{4+\sqrt{8+8\cos 6x}}}
    =224+8(1+cos⁡6x)= \frac{2\sqrt{2}}{\sqrt{4+\sqrt{8(1+\cos 6x)}}}

    আমরা জানি, 1+cos⁡2θ=2cos⁡2θ1+\cos 2\theta = 2\cos^2 \theta
    সুতরাং, 1+cos⁡6x=2cos⁡23x1+\cos 6x = 2\cos^2 3x

    =224+8⋅2cos⁡23x= \frac{2\sqrt{2}}{\sqrt{4+\sqrt{8 \cdot 2\cos^2 3x}}}
    =224+16cos⁡23x= \frac{2\sqrt{2}}{\sqrt{4+\sqrt{16\cos^2 3x}}}
    =224+4cos⁡3x= \frac{2\sqrt{2}}{\sqrt{4+4\cos 3x}}
    =224(1+cos⁡3x)= \frac{2\sqrt{2}}{\sqrt{4(1+\cos 3x)}}
    =224⋅2cos⁡23x2= \frac{2\sqrt{2}}{\sqrt{4 \cdot 2\cos^2 \frac{3x}{2}}}
    =228cos⁡23x2= \frac{2\sqrt{2}}{\sqrt{8\cos^2 \frac{3x}{2}}}
    =2222cos⁡3x2= \frac{2\sqrt{2}}{2\sqrt{2}\cos \frac{3x}{2}}
    =1cos⁡3x2= \frac{1}{\cos \frac{3x}{2}}
    =sec⁡3x2== \sec \frac{3x}{2} = বামপক্ষ। (প্রমাণিত)

  2. খ)

    AI-তৈরি নমুনা উত্তর

    উদ্দীপক হতে পাই,
    p=sin⁡2α,q=sin⁡2β,r=cos⁡2α,s=cos⁡2βp=\sin 2\alpha, q=\sin 2\beta, r=\cos 2\alpha, s=\cos 2\beta

    দেওয়া আছে,
    p+q=c  ⟹  sin⁡2α+sin⁡2β=c…(i)p+q = c \implies \sin 2\alpha + \sin 2\beta = c \quad \dots (i)
    r+s=d  ⟹  cos⁡2α+cos⁡2β=d…(ii)r+s = d \implies \cos 2\alpha + \cos 2\beta = d \quad \dots (ii)

    সমীকরণ (i) থেকে পাই,
    2sin⁡2α+2β2cos⁡2α−2β2=c2\sin\frac{2\alpha+2\beta}{2}\cos\frac{2\alpha-2\beta}{2} = c
      ⟹  2sin⁡(α+β)cos⁡(α−β)=c…(iii)\implies 2\sin(\alpha+\beta)\cos(\alpha-\beta) = c \quad \dots (iii)

    সমীকরণ (ii) থেকে পাই,
    2cos⁡2α+2β2cos⁡2α−2β2=d2\cos\frac{2\alpha+2\beta}{2}\cos\frac{2\alpha-2\beta}{2} = d
      ⟹  2cos⁡(α+β)cos⁡(α−β)=d…(iv)\implies 2\cos(\alpha+\beta)\cos(\alpha-\beta) = d \quad \dots (iv)

    (iii) নং সমীকরণকে (iv) নং সমীকরণ দ্বারা ভাগ করে পাই,
    2sin⁡(α+β)cos⁡(α−β)2cos⁡(α+β)cos⁡(α−β)=cd\frac{2\sin(\alpha+\beta)\cos(\alpha-\beta)}{2\cos(\alpha+\beta)\cos(\alpha-\beta)} = \frac{c}{d}
      ⟹  sin⁡(α+β)cos⁡(α+β)=cd\implies \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{c}{d}
      ⟹  tan⁡(α+β)=cd\implies \tan(\alpha+\beta) = \frac{c}{d}

    আমরা জানি,
    cos⁡2θ=1−tan⁡2θ1+tan⁡2θ\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}

    এখানে θ=α+β\theta = \alpha+\beta হলে,
    cos⁡(2α+2β)=cos⁡2(α+β)\cos(2\alpha+2\beta) = \cos 2(\alpha+\beta)
    =1−tan⁡2(α+β)1+tan⁡2(α+β)= \frac{1-\tan^2(\alpha+\beta)}{1+\tan^2(\alpha+\beta)}
    =1−(cd)21+(cd)2= \frac{1-\left(\frac{c}{d}\right)^2}{1+\left(\frac{c}{d}\right)^2}
    =1−c2d21+c2d2= \frac{1-\frac{c^2}{d^2}}{1+\frac{c^2}{d^2}}
    =d2−c2d2d2+c2d2= \frac{\frac{d^2-c^2}{d^2}}{\frac{d^2+c^2}{d^2}}
    =d2−c2d2+c2= \frac{d^2-c^2}{d^2+c^2}

    ∴cos⁡(2α+2β)=d2−c2d2+c2\therefore \cos(2\alpha+2\beta) = \frac{d^2-c^2}{d^2+c^2} (দেখানো হলো)

  3. গ)

    AI-তৈরি নমুনা উত্তর

    উদ্দীপক হতে পাই,
    p=sin⁡2α,q=sin⁡2β,t=sin⁡2γp=\sin 2\alpha, q=\sin 2\beta, t=\sin 2\gamma

    দেওয়া আছে,
    α+β+γ=π\alpha+\beta+\gamma = \pi
      ⟹  2α+2β+2γ=2π\implies 2\alpha+2\beta+2\gamma = 2\pi
      ⟹  2α+2β=2π−2γ\implies 2\alpha+2\beta = 2\pi - 2\gamma

    বামপক্ষ =p2+q2+t2= p^2+q^2+t^2
    =sin⁡22α+sin⁡22β+sin⁡22γ= \sin^2 2\alpha + \sin^2 2\beta + \sin^2 2\gamma
    =12[2sin⁡22α+2sin⁡22β]+sin⁡22γ= \frac{1}{2} [2\sin^2 2\alpha + 2\sin^2 2\beta] + \sin^2 2\gamma
    =12[(1−cos⁡4α)+(1−cos⁡4β)]+sin⁡22γ= \frac{1}{2} [(1-\cos 4\alpha) + (1-\cos 4\beta)] + \sin^2 2\gamma
    =12[2−(cos⁡4α+cos⁡4β)]+sin⁡22γ= \frac{1}{2} [2 - (\cos 4\alpha + \cos 4\beta)] + \sin^2 2\gamma
    =1−12(cos⁡4α+cos⁡4β)+sin⁡22γ= 1 - \frac{1}{2}(\cos 4\alpha + \cos 4\beta) + \sin^2 2\gamma
    =1−12[2cos⁡4α+4β2cos⁡4α−4β2]+(1−cos⁡22γ)= 1 - \frac{1}{2}\left[2\cos\frac{4\alpha+4\beta}{2}\cos\frac{4\alpha-4\beta}{2}\right] + (1-\cos^2 2\gamma)
    =2−cos⁡(2α+2β)cos⁡(2α−2β)−cos⁡22γ= 2 - \cos(2\alpha+2\beta)\cos(2\alpha-2\beta) - \cos^2 2\gamma
    =2−cos⁡(2π−2γ)cos⁡(2α−2β)−cos⁡22γ= 2 - \cos(2\pi - 2\gamma)\cos(2\alpha-2\beta) - \cos^2 2\gamma [যেহেতু 2α+2β=2π−2γ2\alpha+2\beta = 2\pi-2\gamma]
    =2−cos⁡2γcos⁡(2α−2β)−cos⁡22γ= 2 - \cos 2\gamma\cos(2\alpha-2\beta) - \cos^2 2\gamma [যেহেতু cos⁡(2π−θ)=cos⁡θ\cos(2\pi-\theta) = \cos\theta]
    =2−cos⁡2γ[cos⁡(2α−2β)+cos⁡2γ]= 2 - \cos 2\gamma [\cos(2\alpha-2\beta) + \cos 2\gamma]
    =2−cos⁡2γ[cos⁡(2α−2β)+cos⁡(2π−(2α+2β))]= 2 - \cos 2\gamma [\cos(2\alpha-2\beta) + \cos(2\pi-(2\alpha+2\beta))]
    =2−cos⁡2γ[cos⁡(2α−2β)+cos⁡(2α+2β)]= 2 - \cos 2\gamma [\cos(2\alpha-2\beta) + \cos(2\alpha+2\beta)]
    =2−cos⁡2γ[2cos⁡2αcos⁡2β]= 2 - \cos 2\gamma [2\cos 2\alpha\cos 2\beta]
    =2−2cos⁡2αcos⁡2βcos⁡2γ== 2 - 2\cos 2\alpha\cos 2\beta\cos 2\gamma = ডানপক্ষ।

    ∴p2+q2+t2=2−2cos⁡2αcos⁡2βcos⁡2γ\therefore p^2+q^2+t^2 = 2 - 2\cos 2\alpha\cos 2\beta\cos 2\gamma (দেখানো হলো)

আরও প্রশ্ন অনুশীলন করতে অ্যাপে যাও

অ্যাপ ডাউনলোড করোওয়েবে খুলো

সম্পর্কিত প্রশ্ন